Class Wiki - User contributions [en]

Coupled Oscillator: Jonathan Schreven

2009-12-11T22:41:56Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.1877i\,</math>
:<math>\lambda_4=-1.1877i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>\bar{x}=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>\bar{x}=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.1877it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.1877it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>
:<math>=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

If we take the Laplace transform of the above equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
where
:<math>e^{\bold{\hat{A}}t}=\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}</math>

We then substitute this equation back into
:<math>\bar{x}=\bold{T^{-1}}\bar{z}</math>
and get
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
Notice here that
:<math>e^{\bold{A}t}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}</math>
:<math>e^{\bold{A}t}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}
\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
</math>
You can solve this with a computer program or your calculator and plug it into the equation for '''A'''. I have not listed the answer for this problem here because it is very messy and extremely long. I did calculate it to make sure it is solvable. But if your numbers are easier to work with you would finish by plugging this value into the equation below.
:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-11T00:45:17Z

Jonathan.schreven:

Coupled Oscillator: Jonathan Schreven

2009-12-11T00:20:12Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.1877i\,</math>
:<math>\lambda_4=-1.1877i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>\bar{x}=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>\bar{x}=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.1877it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.1877it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>
:<math>=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

If we take the Laplace transform of the above equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
where
:<math>e^{\bold{\hat{A}}t}=\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}</math>

We then substitute this equation back into
:<math>\bar{x}=\bold{T^{-1}}\bar{z}</math>
and get
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>
where
:<math>e^{\bold{A}t}=</math>

Coupled Oscillator: Jonathan Schreven

2009-12-11T00:01:32Z

Jonathan.schreven: /* Solving */

Coupled Oscillator: Jonathan Schreven

2009-12-11T00:00:42Z

Jonathan.schreven: /* Solving */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.1877i\,</math>
:<math>\lambda_4=-1.1877i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.1877it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.1877it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>
:<math>=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

If we take the Laplace transform of the above equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
where
:<math>e^{\bold{\hat{A}}t}=\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}</math>

We then substitute this equation back into
:<math>\bar{x}=\bold{T^{-1}}\bar{z}</math>
and get
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-11T00:00:16Z

Jonathan.schreven: /* Eigen Values */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.1877i\,</math>
:<math>\lambda_4=-1.1877i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>
:<math>=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

If we take the Laplace transform of the above equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
where
:<math>e^{\bold{\hat{A}}t}=\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}</math>

We then substitute this equation back into
:<math>\bar{x}=\bold{T^{-1}}\bar{z}</math>
and get
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T23:59:12Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>
:<math>=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

If we take the Laplace transform of the above equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
where
:<math>e^{\bold{\hat{A}}t}=\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}</math>

We then substitute this equation back into
:<math>\bar{x}=\bold{T^{-1}}\bar{z}</math>
and get
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T23:49:23Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>
:<math>=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

If we take the Laplace transform of the above equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
where
:<math>e^{\bold{\hat{A}}t}=\begin{bmatrix}
e^{2.6626it} & 0 & 0 & 0 \\
0 & e^{-2.6626it} & 0 & 0 \\
0 & 0 & e^{1.1877it} & 0 \\
0 & 0 & 0 & e^{1.1877it}
\end{bmatrix}</math>

We then substitute this equation back into
:<math>\bar{x}=\bold{T^{-1}}\bar{z}</math>
and get
:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T23:21:24Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
We already know what the matrix A is from our state space equation
:<math>\bold{A}=\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}</math>

And we know that the T-inverse matrix is
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

It then follows that matrix T is
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>

This new equation has the same form as
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>

If we take the Laplace transform of this equation we can come up with the following
:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>

We know the values of T, A, and T^{-1}. If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\hat{\lambda}\bold{I}</math>
:<math>\bold{\hat{A}}=
\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}
=
\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:54:52Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying the inverse of '''T''' to the left side of the equation.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can use another identity that we already know
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>

This new equation has the same form as
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>

If we take the Laplace transform of this equation we can come up with the following
:<math>\bar{z}=e^{\bold{A}t}\bar{z}(0)</math>

If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\lambda\bold{I}</math>
:<math>\bold{\hat{A}}=\begin{bmatrix}
2.6626i & 0 & 0 & 0 \\
0 & -2.6626i & 0 & 0 \\
0 & 0 & 1.1877i & 0 \\
0 & 0 & 0 & 1.1877i
\end{bmatrix}</math>

We also know what '''T''' equals and we can solve it for our case
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for '''T'''
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:48:49Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:Double Oscillator System.JPG|500px||center|Double Mass/Spring Oscillator]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying the inverse of '''T''' to the left side of the equation.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can use another identity that we already know
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>

This new equation has the same form as
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>

If we take the Laplace transform of this equation we can come up with the following
:<math>\bar{z}=e^{\bold{A}t}\bar{z}(0)</math>

If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\lambda\bold{I}</math>
:<math>\bold{\hat{A}}=\begin{bmatrix}
& 0 & 0 & 0 \\
0 & & 0 & 0 \\
0 & 0 & & 0 \\
0 & 0 & 0 &
\end{bmatrix}</math>

We also know what '''T''' equals and we can solve it for our case
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for '''T'''
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:38:10Z

Jonathan.schreven: /* Problem */

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:34:37Z

Jonathan.schreven: /* Problem */

File:Double Oscillator System.JPG

2009-12-10T22:33:17Z

Jonathan.schreven: A simple double mass/spring system.

A simple double mass/spring system.

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:29:39Z

Jonathan.schreven: /* Problem */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

:[[Image:]]

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying the inverse of '''T''' to the left side of the equation.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can use another identity that we already know
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>

This new equation has the same form as
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>

If we take the Laplace transform of this equation we can come up with the following
:<math>\bar{z}=e^{\bold{A}t}\bar{z}(0)</math>

We also know what T equals and we can solve it for our case
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:29:15Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying the inverse of '''T''' to the left side of the equation.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can use another identity that we already know
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>

This new equation has the same form as
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>

If we take the Laplace transform of this equation we can come up with the following
:<math>\bar{z}=e^{\bold{A}t}\bar{z}(0)</math>

We also know what T equals and we can solve it for our case
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T22:26:50Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

This can be rearranged by multiplying the inverse of '''T''' to the left side of the equation.
:<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>

Now we can use another identity that we already know
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>

This new equation has the same form as
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
where
:<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>

If we take the Laplace transform of this equation we can come up with the following
:<math>\bar{z}=e^{\bold{A}t}\bar{x}(0)</math>

We also know what T equals and we can solve it for our case
:<math>\bold{T^{-1}}=[\bar{k_1}|\bar{k_2}|\bar{k_3}|\bar{k_4}]\,</math>
:<math>\bold{T^{-1}}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>\bold{T}=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T21:54:05Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>

This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>

Now we can use another identity that we already know
:<math>\dot{x}=Ax</math>

Combining the two equations we then get
:<math>T^{-1}\dot{z}=AT^{-1}z</math>

Multiplying both sides of the equation on the left by T we get
:<math>\dot{z}=TAT^{-1}z</math>

We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T21:41:53Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>

This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>

Now we can use another identity that we already know
:<math>\dot{x}=Ax</math>

Combining the two equations we then get
:<math>T^{-1}\dot{z}=AT^{-1}z</math>

We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T21:15:14Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>
This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>
We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use another identity that we already know to find the matrix exponential
:<math>\dot{x}=Ax</math>
Then using laplace transforms we can find that
:<math>x=e^{At}x(0)\,</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T21:14:57Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>
This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>
We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Now we can use another identity that we already know to find the matrix exponential
:<math>\dot{x}=Ax</math>
Then using laplace transforms we can find that
:<math>x=e^{At}x(0)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T20:06:10Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>
This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>
We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Next we will use

Coupled Oscillator: Jonathan Schreven

2009-12-10T19:37:08Z

Jonathan.schreven: /* Matrix Exponential */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>
This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>
We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
-1.2657i & -0.4753 & 0.8193i & 0.3077 \\
1.2657i & -0.4753 & -0.8193i & 0.3077 \\
0.6514i & 0.5484 & 0.5031i & 0.4236 \\
-0.6514i & 0.5484 & -0.5031 & 0.4236
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T19:24:05Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>
This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>
We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i & -0.3500i & 0.3500i \\
-0.5722 & -0.5722 & 0.4157 & 0.4157 \\
-0.2783i & 0.2783i & -0.5407i & 0.5407i \\
0.7409 & 0.7409 & 0.6421 & 0.6421
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T18:56:05Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==
In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
:<math>z=Tx\,</math>
This can be rearranged by multiplying the inverse of T to the left side of the equation.
:<math>T^{-1}z=x\,</math>
We also know what T equals and we can solve it for our case
:<math>T^{-1}=[k_1|k_2|k_3|k_4]\,</math>
:<math>T^{-1}=\begin{bmatrix}
0.2149i & -0.2149i \\
-0.5722 & -0.5722 \\
-0.2783i & 0.2783i \\
0.7409 & 0.7409
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T04:01:46Z

Jonathan.schreven: /* Solving */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

== Matrix Exponential ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:58:47Z

Jonathan.schreven: /* Solving */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}e^{2.6626it}+c_2\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}e^{-2.6626it}+c_3\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}e^{1.18766it}+c_4\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}e^{-1.18766it}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:44:38Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1\begin{bmatrix}
0.2149 \\
-0.5722 \\
-0.2783 \\
0.7409
\end{bmatrix}e^{2.6626t}+c_2\begin{bmatrix}
-0.2149 \\
-0.5722 \\
0.2783 \\
0.7409
\end{bmatrix}e^{-2.6626t}+c_3\begin{bmatrix}
-0.3500 \\
0.4157 \\
-0.5407 \\
0.6421
\end{bmatrix}e^{1.18766t}+c_4\begin{bmatrix}
0.3500 \\
0.4157 \\
0.5407 \\
0.6421
\end{bmatrix}e^{-1.18766t}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:40:32Z

Jonathan.schreven: /* Eigen Vectors */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

== Solving ==

We can now plug these eigen vectors and eigen values into the standard equation
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

And our final answer is
:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

Coupled Horizontal Spring Mass Oscillator

2009-12-10T03:32:52Z

Jonathan.schreven: /* Things we know */

=Coupled Oscillator Spring Mass Oscillator: State Space =

==Problem Statement==
Two 4 Kg Weights are suspended between two walls. They are connected by a spring between them with a spring constant k2.
They are connected to the walls by two springs k1 and k3 with k1=k3. m1 is a distance x1 form m2 and m2 is x2 from the wall.

==Solution==

===Things we know===

<math>m_1 = 5 kg \frac{}{}</math>

<math>m_2 = 5 kg \frac{}{}</math>

<math>k_1 = 50 Nm \frac{}{}</math>

<math>k_2 = 100 Nm \frac{}{}</math>

<math>k_3 = 50 Nm \frac{}{}</math>

<math>\text {So now that we have are problem we need to start setting up the equations we need to solve it.}\,</math>

<math>\dot{x_1}=\dot{x_1}</math>

<math>\ddot{x_1}+\frac{k_1+k_2}{m_1}{x_1}-\frac{k_2}{m_1}{x_2}=0</math>

<math>\dot{x_2}=\dot{x_2}</math>

<math>\ddot{x_2}+\frac{k_3+k_2}{m_2}{x_2}-\frac{k_2}{m_2}{x_1}=0</math>

<math>\text {Now we take these equations and put them in a state space model.}\,</math>

<math>\begin{bmatrix} \dot{x_1} \\\ddot{x_1} \\\dot{x_2} \\\ddot{x_2}\end{bmatrix}\,
</math>
=
<math>\begin{bmatrix}0&1&0&0 \\\frac{(k_1+k_2)}{m_1}&0&\frac{-k_1}{m_1}&0 \\0&0&0&1 \\\frac{-k_1}{m_2}&0&\frac{(k_1+k_2)}{m_2}&0
\end{bmatrix}

\begin{bmatrix}x_1 \\\dot{x}_1 \\x_2 \\\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}0\end{bmatrix}
</math>

<math>\text {Now we make the appropriate numerical substitutions.}\,</math>

<math>\begin{bmatrix} \dot{x_1} \\\ddot{x_1} \\\dot{x_2} \\\ddot{x_2}\end{bmatrix}\,
</math>
=
<math>\begin{bmatrix}0&1&0&0 \\\frac{150}{5}&0&\frac{-50}{5}&0 \\0&0&0&1 \\\frac{-50}{5}&0&\frac{150}{5}&0 \end{bmatrix}

\begin{bmatrix}x_1 \\\dot{x}_1 \\x_2 \\\dot{x}_2\end{bmatrix}

+

\begin{bmatrix}0\end{bmatrix}
</math>

<math>\begin{bmatrix} \dot{x_1} \\\ddot{x_1} \\\dot{x_2} \\\ddot{x_2}\end{bmatrix}\,

=

\begin{bmatrix}0&1&0&0 \\30&0&-10&0 \\0&0&0&1 \\-10&0&30&0\end{bmatrix}

\begin{bmatrix}x_1 \\\dot{x}_1 \\x_2 \\\dot{x}_2\end{bmatrix}

+

\begin{bmatrix}0\end{bmatrix}
</math>

<math>\text {So using Maple I was able to obtain the eigenvalues and eigenvectors.}\,</math>

<math>\text {Eigenvalues.}\,</math>

<math>\lambda_1=2\sqrt{10}\,</math>
<math>\lambda_2=-2\sqrt{10}\,</math>
<math>\lambda_3=2\sqrt{5}\,</math>
<math>\lambda_4=-2\sqrt{5}\,</math>

<math>\text {Eigenvectors.}\,</math>

<math>\ {K_1=}\,</math><math>\begin{bmatrix}-1 \\-2\sqrt(10) \\1 \\2\sqrt(10)\end{bmatrix}\,
</math>,<math>\ {K_2=}\,</math><math>\begin{bmatrix}-1 \\2\sqrt(10) \\1 \\-2\sqrt(10)\end{bmatrix}\,
</math>,<math>\ {K_3=}\,</math><math>\begin{bmatrix}1 \\2\sqrt(5) \\1 \\2\sqrt(5)\end{bmatrix}\,
</math>,<math>\ {K_4=}\,</math><math>\begin{bmatrix}1 \\-2\sqrt(5) \\1 \\-2\sqrt(5)\end{bmatrix}\,
</math>

<math>\text {So then the answer is...}\,</math>

<math>\ x=c_1</math><math>\begin{bmatrix}-1 \\-2\sqrt(10) \\1 \\2\sqrt(10)\end{bmatrix}\,</math><math>e^{2\sqrt{10}}+ c_2</math><math>\begin{bmatrix}-1 \\2\sqrt(10) \\1 \\-2\sqrt(10)\end{bmatrix}\,</math><math>e^{2*-2\sqrt{10}}+ c_3</math><math>\begin{bmatrix}1 \\2\sqrt(5) \\1 \\2\sqrt(5)\end{bmatrix}\,</math><math>e^{3*2\sqrt{5}}+ c_4</math><math>\begin{bmatrix}1 \\-2\sqrt(5) \\1 \\-2\sqrt(5)\end{bmatrix}\,</math><math>e^{4*-2\sqrt{5}}\,</math>

==Solve with the Matrix exponential==

<math>\text {So first we need to know what the matrix exponential equation looks like.}\,</math>

<math>\text {it is...}\,</math>

<math>\tilde{x}=e^{\tilde{A}t}\tilde{x(0)}\,</math>

<math>\text {Where A is a matrix}\,</math>

<math>\text {Also }\,</math>

<math>\tilde{z}=\tilde{T}\tilde{x}\,</math>

<math>\tilde{x}=\tilde{T}^{-1}\tilde{z}\,</math>

<math>\text {Where }\,</math>

<math>\tilde{T}^{-1}=\,</math><math>\begin{bmatrix}-1&-1&1&1 \\-2\sqrt(10)&2\sqrt(10)&2\sqrt(5)&-2\sqrt(5) \\1&1&1&1 \\2\sqrt(10)&-2\sqrt(10)&2\sqrt(5)&-2\sqrt(5)\end{bmatrix}\,
</math>

<math>\text {I converted the T matrix to decimal form for make it easier to write up on here }\,</math>

<math>\tilde{T}=\,</math></math><math>\begin{bmatrix}-.25&-.039528&.25&.039528 \\-.25&.039528&.25&-.039528 \\.25&.055902&.25&.055902 \\.25&-.055902&.25&-.055902\end{bmatrix}\,
</math>

<math>\text {and}\,</math>

<math>\hat{A}=\,</math><math>\begin{bmatrix}e^{\lambda_1t}&0&0&0 \\0&e^{\lambda_2t}&0&0 \\0&0&e^{\lambda_3t}&0 \\0&0&0&e^{\lambda_4t}\end{bmatrix}\,
</math>

<math>e^{\hat{A}t}=\,</math><math>\begin{bmatrix}e^{2\sqrt{10}t}&0&0&0 \\0&e^{-2\sqrt{10}t}&0&0 \\0&0&e^{2\sqrt{5}t}&0 \\0&0&0&e^{-2\sqrt{5}t}\end{bmatrix}\,
</math>

<math>\text {Then the next step is}\,</math>

<math>\tilde{z}=e^{\hat{A}t}\tilde{z}(0)\,</math>

<math>\text {So that implies}\,</math>

<math>\tilde{x}=\tilde{T}^{-1}e^{\hat{A}t}\tilde{z}(0)\,</math><math>=\tilde{T}^{-1}e^{\hat{A}t}\tilde{T}^{-1}\tilde{x}(0)\,</math>

<math>\text {Now Simply substitute back in and we have the answer. }\,</math>

<math>\tilde{x}=\,</math><math>\begin{bmatrix}-1&-1&1&1 \\-2\sqrt(10)&2\sqrt(10)&2\sqrt(5)&-2\sqrt(5) \\1&1&1&1 \\2\sqrt(10)&-2\sqrt(10)&2\sqrt(5)&-2\sqrt(5)\end{bmatrix}\begin{bmatrix}e^{2\sqrt{10}t}&0&0&0 \\0&e^{-2\sqrt{10}t}&0&0 \\0&0&e^{2\sqrt{5}t}&0 \\0&0&0&e^{-2\sqrt{5}t}\end{bmatrix}</math><math>\begin{bmatrix}-.25&-.039528&.25&.039528 \\-.25&.039528&.25&-.039528 \\.25&.055902&.25&.055902 \\.25&-.055902&.25&-.055902\end{bmatrix}\tilde{x}(0)\,</math>

----
Created By: Mark Bernet

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:31:11Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.
:<math>k_1=\begin{bmatrix}
0.2149i \\
-0.5722 \\
-0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
-0.2149i \\
-0.5722 \\
0.2783i \\
0.7409
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
-0.3500i \\
0.4157 \\
-0.5407i \\
0.6421
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0.3500i \\
0.4157 \\
0.5407i \\
0.6421
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:22:19Z

Jonathan.schreven: /* Eigen Vectors */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==
Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as <math>k_1,k_2,k_3,k_4\,</math>.

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:16:48Z

Jonathan.schreven: /* Eigen Values */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
4 & 0 & -4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_4=-1.18766i\,</math>

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T03:14:33Z

Jonathan.schreven: /* Equations of Equilibrium */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
{k_2\over {m_2}} & 0 & -{k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
-4 & 0 & 4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=\,</math>
:<math>\lambda_2=\,</math>
:<math>\lambda_3=\,</math>
:<math>\lambda_4=\,</math>

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T02:28:18Z

Jonathan.schreven: /* Eigen Values */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
-4 & 0 & 4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

From this we get
:<math>\lambda_1=\,</math>
:<math>\lambda_2=\,</math>
:<math>\lambda_3=\,</math>
:<math>\lambda_4=\,</math>

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T02:23:42Z

Jonathan.schreven: /* Eigen Values */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=5kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 \\
-4 & 0 & 4 & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Vectors ==

Fall 2009/JonathanS

2009-12-10T02:17:14Z

Jonathan.schreven: /* Asymptotes & Break Points */

== Problem ==

A simple pendulum with a length L = 0.5m is pulled back and released from an initial angle <math>\theta_0 = 12^o</math>. Then it is run with a forcing function of cos(3<math>\theta</math>) Find an equation that gives the angle of the pendulum at any given time '''t'''.

== Solution ==

Assuming no damping and a small angle(<math>\theta < 15^o</math>), the equation for the motion of a simple pendulum can be written as
:<math>{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{g\over \ell}\theta=cos(3\theta).</math>

Substituting values we get
:<math>{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{9.81\over 0.5}\theta=cos(3\theta).</math>

<math>\Rightarrow{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{19.62}\theta=cos(3\theta).</math>

Remember the identities
:<math>\mathcal{L}\{f(t)\}=F(s)=\int_0^{\infty} e^{-st} f(t) \,dt. </math>

:<math>\mathcal{L}\{f^{ ''}(t)\}=s^2F(s)-sf(0)-f^{ '}(0)</math>

Now we can take the Laplace Transform to change the second order differential equation, from the t domain, into a simple linear equation, from the s domain, that's much easier to work with
:<math>\mathcal{L}\bigg\{{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{19.62}\theta\bigg\}=s^2F(s)-sf(0)-f^{ '}(0)+{19.62}\theta=\mathcal{L}\{cos(3\theta)\}</math>

<math>\Rightarrow</math> <math> s^2\boldsymbol{\theta}-s\theta(0)-\theta^{ '}(0)+{19.62}\boldsymbol{\theta}={s\over {s^2+9}} \,</math>

Since we know that <math>\theta(0)=12^o</math> and the initial velocity <math>\theta^{ '}(0)=0</math> we get
:<math>s^2\boldsymbol{\theta}-12s+19.62\boldsymbol{\theta}={s\over {s^2+9}} \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}(s^2+19.62)={s\over {s^2+9}}+12s \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}(s^2+19.62)={s\over {s^2+9}}+{12s(s^2+9)\over {s^2+9}} \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}(s^2+19.62)={(12s^3+109s)\over {s^2+9}} \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}={(12s^3+109s)\over {(s^2+9)(s^2+19.62)}}</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}={12s^3+109s\over {(s^4+28.62s^2+176.58)}}</math>

Now we can take the inverse Laplace Transform to convert our equation back into the time domain

:<math>\mathcal{L}^{-1}\bigg\{{12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg\}=0.094162cos(3t)+11.9058cos(4.42945t)</math>

This will give us the angle (in degrees) of the pendulum at any given time '''t'''.

== Initial Value Theorem ==

We can use the Initial Value Theorem as a check that our initial values for the problem are valid.
:<math>\mathcal{L}\{f^{ '}(t)\}=sF(s)-f(0^-)=\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt</math>

:<math>\begin{alignat}{3}
\lim_{n \to \infty}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to \infty}sF(s)-f(0^-) \\
& = \lim_{n \to \infty}sF(s)-f(0^-)=0 \\
& = \lim_{n \to \infty}sF(s)=f(0^-) \\
\end{alignat}</math>

Below we will use this theorem to check the values for our problem.
:<math>\begin{alignat}{3}
\lim_{n \to \infty}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to \infty}sF(s)-f(0^-)=\lim_{n \to \infty}\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt \\
& = \lim_{n \to \infty}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)-f(0^-)=0 \\
& = \lim_{n \to \infty}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)=f(0^-) \\
12 & = f (0^-) \\
\end{alignat}</math>
This value <math>f(0^-)=12</math> is the initial angle we gave the pendulum so it checks out.

== Final Value Theorem ==
We can use the Final Value Theorem as a check that our final values for the problem are valid.
:<math>\mathcal{L}\{f^{ '}(t)\}=sF(s)-f(\infty)=\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt</math>

:<math>\begin{alignat}{3}
\lim_{n \to 0}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to 0}sF(s)-f(\infty) \\
& = \lim_{n \to 0}sF(s)-f(\infty)=0 \\
& = \lim_{n \to 0}sF(s)=f(\infty) \\
\end{alignat}</math>

Below we will use this theorem to check the values for our problem.
:<math>\begin{alignat}{3}
\lim_{n \to 0}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to 0}sF(s)-f(\infty)=\lim_{n \to 0}\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt \\
& = \lim_{n \to 0}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)-f(\infty)=0 \\
& = \lim_{n \to 0}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)=f(\infty) \\
0 & = f (\infty) \\
\end{alignat}</math>

This is zero because the average angle as time goes to infinity will be zero (halfway between -12 and 12 degrees).

== Bode Plots ==
To find the bode plot we use the function
:<math>H(s)={12s^3+109s\over {(s^4+28.62s^2+176.58)}}</math>

Entering this into a program such as Maple, Octave, or MATLAB will give you a plot that looks like this

[[Image:Bode_Plot.jpg]]

== Asymptotes & Break Points ==

In this section I will describe how you can use the transfer function to quickly estimate what the bode plot will look like.

To find the initial value we can take <math>20log(H(j\omega))</math>. Which gives us
:<math>H(j\omega)={1 \over {\sqrt{\omega^2+1}\sqrt{\omega^2-1}}}</math>
From our original input we know that <math>\omega=3</math> so
:<math>H(j\omega)={1 \over {\sqrt{3^2+1}\sqrt{3^2-1}}}</math>
:<math>H(j\omega)=0.1118\,</math>
Now we can plug that into our log function
:<math>\begin{alignat}{3}
db_0 & = 20log(H(j\omega)) \\
& = 20log(0.1118) \\
& = -19.0 \\
\end{alignat}</math>

If we check that against our bode plot above we can see that this is pretty close. Now we can find the deflections using the transfer function. It will be easier to work with in factored form.
For every "s" we have in the numerator we will increase the slope by 20db/decade. For every "s" we have in the denominator we will decrease the slope by 20db/decade. These changes will occur at the poles which form our break points and asymptotes.

:<math>\boldsymbol{\theta}={2s(12s^2+109) \over {(s^2+9)(2s^2+39.24)}}</math>

We get asymptotes at s=3j and s=4.47j from the denominator. For our purposes we can just ignore the imaginary part. We should also have increases of 20db/decade at s=0 and s=3.01 which we find from the numerator.

== Convolution ==

Convolution, by definition, states that H(s)=Y(s)/X(s) where H(s) is the transfer function, Y(s) is the output function, and X(s) is the input function. We would like to find the output of the system, so for our purposes we will rearrange the above equation like this.
:<math>\begin{alignat}{3}
Y(s) & = H(s)X(s) \\
& = \bigg({12s^3+109s\over {s^4+28.62s^2+176.58}}\bigg)\bigg({s\over {s^2+9}}\bigg) \\
& = {2s^2(12s^2+109)\over {(s^2+9)^2(2s^2+39.24)}} \\
\end{alignat}</math>

== State Space Equations ==

Now we can transfer the functions we already have into state space. This is simply taking our original differential equation and putting it in terms of vectors and matrices. This can make the equations easier to work with.

<math>\begin{bmatrix}
\dot{\theta} \\
\ddot{\theta}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
{-g\over \ell} & 0
\end{bmatrix}

\begin{bmatrix}
\theta \\
\dot{\theta}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}cos(3\theta)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T02:16:51Z

Jonathan.schreven: /* Eigen Values */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>x_1=1m\,</math>
:<math>x_2=2.5m\,</math>
:<math>m_1=10kg\,</math>
:<math>m_2=7kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=20\,{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T02:09:50Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>x_1=1m</math>
:<math>x_2=2.5m</math>
:<math>m_1=10kg</math>
:<math>m_2=7kg</math>
:<math>k_1=25{N\over {m}}</math>
:<math>k_2=20{N\over {m}}</math>

We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:54:50Z

Jonathan.schreven: /* Equations of Equilibrium */

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our four equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:54:18Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Eigen Values ==

== Eigen Vectors ==

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:53:33Z

Jonathan.schreven:

== Problem ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:53:09Z

Jonathan.schreven: /* Introduction */

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:52:51Z

Jonathan.schreven: /* Equations of Equilibrium */

== Introduction ==

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

== Equations of Equilibrium ==

Using F=ma we can then find our equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:51:05Z

Jonathan.schreven: /* Coupled Oscillator System */

== Equations of Equilibrium ==

In this problem I would like to explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Our system might look something like this.

Using F=ma we can then find our equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:50:25Z

Jonathan.schreven: /* Coupled Oscillator System */

== Coupled Oscillator System ==

In this problem I would like to explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Our system might look something like this.

Using F=ma we can then find our equations of equilibrium.
:'''Equation 1'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
F & = ma \\
F & = m\ddot{x} \\
-k_2(x_2-x_1) & = m_2\ddot{x_2} \\
{-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
-{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Now we can put these four equations into the state space form.

:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}

\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:41:06Z

Jonathan.schreven: /* Coupled Oscillator System */

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:27:11Z

Jonathan.schreven: /* Coupled Oscillator System */

Fall 2009/JonathanS

2009-12-10T01:12:22Z

Jonathan.schreven:

== Problem ==

A simple pendulum with a length L = 0.5m is pulled back and released from an initial angle <math>\theta_0 = 12^o</math>. Then it is run with a forcing function of cos(3<math>\theta</math>) Find an equation that gives the angle of the pendulum at any given time '''t'''.

== Solution ==

Assuming no damping and a small angle(<math>\theta < 15^o</math>), the equation for the motion of a simple pendulum can be written as
:<math>{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{g\over \ell}\theta=cos(3\theta).</math>

Substituting values we get
:<math>{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{9.81\over 0.5}\theta=cos(3\theta).</math>

<math>\Rightarrow{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{19.62}\theta=cos(3\theta).</math>

Remember the identities
:<math>\mathcal{L}\{f(t)\}=F(s)=\int_0^{\infty} e^{-st} f(t) \,dt. </math>

:<math>\mathcal{L}\{f^{ ''}(t)\}=s^2F(s)-sf(0)-f^{ '}(0)</math>

Now we can take the Laplace Transform to change the second order differential equation, from the t domain, into a simple linear equation, from the s domain, that's much easier to work with
:<math>\mathcal{L}\bigg\{{\mathrm{d}^2\theta\over \mathrm{d}t^2}+{19.62}\theta\bigg\}=s^2F(s)-sf(0)-f^{ '}(0)+{19.62}\theta=\mathcal{L}\{cos(3\theta)\}</math>

<math>\Rightarrow</math> <math> s^2\boldsymbol{\theta}-s\theta(0)-\theta^{ '}(0)+{19.62}\boldsymbol{\theta}={s\over {s^2+9}} \,</math>

Since we know that <math>\theta(0)=12^o</math> and the initial velocity <math>\theta^{ '}(0)=0</math> we get
:<math>s^2\boldsymbol{\theta}-12s+19.62\boldsymbol{\theta}={s\over {s^2+9}} \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}(s^2+19.62)={s\over {s^2+9}}+12s \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}(s^2+19.62)={s\over {s^2+9}}+{12s(s^2+9)\over {s^2+9}} \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}(s^2+19.62)={(12s^3+109s)\over {s^2+9}} \,</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}={(12s^3+109s)\over {(s^2+9)(s^2+19.62)}}</math>

<math>\Rightarrow</math> <math>\boldsymbol{\theta}={12s^3+109s\over {(s^4+28.62s^2+176.58)}}</math>

Now we can take the inverse Laplace Transform to convert our equation back into the time domain

:<math>\mathcal{L}^{-1}\bigg\{{12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg\}=0.094162cos(3t)+11.9058cos(4.42945t)</math>

This will give us the angle (in degrees) of the pendulum at any given time '''t'''.

== Initial Value Theorem ==

We can use the Initial Value Theorem as a check that our initial values for the problem are valid.
:<math>\mathcal{L}\{f^{ '}(t)\}=sF(s)-f(0^-)=\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt</math>

:<math>\begin{alignat}{3}
\lim_{n \to \infty}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to \infty}sF(s)-f(0^-) \\
& = \lim_{n \to \infty}sF(s)-f(0^-)=0 \\
& = \lim_{n \to \infty}sF(s)=f(0^-) \\
\end{alignat}</math>

Below we will use this theorem to check the values for our problem.
:<math>\begin{alignat}{3}
\lim_{n \to \infty}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to \infty}sF(s)-f(0^-)=\lim_{n \to \infty}\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt \\
& = \lim_{n \to \infty}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)-f(0^-)=0 \\
& = \lim_{n \to \infty}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)=f(0^-) \\
12 & = f (0^-) \\
\end{alignat}</math>
This value <math>f(0^-)=12</math> is the initial angle we gave the pendulum so it checks out.

== Final Value Theorem ==
We can use the Final Value Theorem as a check that our final values for the problem are valid.
:<math>\mathcal{L}\{f^{ '}(t)\}=sF(s)-f(\infty)=\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt</math>

:<math>\begin{alignat}{3}
\lim_{n \to 0}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to 0}sF(s)-f(\infty) \\
& = \lim_{n \to 0}sF(s)-f(\infty)=0 \\
& = \lim_{n \to 0}sF(s)=f(\infty) \\
\end{alignat}</math>

Below we will use this theorem to check the values for our problem.
:<math>\begin{alignat}{3}
\lim_{n \to 0}\mathcal{L}\{f^{ '}(t)\} & = \lim_{n \to 0}sF(s)-f(\infty)=\lim_{n \to 0}\int_{0^-}^{\infty}f^{ '}(t)e^{-st}\, dt \\
& = \lim_{n \to 0}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)-f(\infty)=0 \\
& = \lim_{n \to 0}s\bigg({12s^3+109s\over {(s^4+28.62s^2+176.58)}}\bigg)=f(\infty) \\
0 & = f (\infty) \\
\end{alignat}</math>

This is zero because the average angle as time goes to infinity will be zero (halfway between -12 and 12 degrees).

== Bode Plots ==
To find the bode plot we use the function
:<math>H(s)={12s^3+109s\over {(s^4+28.62s^2+176.58)}}</math>

Entering this into a program such as Maple, Octave, or MATLAB will give you a plot that looks like this

[[Image:Bode_Plot.jpg]]

== Asymptotes & Break Points ==

In this section I will describe how you can use the transfer function to quickly estimate what the bode plot will look like.

To find the initial value we can take <math>20log(H(j\omega))</math>. Which gives us
:<math>H(j\omega)={1 \over {\sqrt{\omega^2+1}\sqrt{\omega^2-1}}}</math>
From our original input we know that <math>\omega=3</math> so
:<math>H(j\omega)={1 \over {\sqrt{3^2+1}\sqrt{3^2-1}}}</math>
:<math>H(j\omega)=0.1118</math>
Now we can plug that into our log function
:<math>\begin{alignat}{3}
db_0 & = 20log(H(j\omega)) \\
& = 20log(0.1118) \\
& = -19.0 \\
\end{alignat}</math>

If we check that against our bode plot above we can see that this is pretty close. Now we can find the deflections using the transfer function. It will be easier to work with in factored form.
For every "s" we have in the numerator we will increase the slope by 20db/decade. For every "s" we have in the denominator we will decrease the slope by 20db/decade. These changes will occur at the poles which form our break points and asymptotes.

:<math>\boldsymbol{\theta}={2s(12s^2+109) \over {(s^2+9)(2s^2+39.24)}}</math>

We get asymptotes at s=3j and s=4.47j from the denominator. For our purposes we can just ignore the imaginary part. We should also have increases of 20db/decade at s=0 and s=3.01 which we find from the numerator.

== Convolution ==

Convolution, by definition, states that H(s)=Y(s)/X(s) where H(s) is the transfer function, Y(s) is the output function, and X(s) is the input function. We would like to find the output of the system, so for our purposes we will rearrange the above equation like this.
:<math>\begin{alignat}{3}
Y(s) & = H(s)X(s) \\
& = \bigg({12s^3+109s\over {s^4+28.62s^2+176.58}}\bigg)\bigg({s\over {s^2+9}}\bigg) \\
& = {2s^2(12s^2+109)\over {(s^2+9)^2(2s^2+39.24)}} \\
\end{alignat}</math>

== State Space Equations ==

Now we can transfer the functions we already have into state space. This is simply taking our original differential equation and putting it in terms of vectors and matrices. This can make the equations easier to work with.

<math>\begin{bmatrix}
\dot{\theta} \\
\ddot{\theta}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
{-g\over \ell} & 0
\end{bmatrix}

\begin{bmatrix}
\theta \\
\dot{\theta}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}cos(3\theta)</math>

Coupled Oscillator: Jonathan Schreven

2009-12-10T01:11:18Z

Jonathan.schreven: /* Coupled Oscillator System */