Class Wiki - User contributions [en]

Coupled Oscillator: horizontal Mass-Spring

2009-12-14T20:55:26Z

Gregory.peterson: /* Matrix Exponential */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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=-3.0937,</math>
:<math>\lambda_2=2.1380i,</math>
:<math>\lambda_3=- 2.1380i,</math>
:<math>\lambda_4=3.0937,</math>

== Eigen Vectors ==
'''Using the equation above and the same given conditions we can plug everything into 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}
0.0520 \\
-0.1609 \\
-0.3031 \\
0.9378
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0.4176i \\
-0.8928 \\
- 0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
- 0.4176i \\
-0.8928 \\
0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
-0.0520 \\
-0.1609 \\
0.3031 \\
0.9378
\end{bmatrix}</math>

==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0.0520 \\-0.1609 \\-0.3031\\0.9378\end{bmatrix}\,</math><math>e^{-3.0937}+ c_2</math><math>\begin{bmatrix}0.4176i \\-0.8928\\- 0.0716i\\0.1532\end{bmatrix}\,</math><math>e^{2.1380i}+ c_3</math><math>\begin{bmatrix}- 0.4176i \\-0.8928\\0.0716i\\0.1532\end{bmatrix}\,</math><math>e^{- 2.1380i}+ c_4</math><math>\begin{bmatrix}-0.0520 \\-0.1609\\0.3031\\0.9378\end{bmatrix}\,
</math><math>e^{3.0937}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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}
0.0520 & 0.4176i & - 0.4176i & -0.0520 \\
-0.1609 & -0.8928 & -0.8928 & -0.1609 \\
-0.3031 & - 0.0716i & 0.0716i & 0.3031 \\
0.9378 & 0.1532 & 0.1532 & 0.9378
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
-0.2914 & 0.0943 & -1.6996 & 0.5493 \\
- 1.2337i & -0.5770 & - 0.2117i & -0.0990 \\
1.2335i & -0.5770 & 0.2116i & -0.0990 \\
0.2914 & 0.0943 & 1.6996 & 0.5493
\end{bmatrix}</math>

'''So taking'''
:<math>\dot{z}=TAT^{-1}z</math>
'''We get the uncoupled matrix of'''
:<math>\dot{z}=\begin{bmatrix}
-3.0937 & 0 & 0 & 0 \\
0 & 2.1380i & 0 & 0 \\
0 & 0 & - 2.1380i & 0 \\
0 & 0 & 0 & 3.0937
\end{bmatrix}</math>

created by Greg Peterson

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T23:17:53Z

Gregory.peterson: /* Matrix Exponential */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T23:10:48Z

Gregory.peterson: /* Matrix Exponential */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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=-3.0937,</math>
:<math>\lambda_2=2.1380i,</math>
:<math>\lambda_3=- 2.1380i,</math>
:<math>\lambda_4=3.0937,</math>

== Eigen Vectors ==
'''Using the equation above and the same given conditions we can plug everything into 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}
0.0520 \\
-0.1609 \\
-0.3031 \\
0.9378
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0.4176i \\
-0.8928 \\
- 0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
- 0.4176i \\
-0.8928 \\
0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
-0.0520 \\
-0.1609 \\
0.3031 \\
0.9378
\end{bmatrix}</math>

==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0.0520 \\-0.1609 \\-0.3031\\0.9378\end{bmatrix}\,</math><math>e^{-3.0937}+ c_2</math><math>\begin{bmatrix}0.4176i \\-0.8928\\- 0.0716i\\0.1532\end{bmatrix}\,</math><math>e^{2.1380i}+ c_3</math><math>\begin{bmatrix}- 0.4176i \\-0.8928\\0.0716i\\0.1532\end{bmatrix}\,</math><math>e^{- 2.1380i}+ c_4</math><math>\begin{bmatrix}-0.0520 \\-0.1609\\0.3031\\0.9378\end{bmatrix}\,
</math><math>e^{3.0937}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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}
0.0520 & 0.4176i & - 0.4176i & -0.0520 \\
-0.1609 & -0.8928 & -0.8928 & -0.1609 \\
-0.3031 & - 0.0716i & 0.0716i & 0.3031 \\
0.9378 & 0.1532 & 0.1532 & 0.9378
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''So taking'''
:<math>\dot{z}=TAT^{-1}z</math>
'''We get the uncoupled matrix of'''
:<math>\dot{z}=\begin{bmatrix}
-3.0937 & 0 & 0 & 0 \\
0 & 2.1380i & 0 & 0 \\
0 & 0 & - 2.1380i & 0 \\
0 & 0 & 0 & 3.0937
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T23:00:47Z

Gregory.peterson: /* So then the answer is... */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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=-3.0937,</math>
:<math>\lambda_2=2.1380i,</math>
:<math>\lambda_3=- 2.1380i,</math>
:<math>\lambda_4=3.0937,</math>

== Eigen Vectors ==
'''Using the equation above and the same given conditions we can plug everything into 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}
0.0520 \\
-0.1609 \\
-0.3031 \\
0.9378
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0.4176i \\
-0.8928 \\
- 0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
- 0.4176i \\
-0.8928 \\
0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
-0.0520 \\
-0.1609 \\
0.3031 \\
0.9378
\end{bmatrix}</math>

==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0.0520 \\-0.1609 \\-0.3031\\0.9378\end{bmatrix}\,</math><math>e^{-3.0937}+ c_2</math><math>\begin{bmatrix}0.4176i \\-0.8928\\- 0.0716i\\0.1532\end{bmatrix}\,</math><math>e^{2.1380i}+ c_3</math><math>\begin{bmatrix}- 0.4176i \\-0.8928\\0.0716i\\0.1532\end{bmatrix}\,</math><math>e^{- 2.1380i}+ c_4</math><math>\begin{bmatrix}-0.0520 \\-0.1609\\0.3031\\0.9378\end{bmatrix}\,
</math><math>e^{3.0937}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:57:43Z

Gregory.peterson: /* So then the answer is... */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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=-3.0937,</math>
:<math>\lambda_2=2.1380i,</math>
:<math>\lambda_3=- 2.1380i,</math>
:<math>\lambda_4=3.0937,</math>

== Eigen Vectors ==
'''Using the equation above and the same given conditions we can plug everything into 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}
0.0520 \\
-0.1609 \\
-0.3031 \\
0.9378
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0.4176i \\
-0.8928 \\
- 0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
- 0.4176i \\
-0.8928 \\
0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
-0.0520 \\
-0.1609 \\
0.3031 \\
0.9378
\end{bmatrix}</math>

==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0 \\0 \\0\\0\end{bmatrix}\,</math><math>e^{-3.0937}+ c_2</math><math>\begin{bmatrix}0 \\0\\0\\0\end{bmatrix}\,</math><math>e^{1}+ c_3</math><math>\begin{bmatrix}0 \\0\\0\\0\end{bmatrix}\,</math><math>e^{0}+ c_4</math><math>\begin{bmatrix}0 \\0\\0\\0\end{bmatrix}\,
</math><math>e^{0}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:50:05Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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=-3.0937,</math>
:<math>\lambda_2=2.1380i,</math>
:<math>\lambda_3=- 2.1380i,</math>
:<math>\lambda_4=3.0937,</math>

== Eigen Vectors ==
'''Using the equation above and the same given conditions we can plug everything into 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}
0.0520 \\
-0.1609 \\
-0.3031 \\
0.9378
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0.4176i \\
-0.8928 \\
- 0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
- 0.4176i \\
-0.8928 \\
0.0716i \\
0.1532
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
-0.0520 \\
-0.1609 \\
0.3031 \\
0.9378
\end{bmatrix}</math>

==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0\sqrt{0}}+ c_2</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_3</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_4</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:46:14Z

Gregory.peterson: /* Eigen Values */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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=-3.0937,</math>
:<math>\lambda_2=2.1380i,</math>
:<math>\lambda_3=- 2.1380i,</math>
:<math>\lambda_4=3.0937,</math>

== Eigen Vectors ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>
==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0\sqrt{0}}+ c_2</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_3</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_4</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:09:44Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>
==So then the answer is...==
'''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>

<math>\ x=c_1</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0\sqrt{0}}+ c_2</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_3</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_4</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:08:03Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>
'''So then the answer is...'''

<math>\ x=c_1</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0\sqrt{0}}+ c_2</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_3</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}+ c_4</math><math>\begin{bmatrix}0 \\0\sqrt(0) \\0 \\0\sqrt(0)\end{bmatrix}\,</math><math>e^{0*0\sqrt{0}}\,</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:06:22Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>
'''So then the answer is...'''

<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>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:05:25Z

Gregory.peterson: /* Matrix Exponential */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

'''Taking the inverse of this we can solve for T'''
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:03:53Z

Gregory.peterson: /* Matrix Exponential */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

== Matrix Exponential ==
'''We now use matrix exponentials to solve the same problem.'''
:<math>z=Tx\,</math>

'''So from the above equation we get this to prove the matrix exponetial works.'''
:<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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:02:12Z

Gregory.peterson: /* Eigen Values */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>
== Eigen Values ==
'''Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.'''
:'''Given'''
:<math>m_1=10kg\,</math>
:<math>m_2=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:01:16Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\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=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''Using the equation above and the same given conditions we can plug everything into 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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:00:41Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\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=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
'''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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T22:00:21Z

Gregory.peterson: /* Eigen Values */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\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=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:59:33Z

Gregory.peterson: /* Matrix Exponential */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\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=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</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 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Taking the inverse of this we can solve for T
:<math>T=\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:57:52Z

Gregory.peterson: /* Eigen Vectors */

=Problem Statement=

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

==State Equations==

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\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=10kg\,</math>
:<math>k_1=25\,{N\over {m}}</math>
:<math>k_2=50\,{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 \\
-5 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 1 \\
2.5 & 0 & 10 & 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 ==
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 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_2=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_3=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>

:<math>k_4=\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</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: horizontal Mass-Spring

2009-12-10T21:57:08Z

Gregory.peterson: /* Eigen Values */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:55:20Z

Gregory.peterson: /* Eigen Values */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:54:55Z

Gregory.peterson: /* Eigen Values */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:54:35Z

Gregory.peterson: /* Eigen Values */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:53:44Z

Gregory.peterson: /* Eigen Values */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:52:23Z

Gregory.peterson: /* Eigen Values */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:52:03Z

Gregory.peterson: /* State Equations */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:51:13Z

Gregory.peterson: /* Problem Statement */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:48:39Z

Gregory.peterson: /* Problem Statement */

===Problem Statement===

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

'''State Equations'''

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{10 kg}&0&\frac{-25 N/m}{10 kg}&0 \\
0&0&0&1 \\
\frac{25 N/m}{10 kg}&0&\frac{(100 N/m)}{10 kg}&0
\end{bmatrix}

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

</math>

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:45:38Z

Gregory.peterson:

===Problem Statement===

'''Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes of the system.'''

[[Image:horizontal spring.jpg]]
'''Initial Conditions:'''
:<math>m_1= 10 kg\,</math>

:<math>m_2 = 10 kg\,</math>

:<math>k1=25 N/m\,</math>

:<math>k2=75 N/m\,</math>

:<math>k3=50 N/m\,</math>

'''Equations for M_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>

'''Equations for M_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>

'''Additional Equations'''
:<math>\dot{x_1}=\dot{x_1}</math>
:<math>\dot{x_2}=\dot{x_2}</math>

'''State Equations'''

<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&0&0&0 \\
0&0&0&0 \\
0&0&0&0 \\
0&0&0&0
\end{bmatrix}

\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}

</math>

'''With the numbers...'''

<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{(-50 N/m)}{15 kg}&0&\frac{-100 N/m}{15 kg}&0 \\
0&0&0&1 \\
\frac{100 N/m}{15 kg}&0&\frac{(250 N/m)}{15 kg}&0
\end{bmatrix}

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

</math>

File:Horizontal spring.jpg

2009-12-10T21:34:15Z

Gregory.peterson:

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:33:53Z

Gregory.peterson: /* Problem Statement */

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:32:53Z

Gregory.peterson:

Coupled Oscillator: horizontal Mass-Spring

2009-12-10T21:28:43Z

Gregory.peterson: New page: =Setup=

=Setup=

Fall 2009

2009-12-10T21:27:59Z

Gregory.peterson: /* HW #12 */

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms:DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

[[Laplace Transforms: Vertical Motion of a Coupled Spring System]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

[[Coupled Oscillator: Coupled Mass-Spring System with Damping]]

[[Coupled Oscillator: Jonathan Schreven]]

[[Coupled Horizontal Spring Mass Oscillator]]

[[Coupled Oscillator: Double Pendulum]]

[[Coupled Oscillator: horizontal Mass-Spring]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Laplace transforms: Critically Damped Spring Mass system

2009-12-10T21:26:07Z

Gregory.peterson: /* ______________________________Break Points__________________________________ */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==Break Points==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

<math>\text {The convolution equation is as follows: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-4(t-t_0)e^{-2t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>

<math>\mathbf {x}(t)=(-cte^{-2t})</math> 

<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>

==State Space==

<math>\text {Using state equatons is just another way to solve a system modeled by an ODE }\,</math>

<math>\text {First we need to add an applied force so u(t)=2N }\,</math>

<math>m=\frac{98.1}{9.81}=10 kg</math>

<math>\text {k=40}\,</math>

<math>\text {C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\ddot{x}(0)=0</math>

<math>\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -k/m & -C/m \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}u(t)</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-12-03T06:31:52Z

Gregory.peterson: /* State Space */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==______________________________Break Points__________________________________==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

<math>\text {The convolution equation is as follows: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-4(t-t_0)e^{-2t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>

<math>\mathbf {x}(t)=(-cte^{-2t})</math> 

<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>

==State Space==

<math>\text {Using state equatons is just another way to solve a system modeled by an ODE }\,</math>

<math>\text {First we need to add an applied force so u(t)=2N }\,</math>

<math>m=\frac{98.1}{9.81}=10 kg</math>

<math>\text {k=40}\,</math>

<math>\text {C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\ddot{x}(0)=0</math>

<math>\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -k/m & -C/m \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}u(t)</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-12-03T06:31:35Z

Gregory.peterson: /* State Space */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==______________________________Break Points__________________________________==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

<math>\text {The convolution equation is as follows: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-4(t-t_0)e^{-2t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>

<math>\mathbf {x}(t)=(-cte^{-2t})</math> 

<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>

==State Space==

<math>\text {Using state equatons is just another way to solve a system modeled by an ODE }\,</math>

<math>\text {First we need to add an applied force so u(t)=2N }\,</math>

<math>m=\frac{98.1}{9.81}=\frac1 10 kg</math>

<math>\text {k=40}\,</math>

<math>\text {C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\ddot{x}(0)=0</math>

<math>\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -k/m & -C/m \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}u(t)</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-12-03T06:30:43Z

Gregory.peterson: /* State Space */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==______________________________Break Points__________________________________==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

<math>\text {The convolution equation is as follows: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-4(t-t_0)e^{-2t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>

<math>\mathbf {x}(t)=(-cte^{-2t})</math> 

<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>

==State Space==

<math>\text {Using state equatons is just another way to solve a system modeled by an ODE }\,</math>

<math>\text {First we need to add an applied force so u(t)=2N }\,</math>

<math>m=\frac{8}{32}=\frac1 4 slugs</math>

<math>\text {k=40}\,</math>

<math>\text {C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\ddot{x}(0)=0</math>

<math>\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -k/m & -C/m \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}u(t)</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-12-03T06:27:27Z

Gregory.peterson: /* Convolution */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==______________________________Break Points__________________________________==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

<math>\text {The convolution equation is as follows: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-4(t-t_0)e^{-2t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>

<math>\mathbf {x}(t)=(-cte^{-2t})</math> 

<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>

==State Space==

<math>\text {Using state equatons is just another way to solve a system modeled by an ODE }\,</math>

<math>\text {First we need to add an applied force so u(t)=2N }\,</math>

<math>m=\frac{8}{32}=\frac1 4 slugs</math>

<math>\text {k=4}\,</math>

<math>\text {C=2}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-3</math>

<math>\ddot{x}(0)=0</math>

<math>\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -k/m & -C/m \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}u(t)</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-12-03T06:21:54Z

Gregory.peterson: /* Convolution */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==______________________________Break Points__________________________________==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

<math>\text {The convolution equation is as follows: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-4(t-t_0)e^{-2t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>

<math>\mathbf {x}(t)=(-cte^{-2t})</math> 

<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T23:07:55Z

Gregory.peterson:

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==______________________________Break Points__________________________________==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

coming soon...?

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T23:06:24Z

Gregory.peterson: /* Given */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}kg</math>

<math>\text {Spring Constant k=40}N/m\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==Break Points ==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

coming soon...?

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T23:05:41Z

Gregory.peterson: /* Applying this to our problem */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}</math>

<math>\text {Spring Constant k=40}\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0.}\,</math>
<math>\text {Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}m</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==Break Points ==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

coming soon...?

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T23:04:26Z

Gregory.peterson:

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}</math>

<math>\text {Spring Constant k=40}\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0. Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}ft</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==Break Points ==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

==Convolution==

coming soon...?

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T23:02:47Z

Gregory.peterson: /* Break Points */

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}</math>

<math>\text {Spring Constant k=40}\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0. Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}ft</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==Break Points ==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T23:02:19Z

Gregory.peterson:

=Using the Laplace Transform to solve a spring mass system that is critically damped=

==Problem Statement==
An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m.
The spring is stretched 4 m and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 2 m/s.
The system contains a damping force of 40 times the initial velocity.

==Solution==

===Given===

<math>m=\frac{98}{9.81}</math>

<math>\text {Spring Constant k=40}\,</math>

<math>\text {Damping Constant C=40}\,</math>

<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-4</math>

<math>\text {Standard equation: }\,</math>

<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac{98}{9.8} \frac{d^2x}{dt^2}=-40x-40\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0</math>

<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]</math> 

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+4\mathbf s{X}(s)+4\mathbf{X}(s)=-4</math> 

<math>\mathbf {X}(s)(s^2+4s+4)=-4</math> 

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{4}{(s+2)^2}]</math> 

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

<math>\mathbf {x}(t)=-4te^{-2t}</math> 

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{4}{(\infty+2)^2}=0\,</math>

<math>\text {So as you can see the value for the initial position will be 0. Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(s+2)^2}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{4}{(0+2)^2}=-\frac 4 {4}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{4}{4}ft</math>

<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>

==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bodeplotlna.jpeg|700px|thumb|left|Fig (1)]]

==Break Points==

<math>\text {Find the Break points using the transfer function.}\,</math>

===Transfer fucntion===

<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math> 

<math>\text {The equation above contains break points but only in the denominator.}\,</math>

<math>\text {There is only the variable s in the denominator so only those types of break point exist}\,</math>

<math>\text {The break points are asymtotes at the point -2 which occurs twice in this particular equation}\,</math>

Created by Greg Peterson

Checked by Mark Bernet

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T22:57:46Z

Gregory.peterson: /* Transfer fucntion */

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T22:57:27Z

Gregory.peterson:

Laplace transforms: Critically Damped Spring Mass system

2009-10-27T22:54:48Z

Gregory.peterson: New page: =Using the Laplace Transform to solve a spring mass system that is critically damped= ==Problem Statement== An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m...

Fall 2009

2009-10-27T22:54:28Z

Gregory.peterson: /* HW # 5 */

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms:DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

File:Bodeplotlna.jpeg

2009-10-27T22:51:49Z

Gregory.peterson: