Latest revision as of 13:55, 14 December 2009

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.

Initial Conditions:

m_{1} = 10 k g

m_{2} = 10 k g

k 1 = 25 N / m

k 2 = 75 N / m

k 3 = 50 N / m

Equations for M_1

\begin{aligned} F & = m a \\ F & = m \ddot{x} \\ - k_{1} x_{1} - k_{2} (x_{1} x_{2}) & = m_{1} \ddot{x_{1}} \\ - \frac{k_{1} x_{1}}{m_{1}} - \frac{k_{2} (x_{1} - x_{2})}{m_{1}} & = m_{1} \ddot{x_{1}} \\ - \frac{k_{1} x_{1}}{m_{1}} - \frac{k_{2} (x_{1} - x_{2})}{m_{1}} & = \ddot{x_{1}} \\ - \frac{k_{1} + k_{2}}{m_{1}} x_{1} + \frac{k_{2}}{m_{1}} x_{2} & = \ddot{x_{1}} \end{aligned}

Equations for M_2

\begin{aligned} F & = m a \\ F & = m \ddot{x} \\ - k_{2} (x_{2} - x_{1}) & = m_{2} \ddot{x_{2}} \\ \frac{- k_{2} (x_{2} - x_{1})}{m_{2}} & = \ddot{x_{2}} \\ - \frac{k_{2}}{m_{2}} x_{2} + \frac{k_{2}}{m_{2}} x_{1} & = \ddot{x_{2}} \end{aligned}

Additional Equations

\dot{x_{1}} = \dot{x_{1}}

\dot{x_{2}} = \dot{x_{2}}

State Equations

$[\begin{matrix} \dot{x_{1}} \\ \ddot{x_{1}} \\ \dot{x_{2}} \\ \ddot{x_{2}} \end{matrix}]$ = $[\begin{matrix} 0 & 1 & 0 & 0 \\ \frac{(k_{1} - k_{2})}{m_{1}} & 0 & \frac{- k_{1}}{m_{1}} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k_{1}}{m_{2}} & 0 & \frac{(k_{1} + k_{2})}{m_{2}} & 0 \end{matrix}] [\begin{matrix} x_{1} \\ {\dot{x}}_{1} \\ x_{2} \\ {\dot{x}}_{2} \end{matrix}] + [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$

With the numbers...

$[\begin{matrix} \dot{x_{1}} \\ \ddot{x_{1}} \\ \dot{x_{2}} \\ \ddot{x_{2}} \end{matrix}]$ = $[\begin{matrix} 0 & 1 & 0 & 0 \\ \frac{(- 50 N / m)}{10 k g} & 0 & \frac{- 25 N / m}{10 k g} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{25 N / m}{10 k g} & 0 & \frac{(100 N / m)}{10 k g} & 0 \end{matrix}] [\begin{matrix} x_{1} \\ {\dot{x}}_{1} \\ x_{2} \\ {\dot{x}}_{2} \end{matrix}]$

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

m_{1} = 10 k g

m_{2} = 10 k g

k_{1} = 25 \frac{N}{m}

k_{2} = 50 \frac{N}{m}

We now have

[\begin{matrix} \dot{x_{1}} \\ \ddot{x_{1}} \\ \dot{x_{2}} \\ \ddot{x_{2}} \end{matrix}] = [\begin{matrix} 0 & 1 & 0 & 0 \\ - 5 & 0 & - 2.5 & 0 \\ 0 & 0 & 0 & 1 \\ 2.5 & 0 & 10 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ \dot{x_{1}} \\ x_{2} \\ \dot{x_{2}} \end{matrix}] + [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}]

From this we get

λ_{1} = - 3.0937,

λ_{2} = 2.1380 i,

λ_{3} = - 2.1380 i,

λ_{4} = 3.0937,

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 $k_{1}, k_{2}, k_{3}, k_{4}$ .

k_{1} = [\begin{matrix} 0.0520 \\ - 0.1609 \\ - 0.3031 \\ 0.9378 \end{matrix}]

k_{2} = [\begin{matrix} 0.4176 i \\ - 0.8928 \\ - 0.0716 i \\ 0.1532 \end{matrix}]

k_{3} = [\begin{matrix} - 0.4176 i \\ - 0.8928 \\ 0.0716 i \\ 0.1532 \end{matrix}]

k_{4} = [\begin{matrix} - 0.0520 \\ - 0.1609 \\ 0.3031 \\ 0.9378 \end{matrix}]

So then the answer is...

We can now plug these eigen vectors and eigen values into the standard equation

x = c_{1} k_{1} e^{λ_{1} t} + c_{2} k_{2} e^{λ_{2} t} + c_{3} k_{3} e^{λ_{3} t} + c_{4} k_{4} e^{λ_{4} t}

$x = c_{1}$ $[\begin{matrix} 0.0520 \\ - 0.1609 \\ - 0.3031 \\ 0.9378 \end{matrix}]$ $e^{- 3.0937} + c_{2}$ $[\begin{matrix} 0.4176 i \\ - 0.8928 \\ - 0.0716 i \\ 0.1532 \end{matrix}]$ $e^{2.1380 i} + c_{3}$ $[\begin{matrix} - 0.4176 i \\ - 0.8928 \\ 0.0716 i \\ 0.1532 \end{matrix}]$ $e^{- 2.1380 i} + c_{4}$ $[\begin{matrix} - 0.0520 \\ - 0.1609 \\ 0.3031 \\ 0.9378 \end{matrix}]$ $e^{3.0937}$

Matrix Exponential

We now use matrix exponentials to solve the same problem.

z = T x

So from the above equation we get this to prove the matrix exponetial works.

\dot{z} = T A T^{- 1} z

We also know what T equals and we can solve it for our case

T^{- 1} = [k_{1} | k_{2} | k_{3} | k_{4}]

T^{- 1} = [\begin{matrix} 0.0520 & 0.4176 i & - 0.4176 i & - 0.0520 \\ - 0.1609 & - 0.8928 & - 0.8928 & - 0.1609 \\ - 0.3031 & - 0.0716 i & 0.0716 i & 0.3031 \\ 0.9378 & 0.1532 & 0.1532 & 0.9378 \end{matrix}]

Taking the inverse of this we can solve for T

T = [\begin{matrix} - 0.2914 & 0.0943 & - 1.6996 & 0.5493 \\ - 1.2337 i & - 0.5770 & - 0.2117 i & - 0.0990 \\ 1.2335 i & - 0.5770 & 0.2116 i & - 0.0990 \\ 0.2914 & 0.0943 & 1.6996 & 0.5493 \end{matrix}]

So taking

\dot{z} = T A T^{- 1} z

We get the uncoupled matrix of

\dot{z} = [\begin{matrix} - 3.0937 & 0 & 0 & 0 \\ 0 & 2.1380 i & 0 & 0 \\ 0 & 0 & - 2.1380 i & 0 \\ 0 & 0 & 0 & 3.0937 \end{matrix}]

created by Greg Peterson

@@ Line 1: / Line 1: @@
-===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.'''
+'''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]]
@@ Line 39: / Line 39: @@
 :<math>\dot{x_2}=\dot{x_2}</math>
-'''State Equations'''
+==State Equations==
 <math>
@@ Line 112: / Line 112: @@
 </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}
+& 1 & 0 & 0 \\
+-5 & 0 & -2.5 & 0 \\
+& 0 & 0 & 1 \\
+.5 & 0 & 10 & 0
+\end{bmatrix}
+\begin{bmatrix}
+{x_1} \\
+\dot{x_1} \\
+{x_2} \\
+\dot{x_2}
+\end{bmatrix}
++
+\begin{bmatrix}
+\\
+\\
+\\
+\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}
+.0520 \\
+-0.1609 \\
+-0.3031 \\
+.9378
+\end{bmatrix}</math>
+:<math>k_2=\begin{bmatrix}
+.4176i \\
+-0.8928 \\
+- 0.0716i \\
+.1532
+\end{bmatrix}</math>
+:<math>k_3=\begin{bmatrix}
+- 0.4176i \\
+-0.8928 \\
+.0716i \\
+.1532
+\end{bmatrix}</math>
+:<math>k_4=\begin{bmatrix}
+-0.0520 \\
+-0.1609 \\
+.3031 \\
+.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}
+.0520 & 0.4176i & - 0.4176i & -0.0520 \\
+-0.1609 & -0.8928 & -0.8928 & -0.1609 \\
+-0.3031 & - 0.0716i & 0.0716i & 0.3031 \\
+.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 \\
+.2335i & -0.5770 & 0.2116i & -0.0990 \\
+.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 \\
+& 2.1380i & 0 & 0 \\
+& 0 & - 2.1380i & 0 \\
+& 0 & 0 & 3.0937
+\end{bmatrix}</math>
+created by Greg Peterson

Coupled Oscillator: horizontal Mass-Spring: Difference between revisions

Latest revision as of 13:55, 14 December 2009

Contents

Problem Statement

State Equations

Eigen Values

Eigen Vectors

So then the answer is...

Matrix Exponential

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Coupled Oscillator: horizontal Mass-Spring: Difference between revisions

Latest revision as of 13:55, 14 December 2009

Problem Statement

State Equations

Eigen Values

Eigen Vectors

So then the answer is...

Matrix Exponential

Navigation menu

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