Coupled Oscillator: Hellie

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. Solve Using the Matrix Exponential

Initial Conditions:

$m_{1}=10kg\,$

$m_{2}=10kg\,$

$k1=100N/m\,$

$k2=150N/m\,$

$k3=100N/m\,$

F=ma

${\ddot {x_{1}}}={\frac {x_{1}(k_{1}-k_{2})}{m_{1}}}-{\frac {x_{2}*k_{1}}{m_{1}}}\,$

${\ddot {x_{2}}}={\frac {x_{2}(k_{1}+k_{2})}{m_{2}}}-{\frac {x_{1}*k_{1}}{m_{2}}}\,$

State Equations

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

With the numbers...

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}\,$ = ${\begin{bmatrix}0&1&0&0\\{\frac {(-50N/m)}{10kg}}&0&{\frac {-100N/m}{10kg}}&0\\0&0&0&1\\{\frac {-100N/m}{10kg}}&0&{\frac {(250N/m)}{10kg}}&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}$

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}\,$ = ${\begin{bmatrix}0&1&0&0\\-5&0&-10&0\\0&0&0&1\\-10&0&25&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}$

Eigenvalues

$\lambda _{1}=-5.29412\,$

$\lambda _{2}=2.83333i\,$

$\lambda _{3}=-2.83333i\,$

$\lambda _{4}=0\,$

Eigenvectors

$k_{1}={\begin{bmatrix}-.05379\\.28475\\.17764\\-.94046\end{bmatrix}}$

$k_{2}={\begin{bmatrix}-.31854i\\.90253\\-.09645i\\.27326\end{bmatrix}}$

$k_{3}={\begin{bmatrix}.31854i\\.90253\\.09645i\\.27326\end{bmatrix}}$

$k_{4}={\begin{bmatrix}-.05379\\-.28475\\.17764\\.94046\end{bmatrix}}$

Standard Equation

$x=c_{1}k_{1}e^{\lambda _{1}t}+c_{2}k_{2}e^{\lambda _{2}t}+c_{3}k_{3}e^{\lambda _{3}t}+c_{4}k_{4}e^{\lambda _{4}t}$

$\ x=c_{1}$${\begin{bmatrix}-.05379\\.28475\\.17764\\-.94046\end{bmatrix}}\,$$e^{-5.29412}+c_{2}\,$${\begin{bmatrix}-.31854i\\.90253\\-.09645i\\.27326\end{bmatrix}}\,$$e^{2.83333i}+c_{3}\,$${\begin{bmatrix}.31854i\\.90253\\.09645i\\.27326\end{bmatrix}}\,$$e^{-2.83333i}+c_{4}\,$${\begin{bmatrix}-.05379\\-.28475\\.17764\\.94046\end{bmatrix}}\,$$e^{0}\,$

Eigenmodes

There are two eigenmodes for the system

1) m1 and m2 oscillating together

2) m1 and m2 oscillating at exactly a half period difference

Matrix Exponential using transformation z=Tx

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

$z=Tx\,$

${\dot {z}}=TAT^{-1}z\,$

${\dot {z}}=\,$ ${\begin{bmatrix}-5.2941&0&0&0\\0&2.833i&0&0\\0&0&-2.83333i&0\\0&0&0&5.2941\end{bmatrix}}\,$ $z\,$

$B=TAT^{-1}={\begin{bmatrix}-5.2941&0&0&0\\0&2.833i&0&0\\0&0&-2.83333i&0\\0&0&0&5.2941\end{bmatrix}}\,$

$z=e^{Bt}z(0)\,$

$e^{Bt}={\begin{bmatrix}e^{-5.2941t}&0&0&0\\0&e^{2.833it}&0&0\\0&0&e^{-2.83333it}&0\\0&0&0&e^{5.2941t}\end{bmatrix}}\,$

$x=T^{-1}z\,$

$x=T^{-1}e^{Bt}Tx(0)\,$

$e^{Pt}=T^{-1}e^{Bt}T\,$

$e^{Pt}=\,$lots of variables

Another way to solve using the Matrix exponential

$e^{At}={\mathcal {L}}^{-1}\left\{[SI-A]^{-1}\right\}\,$

$[SI-A]\,$ = ${\begin{bmatrix}S&1&0&0\\{\frac {(-50N/m)}{15kg}}&S&{\frac {-100N/m}{15kg}}&0\\0&0&S&1\\{\frac {100N/m}{15kg}}&0&{\frac {(250N/m)}{15kg}}&S\end{bmatrix}}$

$[SI-A]^{-1}=\,$ (something too large for my calculator to display or that I want to type out)

${\mathcal {L}}^{-1}\left\{[SI-A]^{-1}\right\}=\,$(something too large for my calculator to display or that I want to type out)

Written by: Andrew Hellie