Coupled Oscillator: Spring Pendulums

Problem Statement

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Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system.

Solution

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By definition, the the state equation is stated as

${\underline {\dot {x}}}={\widehat {A}}\,{\underline {x}}+{\widehat {B}}\,{\underline {u}}$

Now, consider the motion equations described in the Solution section,

$m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})=m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}x_{2}+k_{2}x_{1}=0$

$m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})-k_{3}x_{2}=m_{2}{\ddot {x}}_{2}+k_{2}x_{2}-k_{2}x_{1}-k_{3}x_{2}=0$

Solving for ${\ddot {x}}_{1}$ and ${\ddot {x}}_{2}$ yields,

${\ddot {x}}_{1}=-{\dfrac {k_{1}}{m_{1}}}x_{1}+{\dfrac {k_{2}}{m_{1}}}x_{2}-{\dfrac {k_{2}}{m_{1}}}x_{1}$

${\ddot {x}}_{2}={\dfrac {-k_{2}}{m_{2}}}x_{2}+{\dfrac {k_{2}}{m_{2}}}x_{1}+{\dfrac {k_{3}}{m_{2}}}x_{2}$

Finally, we let $x_{1}{\frac {}{}}$, ${\dot {x}}_{1}{\frac {}{}}$, $x_{2}{\frac {}{}}$, and ${\dot {x_{2}}}{\frac {}{}}$ be the state variables. Thus,

${\begin{bmatrix}{\dot {x}}_{1}\\{\ddot {x}}_{1}\\{\dot {x}}_{2}\\{\ddot {x}}_{2}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\-{\frac {1}{m_{1}}}(k_{1}+k_{2})&0&{\frac {k_{2}}{m_{1}}}&0\\0&0&0&1\\{\frac {k_{2}}{m_{2}}}&0&{\frac {1}{m_{2}}}(k_{3}-k_{2})&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}$

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Created by Kendrick Mensink