Coupled Oscillator: Spring Pendulums

Problem Statement

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

Solution

By definition, the the state equation is stated as

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

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} = - \frac{k_{1}}{m_{1}} x_{1} + \frac{k_{2}}{m_{1}} x_{2} - \frac{k_{2}}{m_{1}} x_{1}

{\ddot{x}}_{2} = \frac{- k_{2}}{m_{2}} x_{2} + \frac{k_{2}}{m_{2}} x_{1} + \frac{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,

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

Created by Kendrick Mensink

Coupled Oscillator: Spring Pendulums

Problem Statement

Solution

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