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

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

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,

$\left[\begin{array}{c}{\dot{x}}_1\\ {\ddot{x}}_1\\ {\dot{x}}_2\\ {\ddot{x}}_2\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ -\frac{1}{m_1}(k_1+k_2)& 0& \frac{k_2}{m_1}& 0\\ 0& 0& 0& 1\\ \frac{k_2}{m_2}& 0& \frac{1}{m_2}(k_3-k_2)& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ {\dot{x}}_1\\ x_2\\ {\dot{x}}_2\end{array}\right]$

---

Created by Kendrick Mensink