Coupled Oscillator: Double Pendulum

By Jimmy Apablaza By Jimmy Apablaza

This problem is described in Page 321-322, Section 7.6 of the A first Course in Differential Equations textbook, 8ED (ISBN 0-534-41878-3).

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Figure 1. Coupled Pendulum.‎

Problem Statement

Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

Assumptions:

• the system oscillates vertically under the influence of gravity.
• the mass of both rod are neligible
• no dumpung forces act on the system
• positive direction to the right.

The system of differential equations describing the motion is nonlinear

$(m_1+m_2)l_1^2{\theta }_1^{\prime \prime }+m_2{l}_1{l}_2{\theta }_2^{\prime \prime }cos({\theta }_1-{\theta }_2)+m_2{l}_1{l}_2({\theta }_2^{\prime }{)}^{2}sin({\theta }_1-{\theta }_2)+(m_1+m_2)l_{1}gsin{\theta }_1=0$

$m_2{l}_1^2{\theta }_2^{\prime \prime }+m_2{l}_1{l}_2{\theta }_1^{\prime \prime }cos({\theta }_1-{\theta }_2)-m_2{l}_1{l}_2({\theta }_1^{\prime }{)}^{2}sin({\theta }_1-{\theta }_2)+m_2{l}_{2}gsin{\theta }_2=0$

In order to linearize these equations, we assume that the displacements ${\theta }_1$ and ${\theta }_2$ are small enough so that $cos({\theta }_1-{\theta }_2)\approx 1$ and $sin({\theta }_1-{\theta }_2)\approx 0$. Thus,

$(m_1+m_2)l_1^2{\theta }_1^{\prime \prime }+m_2{l}_1{l}_2{\theta }_2^{\prime \prime }+(m_1+m_2)l_{1}g{\theta }_1=0$

$m_2{l}_1^2{\theta }_2^{\prime \prime }+m_2{l}_1{l}_2{\theta }_1^{\prime \prime }+m_2{l}_{2}g{\theta }_2=0$

Solution

Since our concern is about the motion functions, we will assign the masses $m_1$ and $m_2$, the rod lenghts $l_1$ and $l_1$, and gravitational force $g$ constants to different variables as follows,

$A=(m_1+m_2)l_1^{2}B=m_2{l}_1{l}_{2}C=(m_1+m_2)l_{1}gD=m_2{l}_1^{2}E=m_2{l}_{2}g$

Hence,

$A{\theta }_1^{{\text{'}}^{\prime }}+B{\theta }_2^{{\text{'}}^{\prime }}+C{\theta }_1=0$

$D{\theta }_2^{{\text{'}}^{\prime }}+B{\theta }_1^{{\text{'}}^{\prime }}+E{\theta }_2=0$

Solving for ${\theta }_1^{{\text{'}}^{\prime }}$ and ${\theta }_2^{{\text{'}}^{\prime }}$ we obtain,

${\theta }_1^{{\text{'}}^{\prime }}=-\left(\frac{B}{A}\right){\theta }_2^{{\text{'}}^{\prime }}-\left(\frac{C}{A}\right){\theta }_1$

${\theta }_2^{{\text{'}}^{\prime }}=-\left(\frac{B}{D}\right){\theta }_1^{{\text{'}}^{\prime }}-\left(\frac{E}{D}\right){\theta }_2$

Therefore,

${\theta }_1^{{\text{'}}^{\prime }}=-\left(\frac{CD}{AD+B^2}\right){\theta }_1-\left(\frac{BE}{AD+B^2}\right){\theta }_2$

${\theta }_2^{{\text{'}}^{\prime }}=\left(\frac{BC}{AD+B^2}\right){\theta }_1-\left(\frac{AE}{AD+B^2}\right){\theta }_2$

State Space

$\left[\begin{array}{c}{\theta }_1^{\text{'}}\\ {\theta }_1^{{\text{'}}^{\prime }}\\ {\theta }_2^{\text{'}}\\ {\theta }_2^{{\text{'}}^{\prime }}\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ \frac{-CD}{AD+B^2}& 0& \frac{-BE}{AD+B^2}& 0\\ 0& 0& 0& 1\\ \frac{BC}{AD+B^2}& 0& \frac{-AE}{AD+B^2}& 0\\ \end{array}\right]\left\{\begin{array}{c}{\theta }_1\\ {\theta }_1^{\text{'}}\\ {\theta }_2\\ {\theta }_2^{\text{'}}\end{array}\right\}$