Coupled Oscillator: Double Pendulum

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

$\left[\begin{array}{c}{\theta }_1^{\text{'}}\\ {\theta }_1^{{\text{'}}^{\prime }}\\ {\theta }_2^{\text{'}}\\ {\theta }_2^{{\text{'}}^{\prime }}\end{array}\right]=\hat{A}\underset{\_}{x}(t)+\hat{B}\underset{\_}{u}(t)=\left[\begin{array}{cccc}0& 1& 0& 0\\ & & & \\ \frac{-CD}{AD+B^2}& 0& \frac{-BE}{AD+B^2}& 0\\ & & & \\ 0& 0& 0& 1\\ & & & \\ \frac{BC}{AD+B^2}& 0& \frac{-AE}{AD+B^2}& 0\\ \end{array}\right]\left\{\begin{array}{c}{\theta }_1\\ {\theta }_1^{\text{'}}\\ {\theta }_2\\ {\theta }_2^{\text{'}}\end{array}\right\}+\hat{0}$

Plugging the constants yields,

$\left[\begin{array}{c}{\theta }_1^{\text{'}}\\ {\theta }_1^{{\text{'}}^{\prime }}\\ {\theta }_2^{\text{'}}\\ {\theta }_2^{{\text{'}}^{\prime }}\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ & & & \\ \frac{-l_1(m_1+m_2)g}{l_1^2(m_1+m_2)+l_2^2{m}_2}& 0& \frac{-l_2^2{m}_{2}g}{l_1(l_1^2(m_1+m_2)+l_2^2{m}_2)}& 0\\ & & & \\ 0& 0& 0& 1\\ & & & \\ \frac{l_2(m_1+m_2)g}{l_1^2(m_1+m_2)+l_2^2{m}_2}& 0& \frac{-l_2(m_1+m_2)g}{l_1^2(m_1+m_2)+l_2^2{m}_2}& 0\\ \end{array}\right]\left\{\begin{array}{c}{\theta }_1\\ {\theta }_1^{\text{'}}\\ {\theta }_2\\ {\theta }_2^{\text{'}}\end{array}\right\}$

Let's plug some numbers. Knowing $g=32$, and assuming that $m_1=3$, $m_2=1$, and $l_1=l_2=16$, the state space matrix becomes,

$\left[\begin{array}{c}{\theta }_1^{\text{'}}\\ {\theta }_1^{{\text{'}}^{\prime }}\\ {\theta }_2^{\text{'}}\\ {\theta }_2^{{\text{'}}^{\prime }}\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ -\frac{8}{5}& 0& -\frac{8}{5}& 0\\ 0& 0& 0& 1\\ \frac{8}{5}& 0& -\frac{8}{5}& 0\\ \end{array}\right]\left\{\begin{array}{c}{\theta }_1\\ {\theta }_1^{\text{'}}\\ {\theta }_2\\ {\theta }_2^{\text{'}}\end{array}\right\}$

Laplace Transform

First, we determine the eigenvalues of the $\hat{A}$ matrix,

$[sI-A]=\left[\begin{array}{cccc}s& 1& 0& 0\\ -\frac{8}{5}& s& -\frac{8}{5}& 0\\ 0& 0& s& 1\\ \frac{8}{5}& 0& -\frac{8}{5}& s\\ \end{array}\right]$

Hence, the inverse matrix (thanks TI-89!) is,

$[sI-A{]}^{-1}=\left[\begin{array}{cccc}\frac{5s(5s^2+8)}{25s^4+80s^2+128}& -\frac{5(5s^2+8)}{25s^4+80s^2+128}& -\frac{40s}{25s^4+80s^2+128}& \frac{40}{25s^4+80s^2+128}\\ \frac{8(5s^2+16)}{25s^4+80s^2+128}& \frac{5s(5s^2+8)}{25s^4+80s^2+128}& -\frac{40s^2}{25s^4+80s^2+128}& -\frac{40s}{25s^4+80s^2+128}\\ \frac{40s}{25s^4+80s^2+128}& -\frac{40}{25s^4+80s^2+128}& \frac{5s(5s^2+8)}{25s^4+80s^2+128}& -\frac{5(5s^2+8)}{25s^4+80s^2+128}\\ -\frac{40s^2}{25s^4+80s^2+128}& \frac{40s}{25s^4+80s^2+128}& \frac{8(5s^2+16)}{25s^4+80s^2+128}& \frac{5s(5s^2+8)}{25s^4+80s^2+128}\\ \end{array}\right]$