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

Laplace Transform

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,

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

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

Solving the Laplace Transform system yeilds,

$A(s^2{\mathrm{\Theta }}_1-s{\theta }_1-{\theta }_1^{\prime })+B(s^2{\mathrm{\Theta }}_2-s{\theta }_2-{\theta }_2^{\prime })+C{\mathrm{\Theta }}_1=0$

$D(s^2{\mathrm{\Theta }}_2-s{\theta }_2-{\theta }_2^{\prime })+B(s^2{\mathrm{\Theta }}_1-s{\theta }_1-{\theta }_1^{\prime })+E{\mathrm{\Theta }}_2=0$

At time $t=0$ we assume that ${\theta }_1(t{)}^{\prime }={\theta }_2(t{)}^{\prime }=0$. Thus, solvin for ${\mathrm{\Theta }}_1$ and ${\mathrm{\Theta }}_2$ we obtain,

${\mathrm{\Theta }}_1=\frac{sA{\theta }_1+sB{\theta }_2-s^{2}B{\mathrm{\Theta }}_2}{(s^{2}A+C)}$

${\mathrm{\Theta }}_2=\frac{sB{\theta }_1+sD{\theta }_2-s^{2}B{\mathrm{\Theta }}_1}{(s^{2}D+E)}$