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).

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,

$A=(m_{1}+m_{2})l_{1}^{2}\quad B=m_{2}l_{1}l_{2}\quad C=(m_{1}+m_{2})l_{1}g\quad D=m_{2}l_{1}^{2}\quad E=m_{2}l_{2}g$

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,

$A(s^{2}{\boldsymbol {\Theta }}_{1}-s\theta _{1}-\theta _{1}^{\prime })+B(s^{2}{\boldsymbol {\Theta }}_{2}-s\theta _{2}-\theta _{2}^{\prime })+C{\boldsymbol {\Theta }}_{1}=0$

$D(s^{2}{\boldsymbol {\Theta }}_{2}-s\theta _{2}-\theta _{2}^{\prime })+B(s^{2}{\boldsymbol {\Theta }}_{1}-s\theta _{1}-\theta _{1}^{\prime })+E{\boldsymbol {\Theta }}_{2}=0$

At time $t=0$ we assume that $\theta _{1}(t)^{\prime }=\theta _{2}(t)^{\prime }=0$. Thus, solvin for ${\boldsymbol {\Theta }}_{1}$ and ${\boldsymbol {\Theta }}_{2}$ we obtain,

${\boldsymbol {\Theta }}_{1}={\dfrac {sA\theta _{1}+sB\theta _{2}-s^{2}B{\boldsymbol {\Theta }}_{2}}{(s^{2}A+C)}}$

${\boldsymbol {\Theta }}_{2}={\dfrac {sB\theta _{1}+sD\theta _{2}-s^{2}B{\boldsymbol {\Theta }}_{1}}{(s^{2}D+E)}}$