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

$(m_{1}+m_{2})l_{1}^{2}\theta _{1}^{\prime \prime }+m_{2}l_{1}l_{2}\theta _{2}^{\prime \prime }cos(\theta _{1}-\theta _{2})+m_{2}l_{1}l_{2}(\theta _{2}^{\prime })^{2}sin(\theta _{1}-\theta _{2})+(m_{1}+m_{2})l_{1}gsin\theta _{1}=0$

$m_{2}l_{1}^{2}\theta _{2}^{\prime \prime }+m_{2}l_{1}l_{2}\theta _{1}^{\prime \prime }cos(\theta _{1}-\theta _{2})-m_{2}l_{1}l_{2}(\theta _{1}^{\prime })^{2}sin(\theta _{1}-\theta _{2})+m_{2}l_{2}gsin\theta _{2}=0$

Derive the system of differential equations describing the straight-line vertical motion of the coupled spring shown in Figure 1. Use Laplace transform to solve the system when $k_{1}=k_{2}=k_{3}=1{\frac {}{}}$, $m_{1}=m_{2}=1{\frac {}{}}$, and $x_{1}(0)=0{\frac {}{}}$, $x'1(0)=-1{\frac {}{}}$, $x_{2}(0)=0{\frac {}{}}$, and $x'_{2}(0)=1{\frac {}{}}$.