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} θ_{1}^{''} + m_{2} l_{1} l_{2} θ_{2}^{''} c o s (θ_{1} - θ_{2}) + m_{2} l_{1} l_{2} (θ_{2}^{'})^{2} s i n (θ_{1} - θ_{2}) + (m_{1} + m_{2}) l_{1} g s i n θ_{1} = 0

m_{2} l_{1}^{2} θ_{2}^{''} + m_{2} l_{1} l_{2} θ_{1}^{''} c o s (θ_{1} - θ_{2}) - m_{2} l_{1} l_{2} (θ_{1}^{'})^{2} s i n (θ_{1} - θ_{2}) + m_{2} l_{2} g s i n θ_{2} = 0

In order to linearize these equations, we assume that the displacements $θ_{1}$ and $θ_{2}$ are small enough so that $c o s (θ_{1} - θ_{2}) \approx 1$ and $s i n (θ_{1} - θ_{2}) \approx 0$ . Thus,

(m_{1} + m_{2}) l_{1}^{2} θ_{1}^{''} + m_{2} l_{1} l_{2} θ_{2}^{''} + (m_{1} + m_{2}) l_{1} g θ_{1} = 0

m_{2} l_{1}^{2} θ_{2}^{''} + m_{2} l_{1} l_{2} θ_{1}^{''} + m_{2} l_{2} g θ_{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} g D = m_{2} l_{1}^{2} E = m_{2} l_{2} g

Hence,

A θ_{1}^{''} + B θ_{2}^{''} + C θ_{1} = 0

D θ_{2}^{''} + B θ_{1}^{''} + E θ_{2} = 0

Coupled Oscillator: Double Pendulum

Contents

Problem Statement

Solution

Laplace Transform

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Coupled Oscillator: Double Pendulum

Problem Statement

Solution

Laplace Transform

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