Coupled Oscillator: Double Pendulum: Difference between revisions

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: <math>A\theta_1^{\prime\prime} + B\theta_2^{\prime\prime} + C\theta_1 = 0</math>
: <math>A\theta_1^{\prime\prime} + B\theta_2^{\prime\prime} + C\theta_1 = 0</math>
: <math>D\theta_2^{\prime\prime} + B\theta_1^{\prime\prime} + E\theta_2 = 0</math>
: <math>D\theta_2^{\prime\prime} + B\theta_1^{\prime\prime} + E\theta_2 = 0</math>
Solving the Laplace Transform system yeilds,
: <math>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</math>
: <math>\boldsymbol{\Theta}_1(s^2A+C)-s\theta_1-\theta_1^{\prime}) + B(s^2\boldsymbol{\Theta}_2-s\theta_2-\theta_2^{\prime}) + C\boldsymbol{\Theta}_1 = 0</math>
: <math>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</math>

Revision as of 22:15, 6 December 2009

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

(m1+m2)l12θ1+m2l1l2θ2cos(θ1θ2)+m2l1l2(θ2)2sin(θ1θ2)+(m1+m2)l1gsinθ1=0
m2l12θ2+m2l1l2θ1cos(θ1θ2)m2l1l2(θ1)2sin(θ1θ2)+m2l2gsinθ2=0

In order to linearize these equations, we assume that the displacements θ1 and θ2 are small enough so that cos(θ1θ2)1 and sin(θ1θ2)0. Thus,

(m1+m2)l12θ1+m2l1l2θ2+(m1+m2)l1gθ1=0
m2l12θ2+m2l1l2θ1+m2l2gθ2=0

Solution

Laplace Transform

Since our concern is about the motion functions, we will assign the masses m1 and m2, the rod lenghts l1 and l1, and gravitational force g constants to different variables as follows,

A=(m1+m2)l12B=m2l1l2C=(m1+m2)l1gD=m2l12E=m2l2g

Hence,

Aθ1+Bθ2+Cθ1=0
Dθ2+Bθ1+Eθ2=0

Solving the Laplace Transform system yeilds,

A(s2Θ1sθ1θ1)+B(s2Θ2sθ2θ2)+CΘ1=0
Θ1(s2A+C)sθ1θ1)+B(s2Θ2sθ2θ2)+CΘ1=0
D(s2Θ2sθ2θ2)+B(s2Θ1sθ1θ1)+EΘ2=0