Revision as of 21:29, 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

(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

@@ Line 21: / Line 21: @@
 : <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>
-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 <math>k_1=k_2=k_3=1 \frac{}{}</math>, <math>m_1=m_2=1 \frac{}{}</math>, and <math>x_1(0)=0 \frac{}{}</math>, <math>x'1(0)=-1 \frac{}{}</math>, <math>x_2(0)=0 \frac{}{}</math>, and <math>x'_2(0)=1 \frac{}{}</math>.
+In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,
+: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
+: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

Coupled Oscillator: Double Pendulum: Difference between revisions

Revision as of 21:29, 6 December 2009

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Coupled Oscillator: Double Pendulum: Difference between revisions

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