Revision as of 00:11, 7 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

Solution

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

Solving for $t h e t a_{1}^{'^{'}}$ and $t h e t a_{2}^{'^{'}}$ we obtain,

θ_{1}^{'^{'}} = - (\frac{B}{A}) θ_{2}^{'^{'}} - (\frac{C}{A}) θ_{1}

θ_{2}^{'^{'}} = - (\frac{B}{D}) θ_{1}^{'^{'}} - (\frac{E}{D}) θ_{2}

Therefore,

θ_{1}^{'^{'}} = - (\frac{C D}{A D + B^{2}}) θ_{1} - (\frac{B E}{A D + B^{2}}) θ_{2}

θ_{2}^{'^{'}} = (\frac{B C}{A D + B^{2}}) θ_{1} - (\frac{A E}{A D + B^{2}}) θ_{2}

@@ Line 29: / Line 29: @@
 == Solution ==
-=== Laplace Transform ===
 Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,
@@ Line 36: / Line 35: @@
 Hence,
-: <math>A\theta_1^{\prime\prime} + B\theta_2^{\prime\prime} + C\theta_1 = 0</math>
+: <math>A\theta_1^{''} + B\theta_2^{''} + 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^{''} + B\theta_1^{''} + E\theta_2 = 0</math>
-Solving the Laplace Transform system yeilds,
+Solving for <math>theta_1^{''}</math> and <math>theta_2^{''}</math> we obtain,
-: <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>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
+: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</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>
+Therefore,
-At time <math>t=0</math> we assume that <math>\theta_1(t)^{\prime}=\theta_2(t)^{\prime}=0</math>. Thus, solvin for <math>\boldsymbol{\Theta}_1</math> and <math>\boldsymbol{\Theta}_2</math> we obtain,
-: <math>\boldsymbol{\Theta}_1=\dfrac{sA\theta_1+sB\theta_2-s^2B\boldsymbol{\Theta}_2}{(s^2A+C)}</math>
-: <math>\boldsymbol{\Theta}_2=\dfrac{sB\theta_1+sD\theta_2-s^2B\boldsymbol{\Theta}_1}{(s^2D+E)}</math>
+: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
+: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>
 === State Space ===

Coupled Oscillator: Double Pendulum: Difference between revisions

Revision as of 00:11, 7 December 2009

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Problem Statement

Solution

State Space

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Coupled Oscillator: Double Pendulum: Difference between revisions

Revision as of 00:11, 7 December 2009

Problem Statement

Solution

State Space

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