Coupled Oscillator: Spring Pendulums: Difference between revisions
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===Solution=== |
===Solution=== |
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By definition, the the state equation is stated as |
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<math> |
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\underline{\dot{x}} = \widehat{A} \, \underline{x} + \widehat{B} \, \underline{u} |
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</math> |
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Now, consider the motion equations described in the Solution section, |
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<math>m_1\ddot{x}_1+k_1x_1-k_2(x_2-x_1)=m_1\ddot{x}_1+k_1x_1-k_2x_2+k_2x_1=0</math> |
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<math>m_2\ddot{x}_2+k_2(x_2-x_1)-k_3x_2=m_2\ddot{x}_2+k_2x_2-k_2x_1-k_3x_2=0</math> |
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Solving for <math>\ddot{x}_1</math> and <math>\ddot{x}_2</math> yields, |
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:<math>\ddot{x}_1=-\dfrac{k_1}{m_1}x_1+\dfrac{k_2}{m_1}x_2-\dfrac{k_2}{m_1}x_1</math> |
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:<math>\ddot{x}_2=\dfrac{-k_2}{m_2}x_2+\dfrac{k_2}{m_2}x_1+\dfrac{k_3}{m_2}x_2</math> |
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Finally, we let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables. Thus, |
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<math> |
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\begin{bmatrix} |
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\dot{x}_1 \\ |
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\ddot{x}_1 \\ |
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\dot{x}_2 \\ |
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\ddot{x}_2 |
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\end{bmatrix} |
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= |
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\begin{bmatrix} |
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0 & 1 & 0 & 0 \\ |
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-\frac{1}{m_1}(k_1+k_2) & 0 & \frac{k_2}{m_1} & 0 \\ |
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0 & 0 & 0 & 1 \\ |
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\frac{k_2}{m_2} & 0 & \frac{1}{m_2}(k_3-k_2) & 0 |
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\end{bmatrix} |
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\begin{bmatrix} |
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x_1 \\ |
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\dot{x}_1 \\ |
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x_2 \\ |
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\dot{x}_2 |
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\end{bmatrix} |
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</math> |
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Latest revision as of 17:56, 13 December 2009
Problem Statement
Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system.
Solution
By definition, the the state equation is stated as
Now, consider the motion equations described in the Solution section,
Solving for and yields,
Finally, we let , , , and be the state variables. Thus,
Created by Kendrick Mensink