Coupled Oscillator: Jonathan Schreven: Difference between revisions

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== Eigen Values ==
== Eigen Values ==
Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.
:'''Given'''
:<math>x_1=1m</math>
:<math>x_2=2.5m</math>
:<math>m_1=10kg</math>
:<math>m_2=7kg</math>
:<math>k_1=25{N\over {m}}</math>
:<math>k_2=20{N\over {m}}</math>


We now have
:<math>\begin{bmatrix}
\dot{x_1} \\
\ddot{x_1} \\
\dot{x_2} \\
\ddot{x_2}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-{(k_1+k_2)\over {m_1}} & 0 & {k_2\over {m_1}} & 0 \\
0 & 0 & 0 & 1 \\
-{k_2\over {m_2}} & 0 & {k_2\over {m_2}} & 0
\end{bmatrix}
\begin{bmatrix}
{x_1} \\
\dot{x_1} \\
{x_2} \\
\dot{x_2}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}</math>


== Eigen Vectors ==
== Eigen Vectors ==

Revision as of 19:09, 9 December 2009

Problem

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

Equations of Equilibrium

Using F=ma we can then find our four equations of equilibrium.

Equation 1
F=maF=mx¨k1x1k2(x1x2)=m1x1¨k1x1m1k2(x1x2)m1=m1x1¨k1x1m1k2(x1x2)m1=x1¨k1+k2m1x1+k2m1x2=x1¨
Equation 2
F=maF=mx¨k2(x2x1)=m2x2¨k2(x2x1)m2=x2¨k2m2x2+k2m2x1=x2¨
Equation 3
x1˙=x1˙
Equation 4
x2˙=x2˙


Now we can put these four equations into the state space form.

[x1˙x1¨x2˙x2¨]=[0100(k1+k2)m10k2m100001k2m20k2m20][x1x1˙x2x2˙]+[0000]

Eigen Values

Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.

Given
x1=1m
x2=2.5m
m1=10kg
m2=7kg
k1=25Nm
k2=20Nm

We now have

[x1˙x1¨x2˙x2¨]=[0100(k1+k2)m10k2m100001k2m20k2m20][x1x1˙x2x2˙]+[0000]

Eigen Vectors