Coupled Oscillator: horizontal Mass-Spring: Difference between revisions

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=
=
\begin{bmatrix}
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
-4.5 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
4 & 0 & -4 & 0
0 & 0 & 0 & 0
\end{bmatrix}
\end{bmatrix}


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From this we get
From this we get
:<math>\lambda_1=2.6626i\,</math>
:<math>\lambda_1=\,</math>
:<math>\lambda_2=-2.6626i\,</math>
:<math>\lambda_2=\,</math>
:<math>\lambda_3=1.18766i\,</math>
:<math>\lambda_3=\,</math>
:<math>\lambda_4=-1.18766i\,</math>
:<math>\lambda_4=\,</math>

Revision as of 14:53, 10 December 2009

Problem Statement

Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.

 

Initial Conditions:

m1=10kg
m2=10kg
k1=25N/m
k2=75N/m
k3=50N/m

Equations for M_1

F=maF=mx¨k1x1k2(x1x2)=m1x1¨k1x1m1k2(x1x2)m1=m1x1¨k1x1m1k2(x1x2)m1=x1¨k1+k2m1x1+k2m1x2=x1¨

Equations for M_2

F=maF=mx¨k2(x2x1)=m2x2¨k2(x2x1)m2=x2¨k2m2x2+k2m2x1=x2¨

Additional Equations

x1˙=x1˙
x2˙=x2˙

State Equations

[x1˙x1¨x2˙x2¨] = [0100(k1k2)m10k1m100001k1m20(k1+k2)m20][x1x˙1x2x˙2]+[0000000000000000][0000]

With the numbers...


[x1˙x1¨x2˙x2¨] = [0100(50N/m)10kg025N/m10kg0000125N/m10kg0(100N/m)10kg0][x1x˙1x2x˙2]

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
m1=10kg
m2=10kg
k1=25Nm
k2=50Nm

We now have

[x1˙x1¨x2˙x2¨]=[0000000000000000][x1x1˙x2x2˙]+[0000]

From this we get

λ1=
λ2=
λ3=
λ4=