Coupled Horizontal Spring Mass Oscillator: Difference between revisions

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</math>,<math>\ {K_4=}\,</math><math>\begin{bmatrix}1 \\-2\sqrt(5) \\1 \\-2\sqrt(5)\end{bmatrix}\,
</math>,<math>\ {K_4=}\,</math><math>\begin{bmatrix}1 \\-2\sqrt(5) \\1 \\-2\sqrt(5)\end{bmatrix}\,
</math>
</math>
<math>\text {So then the answer is...}\,</math>
<math>\ x=c_1</math><math>\begin{bmatrix}-1 \\-2\sqrt(10) \\1 \\2\sqrt(10)\end{bmatrix}\,</math><math>e^{2\sqrt{10}}+ c_2</math><math>\begin{bmatrix}-1 \\2\sqrt(10) \\1 \\-2\sqrt(10)\end{bmatrix}\,</math><math>e^{-2\sqrt{10}}+ c_3</math><math>\begin{bmatrix}1 \\2\sqrt(5) \\1 \\2\sqrt(5)\end{bmatrix}\,</math><math>e^{2\sqrt{5}}</math>

Revision as of 19:32, 8 December 2009

Coupled Oscillator Spring Mass Oscillator: State Space

Problem Statement

Two 4 Kg Weights are suspended between two walls. They are connected by a spring between them with a spring constant k2. They are connected to the walls by two springs k1 and k3 with k1=k3. m1 is a distance x1 form m2 and m2 is x2 from the wall.


Solution

Things we know

m1=5kg

m2=5kg

k1=50Nm

k2=100Nm

k3=50Nm

So now that we have are problem we need to start setting up the equations we need to solve it.

x1˙=x1˙

x1¨+k1+k2m1x1−k2m1x2=0

x2˙=x2˙

x2¨+k3+k2m2x2−k2m2x1=0

Now we take these equations and put them in a state space model.

[x1˙x1¨x2˙x2¨] = [0100(k1+k2)m10−k1m100001−k1m20(k1+k2)m20][x1x˙1x2x˙2]+[0]

Now we make the appropriate numerical substitutions.


[x1˙x1¨x2˙x2¨] = [010015050−50500001−505015050][x1x˙1x2x˙2]+[0]



[x1˙x1¨x2˙x2¨]=[0100300−1000001−100300][x1x˙1x2x˙2]+[0]

So using maple i was able to obtain the eigenvalues and eigenvectors.

Eigenvalues.


λ1=210 λ2=−210 λ3=25 λ4=−25

Eigenvectors.

K1=[−1−2(10)12(10)],K2=[−12(10)1−2(10)],K3=[12(5)12(5)],K4=[1−2(5)1−2(5)]

So then the answer is...

x=c1[−1−2(10)12(10)]e210+c2[−12(10)1−2(10)]e−210+c3[12(5)12(5)]e25