Coupled Oscillator: Coupled Mass-Spring System with Damping: Difference between revisions

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<math> \underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t) </math>
<math> \underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t) </math>




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\begin{bmatrix}
\begin{bmatrix}
l_1  \\
0 \\
l_2 \\
\end{bmatrix}
\end{bmatrix}



Revision as of 11:36, 3 December 2009

Problem Statement

For the below system set up a set of state variable equations, and then solve using Laplace transformations. Assume all motion takes place in the vertical directions.

Fig. 1

Solution

Initial Values

For the upper mass:

m1=80kg

k1=60000Nm

b1=0.1

And for the lower mass:

m2=400kg

k2=120000Nm

b2=0.2

Find the Force Equations

First we need to sum forces in the y-direction for each block.

For mass 1:

+Fy1=m1x¨1m1x¨1=2b1x˙1k1x1+k2(x2x1)+m1g


For mass 2:

+Fy2=m2x¨2m2x¨2=2b2x˙2k2(x2x1)+m2g


So if we put the equations above into the correct form we have:


x¨1=2b1m1x˙1(k1m1+k2m2)x1k2m2x2+g

and

x¨2=2b2m2x˙2k2m2x2+k2m2x1+g

State Space Equation

The general form for the state equation is as shown below:


x˙_(t)=A^x_(t)+B^u_(t)


Where M^ denotes a matrix and v_ denotes a vector.


If we let x1, x˙1, x2, and x2˙ be the state variables, then


[x˙1x¨1x˙2x¨2]= [0100(k1m1+k2m2)2b1m1k1m100001k2m20k2m22b2m2][x1x˙1x2x˙2]+[00000000][0]


We don't need to include gravity here if we allow are initial conditions for the spring to be zero with gravity accounted for.