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

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For mass 1:
For mass 1:


<math> + \uparrow \sum F_{y_1} = m_1 \ddot{x}_1  \Rightarrow\ m_1 \ddot{x}_1=-2b_1\dot{x}_1-k_1s_1+k_2s_2+m_1g</math>
<math> + \uparrow \sum F_{y_1} = m_1 \ddot{x}_1  \Rightarrow\ m_1 \ddot{x}_1=-2b_1\dot{x}_1-k_1x_1+k_2(x_2-x_1)+m_1g</math>




For mass 2:
For mass 2:


<math> + \uparrow \sum F_{y_2} = m_2 \ddot{x}_2  \Rightarrow\ m_2 \ddot{x}_2=-2b_2\dot{x}_2-k_2s_2+m_2g</math>
<math> + \uparrow \sum F_{y_2} = m_2 \ddot{x}_2  \Rightarrow\ m_2 \ddot{x}_2=-2b_2\dot{x}_2-k_2(x_2-x_1)+m_2g</math>
 
 
For the cases above
 
<math> s_1=(l_1+x_1)\, </math> and <math> s_2=(l_2+x_2)\, </math>
 
where l is the unstretched length of the spring and x is the displacement of the spring.





Revision as of 11:25, 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

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˙1k1m1l1k1m1x1+k2m2l1+k2m2x2+g

and

x¨2=2b2m2x˙2k2m2l2k2m2x2+g

State Space Equation

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


x˙_(t)=A^x_(t)+C^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]= [0100k1m12b1m1k1m10000100k2m22b2m2][x1x˙1x2x˙2]+[00000000][l1l2]


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