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

From Class Wiki
Jump to navigation Jump to search
No edit summary
Line 82: Line 82:
\begin{bmatrix}
\begin{bmatrix}
  0 & 1 & 0 & 0 \\
  0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
-\frac{k_1}{m_1} & -\frac{2b_1}{m_1} & \frac{k_1}{m_1}  & 0 \\
  0 & 0 & 0 & 1 \\
  0 & 0 & 0 & 1 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & -\frac{k_2}{m_2} & -\frac{2b_2}{m_2}\\
\end{bmatrix}
\end{bmatrix}


Line 104: Line 104:


\begin{bmatrix}
\begin{bmatrix}
L_1 \\
l_1 \\
L_2 \\
l_2 \\
\end{bmatrix}
\end{bmatrix}


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

Revision as of 12:57, 1 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˙1k1s1+k2s2+m1g


For mass 2:

+Fy2=m2x¨2m2x¨2=2b2x˙2k2s2+m2g


For the cases above

s1=(l1+x1) and s2=(l2+x2)

where l is the unstretched length of the spring and x is the displacement of the spring.


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.