Coupled Oscillator: Coupled Mass-Spring System with Damping

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:

$m_{1}=80kg{\frac {}{}}$

$k_{1}=60000{\frac {N}{m}}$

$b_{1}=0.1{\frac {}{}}$

And for the lower mass:

$m_{2}=400kg{\frac {}{}}$

$k_{2}=120000{\frac {N}{m}}$

$b_{2}=0.2{\frac {}{}}$

Find the Force Equations

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

For mass 1:

$+\uparrow \sum F_{y_{1}}=m_{1}{\ddot {x}}_{1}\Rightarrow \ m_{1}{\ddot {x}}_{1}=-2b_{1}{\dot {x}}_{1}-k_{1}s_{1}+k_{2}s_{2}+m_{1}g$

For mass 2:

$+\uparrow \sum F_{y_{2}}=m_{2}{\ddot {x}}_{2}\Rightarrow \ m_{2}{\ddot {x}}_{2}=-2b_{2}{\dot {x}}_{3}-k_{2}s_{2}+m_{2}g$

For the cases above

$s_{1}=(l_{1}+x_{1})\,$ and $s_{2}=(l_{2}+x_{2})\,$

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:

${\ddot {x}}_{1}=-{\frac {2b_{1}}{m_{1}}}{\dot {x}}_{1}-{\frac {k_{1}}{m_{1}}}l_{1}-{\frac {k_{1}}{m_{1}}}x_{1}+{\frac {k_{2}}{m_{2}}}l_{1}+{\frac {k_{2}}{m_{2}}}x_{2}+g$

and

${\ddot {x}}_{2}=-{\frac {2b_{2}}{m_{2}}}{\dot {x}}_{2}-{\frac {k_{2}}{m_{2}}}l_{2}-{\frac {k_{2}}{m_{2}}}x_{2}+g$