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. Assume all motion takes place in the vertical directions.

Fig. 1

Solution

Initial Values

For the upper mass:

$m_{1}=2kg{\frac {}{}}$

$k_{1}=10{\frac {N}{m}}$

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

And for the lower mass:

$m_{2}=4kg{\frac {}{}}$

$k_{2}=12{\frac {N}{m}}$

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

Find the Force Equations

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

For mass 1:

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

For mass 2:

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

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

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

and

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

Note: Many people include the original length of the springs in the above equations, however if we take our initial reference point to be where the mass is in equilibrium, we do not need to include the initial spring length or gravity as an imput.

State Space Equation

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

${\underline {\dot {x}}}={\widehat {A}}\,{\underline {x}}+{\widehat {B}}\,{\underline {u}}$

Where ${\widehat {M}}$ denotes a matrix and ${\underline {v}}$ denotes a vector.

If we let $x_{1}{\frac {}{}}$, ${\dot {x}}_{1}{\frac {}{}}$, $x_{2}{\frac {}{}}$, and ${\dot {x_{2}}}{\frac {}{}}$ be the state variables, then

${\begin{bmatrix}{\dot {x}}_{1}\\{\ddot {x}}_{1}\\{\dot {x}}_{2}\\{\ddot {x}}_{2}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\-({\frac {k_{1}}{m_{1}}}+{\frac {k_{2}}{m_{1}}})&-({\frac {2b_{1}}{m_{1}}}+{\frac {2b_{2}}{m_{1}}})&{\frac {k_{2}}{m_{1}}}&{\frac {2b_{1}}{m_{1}}}\\0&0&0&1\\{\frac {k_{2}}{m_{2}}}&{\frac {2b_{2}}{m_{2}}}&-{\frac {k_{2}}{m_{2}}}&-{\frac {2b_{2}}{m_{2}}}\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}+{\begin{bmatrix}0\\0\\0\\0\\\end{bmatrix}}{\begin{bmatrix}u\\\end{bmatrix}}$

Now we need to set up our output. The below is the general form of the output.

$y={\widehat {C}}\,{\underline {x}}+{\widehat {D}}\,{\underline {u}}$

If we want to output both positions and velocities of both masses we will use:

$y={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}+{\begin{bmatrix}0\end{bmatrix}}{\begin{bmatrix}u\\\end{bmatrix}}$

Eigenvalues

If we solve the A matrix for the eigenvalues using Matlab we get:

$\lambda _{1}=-0.1729+3.5780i\,$

$\lambda _{2}=-0.1729-3.5780i\,$

$\lambda _{3}=-0.0271+1.0808i\,$

$\lambda _{4}=-0.0271-1.0808i\,$

Eigenvectors

Now that we have our eigenvalues, we can use them to find the eigenvectors. Again using Matlab we get:

$k_{1}={\begin{bmatrix}0.0124+0.2568i\\-0.9211\\-0.0015-0.0786i\\0.2814+0.0084i\end{bmatrix}}$

$k_{2}={\begin{bmatrix}0.0124-0.2568i\\-0.9211\\-0.0015+0.0786i\\0.2814-0.0084i\end{bmatrix}}$

$k_{3}={\begin{bmatrix}-0.0121-0.3535i\\0.3824-0.0034i\\-0.0145-0.5794i\\0.6266\end{bmatrix}}$

$k_{4}={\begin{bmatrix}-0.0121+0.3535i\\0.3824+0.0034i\\-0.0145-0.5794i\\0.6266\end{bmatrix}}$

Diagonalized A Matrix

With the above eigenvectors we also get the diagonalized A matrix which we will denote as D:

${\mathbf {D}}={\begin{bmatrix}-0.1729+3.5780i&0&0&0\\0&-0.1729-3.5780i&0&0\\0&0&-0.0271+1.0808i&0\\0&0&0&-0.0271-1.0808i\end{bmatrix}}$

Solving

If we take the Laplace transform of the above D matrix and use the below equation:

${\bar {z}}=e^{{\mathbf {D}}t}{\bar {z}}(0)$

where

$e^{{\mathbf {D}}t}={\begin{bmatrix}e^{(-0.1729+3.5780i)t}&0&0&0\\0&e^{(-0.1729-3.5780i)t}&0&0\\0&0&e^{(-0.0271+1.0808i)t}&0\\0&0&0&e^{(-0.0271-1.0808i)t}\end{bmatrix}}$