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

We know that the T inverse matrix is just made up from the eigenvector columns, so:

${\mathbf {T^{-1}}}={\begin{bmatrix}0.0124+0.2568i&0.0124-0.2568i&-0.0121-0.3535i&-0.0121+0.3535i\\-0.9211&-0.9211&0.3824-0.0034i&0.3824+0.0034i\\-0.0015-0.0786i&-0.0015+0.0786i&-0.0145-0.5794i&-0.0145+0.5794i\\0.2814+0.0084i&0.2814-0.0084i&0.6266&0.6266\end{bmatrix}}$

It then follows that the T matix is:

${\mathbf {T}}={\begin{bmatrix}-0.0070-1.6408i&-0.4576-0.0222i&0.0070+1.0012i&0.2793+0.0050i\\-0.0070+1.6408i&-0.4576+0.0222i&0.0070-1.0012i&0.2793-0.0050i\\-0.0188+0.2221i&0.2052+0.0070i&0.0103+0.7274i&0.6726+0.0169i\\-0.0188-0.2221i&0.2052-0.0070i&0.0103-0.7274i&0.6726-0.0169i\end{bmatrix}}$

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.

${\bar {z}}={\mathbf {T}}{\bar {x}}\,$

This can be rearranged by multiplying T-inverse to the left side of the equations.

${\mathbf {T^{-1}}}{\bar {z}}={\bar {x}}\,$

Now we can bring in the standard form of a state space equation

${\dot {\bar {x}}}={\mathbf {A}}{\bar {x}}$

Combining the two equations we then get

${\mathbf {T^{-1}}}{\dot {\bar {z}}}={\mathbf {AT^{-1}}}{\bar {z}}$

Multiplying both sides of the equation on the left by T we get

${\dot {\bar {z}}}={\mathbf {TAT^{-1}}}{\bar {z}}$

${\dot {\bar {z}}}={\mathbf {\hat {A}}}{\bar {z}}$

where

${\mathbf {D}}={\mathbf {TAT^{-1}}}$

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

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

where

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

We then substitute this equation back into

${\bar {x}}={\mathbf {T^{-1}}}{\bar {z}}$

and get

${\bar {x}}={\mathbf {T^{-1}}}e^{{\mathbf {D}}t}{\bar {z}}(0)$

${\bar {x}}={\mathbf {T^{-1}}}e^{{\mathbf {D}}t}{\mathbf {T}}{\bar {x}}(0)$

Notice here that

$e^{{\mathbf {A}}t}={\mathbf {T^{-1}}}e^{{\mathbf {D}}t}{\mathbf {T}}$

Which gives us

${\begin{bmatrix}0.0124+0.2568i&0.0124-0.2568i&-0.0121-0.3535i&-0.0121+0.3535i\\-0.9211&-0.9211&0.3824-0.0034i&0.3824+0.0034i\\-0.0015-0.0786i&-0.0015+0.0786i&-0.0145-0.5794i&-0.0145+0.5794i\\0.2814+0.0084i&0.2814-0.0084i&0.6266&0.6266\end{bmatrix}}$

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

${\begin{bmatrix}-0.0070-1.6408i&-0.4576-0.0222i&0.0070+1.0012i&0.2793+0.0050i\\-0.0070+1.6408i&-0.4576+0.0222i&0.0070-1.0012i&0.2793-0.0050i\\-0.0188+0.2221i&0.2052+0.0070i&0.0103+0.7274i&0.6726+0.0169i\\-0.0188-0.2221i&0.2052-0.0070i&0.0103-0.7274i&0.6726-0.0169i\end{bmatrix}}$

Note: All of the above matrices are supposed to be multiplied in the horizontal, but they are put in the vertical to save space so that you don't need to scroll over.