Coupled Oscillator: horizontal Mass-Spring

Problem Statement

Write up on the Wiki a solution of a coupled oscillator problem like the coupled pendulum. Use State Space methods. Describe the eigenmodes and eigenvectors of the system.

Initial Conditions:

$m_{1}=10kg\,$

$m_{2}=10kg\,$

$k1=25N/m\,$

$k2=75N/m\,$

$k3=50N/m\,$

Equations for M_1

${\begin{alignedat}{3}F&=ma\\F&=m{\ddot {x}}\\-k_{1}x_{1}-k_{2}(x_{1}x_{2})&=m_{1}{\ddot {x_{1}}}\\-{k_{1}x_{1} \over {m_{1}}}-{k_{2}(x_{1}-x_{2}) \over {m_{1}}}&=m_{1}{\ddot {x_{1}}}\\-{k_{1}x_{1} \over {m_{1}}}-{k_{2}(x_{1}-x_{2}) \over {m_{1}}}&={\ddot {x_{1}}}\\-{k_{1}+k_{2} \over {m_{1}}}x_{1}+{k_{2} \over {m_{1}}}x_{2}&={\ddot {x_{1}}}\\\end{alignedat}}$

Equations for M_2

${\begin{alignedat}{3}F&=ma\\F&=m{\ddot {x}}\\-k_{2}(x_{2}-x_{1})&=m_{2}{\ddot {x_{2}}}\\{-k_{2}(x_{2}-x_{1}) \over {m_{2}}}&={\ddot {x_{2}}}\\-{k_{2} \over {m_{2}}}x_{2}+{k_{2} \over {m_{2}}}x_{1}&={\ddot {x_{2}}}\\\end{alignedat}}$

Additional Equations

${\dot {x_{1}}}={\dot {x_{1}}}$

${\dot {x_{2}}}={\dot {x_{2}}}$

State Equations

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

With the numbers...

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}\,$ = ${\begin{bmatrix}0&1&0&0\\{\frac {(-50N/m)}{10kg}}&0&{\frac {-25N/m}{10kg}}&0\\0&0&0&1\\{\frac {25N/m}{10kg}}&0&{\frac {(100N/m)}{10kg}}&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}$

Eigen Values

Once you have your equations of equilibrium in matrix form you can plug them into MATLAB which will give you the eigen values automatically.

Given

$m_{1}=10kg\,$

$m_{2}=10kg\,$

$k_{1}=25\,{N \over {m}}$

$k_{2}=50\,{N \over {m}}$

We now have

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\-5&0&-2.5&0\\0&0&0&1\\2.5&0&10&0\end{bmatrix}}{\begin{bmatrix}{x_{1}}\\{\dot {x_{1}}}\\{x_{2}}\\{\dot {x_{2}}}\end{bmatrix}}+{\begin{bmatrix}0\\0\\0\\0\end{bmatrix}}$

From this we get

$\lambda _{1}=-3.0937,$

$\lambda _{2}=2.1380i,$

$\lambda _{3}=-2.1380i,$

$\lambda _{4}=3.0937,$

Eigen Vectors

Using the equation above and the same given conditions we can plug everything into MATLAB and get the eigen vectors which we will denote as $k_{1},k_{2},k_{3},k_{4}\,$.

$k_{1}={\begin{bmatrix}0.0520\\-0.1609\\-0.3031\\0.9378\end{bmatrix}}$

$k_{2}={\begin{bmatrix}0.4176i\\-0.8928\\-0.0716i\\0.1532\end{bmatrix}}$

$k_{3}={\begin{bmatrix}-0.4176i\\-0.8928\\0.0716i\\0.1532\end{bmatrix}}$

$k_{4}={\begin{bmatrix}-0.0520\\-0.1609\\0.3031\\0.9378\end{bmatrix}}$

So then the answer is...

We can now plug these eigen vectors and eigen values into the standard equation

$x=c_{1}k_{1}e^{\lambda _{1}t}+c_{2}k_{2}e^{\lambda _{2}t}+c_{3}k_{3}e^{\lambda _{3}t}+c_{4}k_{4}e^{\lambda _{4}t}$

$\ x=c_{1}$${\begin{bmatrix}0.0520\\-0.1609\\-0.3031\\0.9378\end{bmatrix}}\,$$e^{-3.0937}+c_{2}$${\begin{bmatrix}0.4176i\\-0.8928\\-0.0716i\\0.1532\end{bmatrix}}\,$$e^{2.1380i}+c_{3}$${\begin{bmatrix}-0.4176i\\-0.8928\\0.0716i\\0.1532\end{bmatrix}}\,$$e^{-2.1380i}+c_{4}$${\begin{bmatrix}-0.0520\\-0.1609\\0.3031\\0.9378\end{bmatrix}}\,$$e^{3.0937}\,$

Matrix Exponential

We now use matrix exponentials to solve the same problem.

$z=Tx\,$

So from the above equation we get this to prove the matrix exponetial works.

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

We also know what T equals and we can solve it for our case

$T^{-1}=[k_{1}|k_{2}|k_{3}|k_{4}]\,$

$T^{-1}={\begin{bmatrix}0.0520&0.4176i&-0.4176i&-0.0520\\-0.1609&-0.8928&-0.8928&-0.1609\\-0.3031&-0.0716i&0.0716i&0.3031\\0.9378&0.1532&0.1532&0.9378\end{bmatrix}}$

Taking the inverse of this we can solve for T

$T={\begin{bmatrix}-0.2914&0.0943&-1.6996&0.5493\\-1.2337i&-0.5770&-0.2117i&-0.0990\\1.2335i&-0.5770&0.2116i&-0.0990\\0.2914&0.0943&1.6996&0.5493\end{bmatrix}}$

So taking

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

We get the uncoupled matrix of

${\dot {z}}={\begin{bmatrix}-3.0937&0&0&0\\0&2.1380i&0&0\\0&0&-2.1380i&0\\0&0&0&3.0937\end{bmatrix}}$

created by Greg Peterson