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}

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