Coupled Oscillator: Jonathan Schreven

Problem

In this problem we will explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Here's a picture of what we are working with.

Equations of Equilibrium

Using F=ma we can then find our four equations of equilibrium.

Equation 1

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

Equation 2

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

Equation 3

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

Equation 4

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

Now we can put these four equations into the state space form.

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

Eigen Values

Once you have your equations of equilibrium in matrix form you can plug them into a calculator or a computer program that will give you the eigen values automatically. This saves you a lot of hand work. Here's what you should come up with for this particular problem given these initial conditions.

Given

$m_{1}=10kg\,$

$m_{2}=5kg\,$

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

$k_{2}=20\,{N \over {m}}$

We now have

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\-4.5&0&2&0\\0&0&0&1\\4&0&-4&0\end{bmatrix}}{\begin{bmatrix}{x_{1}}\\{\dot {x_{1}}}\\{x_{2}}\\{\dot {x_{2}}}\end{bmatrix}}+{\begin{bmatrix}0\\0\\0\\0\end{bmatrix}}$

From this we get

$\lambda _{1}=2.6626i\,$

$\lambda _{2}=-2.6626i\,$

$\lambda _{3}=1.18766i\,$

$\lambda _{4}=-1.18766i\,$

Eigen Vectors

Using the equation above and the same given conditions we can plug everything to a calculator or computer program like MATLAB and get the eigen vectors which we will denote as $k_{1},k_{2},k_{3},k_{4}\,$.

$k_{1}={\begin{bmatrix}0.2149i\\-0.5722\\-0.2783i\\0.7409\end{bmatrix}}$

$k_{2}={\begin{bmatrix}-0.2149i\\-0.5722\\0.2783i\\0.7409\end{bmatrix}}$

$k_{3}={\begin{bmatrix}-0.3500i\\0.4157\\-0.5407i\\0.6421\end{bmatrix}}$

$k_{4}={\begin{bmatrix}0.3500i\\0.4157\\0.5407i\\0.6421\end{bmatrix}}$

Solving

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

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

And our final answer is

$x=c_{1}{\begin{bmatrix}0.2149i\\-0.5722\\-0.2783i\\0.7409\end{bmatrix}}e^{2.6626it}+c_{2}{\begin{bmatrix}-0.2149i\\-0.5722\\0.2783i\\0.7409\end{bmatrix}}e^{-2.6626it}+c_{3}{\begin{bmatrix}-0.3500i\\0.4157\\-0.5407i\\0.6421\end{bmatrix}}e^{1.18766it}+c_{4}{\begin{bmatrix}0.3500i\\0.4157\\0.5407i\\0.6421\end{bmatrix}}e^{-1.18766it}$

Matrix Exponential

In this section we will use matrix exponentials to solve the same problem. First we start with this identity.

$z=Tx\,$

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

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

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

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

$T^{-1}={\begin{bmatrix}0.2149i&-0.2149i&-0.3500i&0.3500i\\-0.5722&-0.5722&0.4157&0.4157\\-0.2783i&0.2783i&-0.5407i&0.5407i\\0.7409&0.7409&0.6421&0.6421\end{bmatrix}}$