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

$\lambda _{1}=\,$

$\lambda _{2}=\,$

$\lambda _{3}=\,$

$\lambda _{4}=\,$