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.

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Equations of Equilibrium

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

Equation 1

$\begin{array}{rl}F& =ma\\ F& =m\ddot{x}\\ -k_1{x}_1-k_2(x_1{x}_2)& =m_1\ddot{x_1}\\ -\frac{k_1{x}_1}{m_1}-\frac{k_2(x_1-x_2)}{m_1}& =m_1\ddot{x_1}\\ -\frac{k_1{x}_1}{m_1}-\frac{k_2(x_1-x_2)}{m_1}& =\ddot{x_1}\\ -\frac{k_1+k_2}{m_1}{x}_1+\frac{k_2}{m_1}{x}_2& =\ddot{x_1}\\ \end{array}$

Equation 2

$\begin{array}{rl}F& =ma\\ F& =m\ddot{x}\\ -k_2(x_2-x_1)& =m_2\ddot{x_2}\\ \frac{-k_2(x_2-x_1)}{m_2}& =\ddot{x_2}\\ -\frac{k_2}{m_2}{x}_2+\frac{k_2}{m_2}{x}_1& =\ddot{x_2}\\ \end{array}$

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.

$\left[\begin{array}{c}\dot{x_1}\\ \ddot{x_1}\\ \dot{x_2}\\ \ddot{x_2}\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ -\frac{(k_1+k_2)}{m_1}& 0& \frac{k_2}{m_1}& 0\\ 0& 0& 0& 1\\ \frac{k_2}{m_2}& 0& -\frac{k_2}{m_2}& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ \dot{x_1}\\ x_2\\ \dot{x_2}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$

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

$m_1=10kg$

$m_2=5kg$

$k_1=25\frac{N}{m}$

$k_2=20\frac{N}{m}$

We now have

$\left[\begin{array}{c}\dot{x_1}\\ \ddot{x_1}\\ \dot{x_2}\\ \ddot{x_2}\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ -4.5& 0& 2& 0\\ 0& 0& 0& 1\\ 4& 0& -4& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ \dot{x_1}\\ x_2\\ \dot{x_2}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$

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 $k_1,k_2,k_3,k_4$.

$k_1=\left[\begin{array}{c}0.2149i\\ -0.5722\\ -0.2783i\\ 0.7409\end{array}\right]$

$k_2=\left[\begin{array}{c}-0.2149i\\ -0.5722\\ 0.2783i\\ 0.7409\end{array}\right]$

$k_3=\left[\begin{array}{c}-0.3500i\\ 0.4157\\ -0.5407i\\ 0.6421\end{array}\right]$

$k_4=\left[\begin{array}{c}0.3500i\\ 0.4157\\ 0.5407i\\ 0.6421\end{array}\right]$

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

$x=c_1\left[\begin{array}{c}0.2149i\\ -0.5722\\ -0.2783i\\ 0.7409\end{array}\right]e^{2.6626it}+c_2\left[\begin{array}{c}-0.2149i\\ -0.5722\\ 0.2783i\\ 0.7409\end{array}\right]e^{-2.6626it}+c_3\left[\begin{array}{c}-0.3500i\\ 0.4157\\ -0.5407i\\ 0.6421\end{array}\right]e^{1.18766it}+c_4\left[\begin{array}{c}0.3500i\\ 0.4157\\ 0.5407i\\ 0.6421\end{array}\right]e^{-1.18766it}$

Matrix Exponential

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

$\overline{z}=\mathbf{T}\overline{x}$

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

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

Now we can use another identity that we already know

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

Combining the two equations we then get

${\mathbf{T}}^{\mathbf{-}\mathbf{1}}\dot{\overline{z}}=\mathbf{A}{\mathbf{T}}^{\mathbf{-}\mathbf{1}}\overline{z}$

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

$\dot{\overline{z}}=\mathbf{T}\mathbf{A}{\mathbf{T}}^{\mathbf{-}\mathbf{1}}\overline{z}$

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

This new equation has the same form as

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

where

$\hat{\mathbf{A}}=\mathbf{T}\mathbf{A}{\mathbf{T}}^{\mathbf{-}\mathbf{1}}$

If we take the Laplace transform of this equation we can come up with the following

$\overline{z}=e^{\mathbf{A}t}\overline{z}(0)$

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

${\mathbf{T}}^{\mathbf{-}\mathbf{1}}=[\overline{k_1}|\overline{k_2}|\overline{k_3}|\overline{k_4}]$

${\mathbf{T}}^{\mathbf{-}\mathbf{1}}=\left[\begin{array}{cccc}0.2149i& -0.2149i& -0.3500i& 0.3500i\\ -0.5722& -0.5722& 0.4157& 0.4157\\ -0.2783i& 0.2783i& -0.5407i& 0.5407i\\ 0.7409& 0.7409& 0.6421& 0.6421\end{array}\right]$

Taking the inverse of this we can solve for T

$\mathbf{T}=\left[\begin{array}{cccc}-1.2657i& -0.4753& 0.8193i& 0.3077\\ 1.2657i& -0.4753& -0.8193i& 0.3077\\ 0.6514i& 0.5484& 0.5031i& 0.4236\\ -0.6514i& 0.5484& -0.5031& 0.4236\end{array}\right]$