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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Double Mass/Spring Oscillator

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

${\lambda }_1=2.6626i$

${\lambda }_2=-2.6626i$

${\lambda }_3=1.1877i$

${\lambda }_4=-1.1877i$

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

$\overline{x}=c_1{k}_1{e}^{{\lambda }_{1}t}+c_2{k}_2{e}^{{\lambda }_{2}t}+c_3{k}_3{e}^{{\lambda }_{3}t}+c_4{k}_4{e}^{{\lambda }_{4}t}$

And our final answer is

$\overline{x}=c_1\left[\begin{array}{c}0.2149i\\ -0.5722\\ -0.2783i\\ 0.7409\end{array}\right]e^{2.6626it}+c_2\left[\begin{array}{c}-0.2149i\\ -0.5722\\ 0.2783i\\ 0.7409\end{array}\right]e^{-2.6626it}+c_3\left[\begin{array}{c}-0.3500i\\ 0.4157\\ -0.5407i\\ 0.6421\end{array}\right]e^{1.1877it}+c_4\left[\begin{array}{c}0.3500i\\ 0.4157\\ 0.5407i\\ 0.6421\end{array}\right]e^{-1.1877it}$

Matrix Exponential

We already know what the matrix A is from our state space equation

$\mathbf{A}=\left[\begin{array}{cccc}0& 1& 0& 0\\ -4.5& 0& 2& 0\\ 0& 0& 0& 1\\ 4& 0& -4& 0\end{array}\right]$

And we know that the T-inverse matrix is

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

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

It then follows that matrix T is

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

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.

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

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

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

Now we can bring in the standard form of a state space equation

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

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

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

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

where

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

$\hat{\mathbf{A}}=\left[\begin{array}{cccc}-1.2657i& -0.4753& 0.8193i& 0.3077\\ 1.2657i& -0.4753& -0.8193i& 0.3077\\ 0.6514i& 0.5484& 0.5031i& 0.4236\\ -0.6514i& 0.5484& -0.5031& 0.4236\end{array}\right]\left[\begin{array}{cccc}0& 1& 0& 0\\ -\frac{(k_1+k_2)}{m_1}& 0& \frac{k_2}{m_1}& 0\\ 0& 0& 0& 1\\ \frac{k_2}{m_2}& 0& -\frac{k_2}{m_2}& 0\end{array}\right]\left[\begin{array}{cccc}0.2149i& -0.2149i& -0.3500i& 0.3500i\\ -0.5722& -0.5722& 0.4157& 0.4157\\ -0.2783i& 0.2783i& -0.5407i& 0.5407i\\ 0.7409& 0.7409& 0.6421& 0.6421\end{array}\right]$

$=\left[\begin{array}{cccc}2.6626i& 0& 0& 0\\ 0& -2.6626i& 0& 0\\ 0& 0& 1.1877i& 0\\ 0& 0& 0& 1.1877i\end{array}\right]$

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

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

where

$e^{\hat{\mathbf{A}}t}=\left[\begin{array}{cccc}{e}^{2.6626it}& 0& 0& 0\\ 0& e^{-2.6626it}& 0& 0\\ 0& 0& e^{1.1877it}& 0\\ 0& 0& 0& e^{1.1877it}\end{array}\right]$

We then substitute this equation back into

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

and get

$\overline{x}={\mathbf{T}}^{\mathbf{-}\mathbf{1}}{e}^{\hat{\mathbf{A}}t}\overline{z}(0)$

$\overline{x}={\mathbf{T}}^{\mathbf{-}\mathbf{1}}{e}^{\hat{\mathbf{A}}t}\mathbf{T}\overline{x}(0)$

Notice here that

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

$e^{\mathbf{A}t}=\left[\begin{array}{cccc}0.2149i& -0.2149i& -0.3500i& 0.3500i\\ -0.5722& -0.5722& 0.4157& 0.4157\\ -0.2783i& 0.2783i& -0.5407i& 0.5407i\\ 0.7409& 0.7409& 0.6421& 0.6421\end{array}\right]\left[\begin{array}{cccc}{e}^{2.6626it}& 0& 0& 0\\ 0& e^{-2.6626it}& 0& 0\\ 0& 0& e^{1.1877it}& 0\\ 0& 0& 0& e^{1.1877it}\end{array}\right]\left[\begin{array}{cccc}-1.2657i& -0.4753& 0.8193i& 0.3077\\ 1.2657i& -0.4753& -0.8193i& 0.3077\\ 0.6514i& 0.5484& 0.5031i& 0.4236\\ -0.6514i& 0.5484& -0.5031& 0.4236\end{array}\right]$

You can solve this with a computer program or your calculator and plug it into the equation for A. I have not listed the answer for this problem here because it is very messy and extremely long. I did calculate it to make sure it is solvable. But if your numbers are easier to work with you would finish by plugging this value into the equation below.

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