Coupled Horizontal Spring Mass Oscillator

Coupled Oscillator Spring Mass Oscillator: State Space

Problem Statement

Two 4 Kg Weights are suspended between two walls. They are connected by a spring between them with a spring constant k2. They are connected to the walls by two springs k1 and k3 with k1=k3. m1 is a distance x1 form m2 and m2 is x2 from the wall.

Solution

Things we know

$m_1=5kg\frac{}{}$

$m_2=5kg\frac{}{}$

$k_1=50Nm\frac{}{}$

$k_2=100Nm\frac{}{}$

$k_3=50Nm\frac{}{}$

$\text{So now that we have are problem we need to start setting up the equations we need to solve it.}$

$\dot{x_1}=\dot{x_1}$

$\ddot{x_1}+\frac{k_1+k_2}{m_1}{x}_1-\frac{k_2}{m_1}{x}_2=0$

$\dot{x_2}=\dot{x_2}$

$\ddot{x_2}+\frac{k_3+k_2}{m_2}{x}_2-\frac{k_2}{m_2}{x}_1=0$

$\text{Now we take these equations and put them in a state space model.}$

$\left[\begin{array}{c}\dot{x_1}\\ \ddot{x_1}\\ \dot{x_2}\\ \ddot{x_2}\end{array}\right]$ = $\left[\begin{array}{cccc}0& 1& 0& 0\\ \frac{(k_1+k_2)}{m_1}& 0& \frac{-k_1}{m_1}& 0\\ 0& 0& 0& 1\\ \frac{-k_1}{m_2}& 0& \frac{(k_1+k_2)}{m_2}& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ {\dot{x}}_1\\ x_2\\ {\dot{x}}_2\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]$

$\text{Now we make the appropriate numerical substitutions.}$

$\left[\begin{array}{c}\dot{x_1}\\ \ddot{x_1}\\ \dot{x_2}\\ \ddot{x_2}\end{array}\right]$ = $\left[\begin{array}{cccc}0& 1& 0& 0\\ \frac{150}{5}& 0& \frac{-50}{5}& 0\\ 0& 0& 0& 1\\ \frac{-50}{5}& 0& \frac{150}{5}& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ {\dot{x}}_1\\ x_2\\ {\dot{x}}_2\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]$

$\left[\begin{array}{c}\dot{x_1}\\ \ddot{x_1}\\ \dot{x_2}\\ \ddot{x_2}\end{array}\right]=\left[\begin{array}{cccc}0& 1& 0& 0\\ 30& 0& -10& 0\\ 0& 0& 0& 1\\ -10& 0& 30& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ {\dot{x}}_1\\ x_2\\ {\dot{x}}_2\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]$

$\text{So using maple i was able to obtain the eigenvalues and eigenvectors.}$

$\text{Eigenvalues.}$

${\lambda }_1=2\sqrt{10}$ ${\lambda }_2=-2\sqrt{10}$ ${\lambda }_3=2\sqrt{5}$ ${\lambda }_4=-2\sqrt{5}$

$\text{Eigenvectors.}$

${K}_1=$$\left[\begin{array}{c}-1\\ -2\sqrt{(}10)\\ 1\\ 2\sqrt{(}10)\end{array}\right]$,${K}_2=$$\left[\begin{array}{c}-1\\ 2\sqrt{(}10)\\ 1\\ -2\sqrt{(}10)\end{array}\right]$,${K}_3=$$\left[\begin{array}{c}1\\ 2\sqrt{(}5)\\ 1\\ 2\sqrt{(}5)\end{array}\right]$,${K}_4=$$\left[\begin{array}{c}1\\ -2\sqrt{(}5)\\ 1\\ -2\sqrt{(}5)\end{array}\right]$

$\text{So then the answer is...}$

$x=c_1$$\left[\begin{array}{c}-1\\ -2\sqrt{(}10)\\ 1\\ 2\sqrt{(}10)\end{array}\right]$$e^{2\sqrt{10}}+c_2$$\left[\begin{array}{c}-1\\ 2\sqrt{(}10)\\ 1\\ -2\sqrt{(}10)\end{array}\right]$$e^{2*-2\sqrt{10}}+c_3$$\left[\begin{array}{c}1\\ 2\sqrt{(}5)\\ 1\\ 2\sqrt{(}5)\end{array}\right]$$e^{3*2\sqrt{5}}+c_4$$\left[\begin{array}{c}1\\ -2\sqrt{(}5)\\ 1\\ -2\sqrt{(}5)\end{array}\right]$$e^{4*-2\sqrt{5}}$

Solve with the Matrix exponential

$\text{So first we need to know what the matrix exponential equation looks like.}$

$\text{it is...}$

$\tilde{x}=e^{\tilde{A}t}\widetilde{x(0)}$

$\text{Where A is a matrix}$

$\text{Also }$

$\tilde{z}=\tilde{T}\tilde{x}$

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

$\text{Where }$

${\tilde{T}}^{-1}=$$\left[\begin{array}{cccc}-1& -1& 1& 1\\ -2\sqrt{(}10)& 2\sqrt{(}10)& 2\sqrt{(}5)& -2\sqrt{(}5)\\ 1& 1& 1& 1\\ 2\sqrt{(}10)& -2\sqrt{(}10)& 2\sqrt{(}5)& -2\sqrt{(}5)\end{array}\right]$

$\text{I converted the T matrix to decimal form for make it easier to write up on here }$

$\tilde{T}=$</math>$\left[\begin{array}{cccc}-.25& -.039528& .25& .039528\\ -.25& .039528& .25& -.039528\\ .25& .055902& .25& .055902\\ .25& -.055902& .25& -.055902\end{array}\right]$

$\text{and}$

$\hat{A}=$$\left[\begin{array}{cccc}{e}^{{\lambda }_{1}t}& 0& 0& 0\\ 0& e^{{\lambda }_{2}t}& 0& 0\\ 0& 0& e^{{\lambda }_{3}t}& 0\\ 0& 0& 0& e^{{\lambda }_{4}t}\end{array}\right]$

$e^{\hat{A}t}=$$\left[\begin{array}{cccc}{e}^{2\sqrt{10}t}& 0& 0& 0\\ 0& e^{-2\sqrt{10}t}& 0& 0\\ 0& 0& e^{2\sqrt{5}t}& 0\\ 0& 0& 0& e^{-2\sqrt{5}t}\end{array}\right]$

$\text{Then the next step is}$

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

$\text{So that implies}$

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

$\text{Now Simply substitute back in and we have the answer. }$

$\tilde{x}=$$\left[\begin{array}{cccc}-1& -1& 1& 1\\ -2\sqrt{(}10)& 2\sqrt{(}10)& 2\sqrt{(}5)& -2\sqrt{(}5)\\ 1& 1& 1& 1\\ 2\sqrt{(}10)& -2\sqrt{(}10)& 2\sqrt{(}5)& -2\sqrt{(}5)\end{array}\right]\left[\begin{array}{cccc}{e}^{2\sqrt{10}t}& 0& 0& 0\\ 0& e^{-2\sqrt{10}t}& 0& 0\\ 0& 0& e^{2\sqrt{5}t}& 0\\ 0& 0& 0& e^{-2\sqrt{5}t}\end{array}\right]$$\left[\begin{array}{cccc}-.25& -.039528& .25& .039528\\ -.25& .039528& .25& -.039528\\ .25& .055902& .25& .055902\\ .25& -.055902& .25& -.055902\end{array}\right]\tilde{x}(0)$

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Created By: Mark Bernet