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Let $$m_0\theta''+k\theta' +\frac{m_0g}{l}\theta=0$$ and let $$y(t)=[\theta(t), \theta'(t)]^T$$ Find matrix A such that $y'=Ay$ and it's eigenvalues.

My guess was to define: $x_1=\theta, x_2=\theta'$,

Hence, $x_1'=x_2=\theta'$ and $x_2=\theta''=-(\frac{g}{l}x_1+\frac{k}{m_0}x_2)$ So we obtain: $$x(t)=[x_1, x_2]^T$$

And we obtain matrix A: $$A=\begin{bmatrix} 1 & 0 \\ \frac{g}{l} & \frac{k}{m_0} \\ \end{bmatrix}$$

And $$x'(t)=Ax(t)$$

But I am not sure whether this matrix is correct, as the characteristic polynomial makes little sense in this case. Hence, I believe that the first line of the matrix is incorrect, but could anyone point me out where I made the mistake?

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  • $\begingroup$ Multiply out your proposed $A x$ and see if it matches the equations you got. $\endgroup$
    – Batman
    Dec 16 '15 at 23:50
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$x_1 = \theta$, $x_2 = \theta'$ is a good choice.

Then, $x_1' = x_2$ and $x_2' = \theta'' = -\left(\frac{g}{l} x_1 + \frac{k}{m_0} x_2 \right)$

Thus, you have $\vec{x} = [x_1,x_2]^T$ and $A = \begin{bmatrix} 0 & 1 \\ -\frac{g}{l} & -\frac{k}{m_0} \end{bmatrix}$ and $\vec{x}' = A \vec{x}$.

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