Question

\left[\begin{array}{ll}

9 & 0 \\

0 & 1

\end{array}\right] \ddot{\mathbf{x}}(t)+\left[\begin{array}{cc}

27 & -3 \\

-3 & 3

\end{array}\right] \mathbf{x}(t)=\mathbf{0} \mathbf{x}(0)=\left[\begin{array}{c}

-\frac{1}{3} \\

1

\end{array}\right] \dot{\mathbf{x}}(0)=\mathbf{0}

Fig: 1

Fig: 2

Fig: 3