Answer
$$
\tau=0.487 \mathrm{~s}
$$
Work Step by Step
$$
\begin{aligned}
E & =T+V \\
& =\frac{1}{2}(3)(0.3 \dot{\theta})^2+\frac{1}{2}(500)\left(\delta_{s t}+0.3 \theta\right)^2-3(9.81)(0.3 \theta) \\
E & =\dot{\theta}\left[\left(3(0.3)^2 \ddot{\theta}+500\left(\delta_{s t}+0.3 \theta\right)(0.3)-3(9.81)(0.3)\right]=0\right.
\end{aligned}
$$
By statics,
$$
\begin{aligned}
& T(0.3)=3(9.81)(0.3) \\
& T=3(9.81) \mathrm{N} \\
& \delta_{s t}=\frac{3(9.81)}{500}
\end{aligned}
$$
Thus,
$$
\begin{aligned}
& 3(0.3)^2 \ddot{\theta}+500(0.3)^2 \theta=0 \\
& \ddot{\theta}+166.67 \theta=0 \\
& \omega_n=\sqrt{166.67}=12.91 \mathrm{rad} / \mathrm{s} \\
& \tau=\frac{2 \pi}{\omega_n}=\frac{2 \pi}{12.91}=0.487 \mathrm{~s}
\end{aligned}
$$
