Answer
$$
\omega_n=3.45 \mathrm{rad} / \mathrm{s}
$$
Work Step by Step
$T_O$ is the equilibrium force.
$$
T_O=\frac{6(3)}{2}=9 \mathrm{lb}
$$
Thus, for small $\theta$
$$
↻+\Sigma M_O=I_O \alpha ; \quad 6(3)-[9+5(2) \theta](2)=\left(\frac{6}{32.2}\right)(3 \ddot{\theta})(3)
$$
Thus
$$
\begin{aligned}
& \ddot{\theta}+11.926 \theta=0 \\
& \omega_n=\sqrt{11.926}=3.45 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
