= Solution
Take the positive coordinate from $O$ toward $S$. The <centre of mass> of the remaining material has coordinate
$$
x_X=\frac{m(0)-(m/4)(a/2)}{3m/4}=-\frac a6.
$$
Thus $SX=a+a/6=7a/6$. If $\phi$ is the small angular displacement from the stable hanging orientation, gravity gives <torque> $-(3m/4)g(7a/6)\sin\phi$ about the suspension point. The <physical pendulum> equation is
$$
I_S\ddot\phi=-\frac{7mga}{8}\sin\phi.
$$
Using the <small-angle approximation> $\sin\phi\simeq\phi$ gives <simple harmonic motion> with
$$
\boxed{\omega^2=\frac{(3m/4)g(7a/6)}{45ma^2/32}
=\frac{28g}{45a},\qquad \omega=\sqrt{\frac{28g}{45a}}.}
$$
Both the reduced <mass> and displaced <centre of mass> matter; changing only the <moment of inertia> would give an incorrect <frequency>.
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