= Solution
The <parallel axis theorem> gives $I=I_{\mathrm{CM}}+Md^2$ for parallel axes separated by $d$, with the reference axis through the <centre of mass>. For a plane lamina, the <perpendicular axis theorem> gives $I_z=I_x+I_y$ for mutually perpendicular axes meeting at one point, with $x,y$ in the plane.
Treat the removed material as negative <mass>. With the hole's centre at $b\hat{\mathbf x}$, its <moment of inertia> about $O$ is $m(a^2/2+b^2)$. The remaining <moment of inertia> and <centre of mass> are
$$
\boxed{I_1=\frac12MR^2-m\left(\frac12a^2+b^2\right),\qquad\mathbf r_{\mathrm{CM},1}=-\frac{mb}{M-m}\hat{\mathbf x}.}
$$
For a uniform original disk, $m/M=a^2/R^2$. Write $d_1=mb/(M-m)$ and measure $\phi$ from the downward equilibrium direction of its <centre of mass>. The <physical pendulum> then has <Lagrangian>
$$
L_1=\frac12I_1\dot\phi^2+(M-m)gd_1\cos\phi=\frac12I_1\dot\phi^2+mbg\cos\phi.
$$
The <Euler-Lagrange equation> is $I_1\ddot\phi+mbg\sin\phi=0$. For small <oscillations>, $\sin\phi\simeq\phi$, giving
$$
\boxed{\omega_1=\sqrt{\frac{mbg}{I_1}},\qquad f_1=\frac{\omega_1}{2\pi}.}
$$
Now add the removed small disk on the opposite side, with its centre a distance $c=R-a$ from $O$ so that it is internally tangent to the large disk. The total mass returns to $M$, while its first moment about $O$ is $-m(b+c)\hat{\mathbf x}$. Thus
$$
I_2=I_1+m\left(\frac12a^2+c^2\right)=\frac12MR^2+m(c^2-b^2),\qquad d_2=\frac{m(b+c)}M.
$$
Replacing both the inertia and gravitational restoring moment gives
$$
\boxed{\omega_2=\sqrt{\frac{mg(b+R-a)}{\tfrac12MR^2+m[(R-a)^2-b^2]}},\qquad f_2=\frac{\omega_2}{2\pi}.}
$$
The same first-moment method applies to a general <perforated-disc pendulum>. The added disk is attached to the lamina and shares its rotation; treating it as a massless patch or updating only the inertia would omit its restoring <torque>.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-perforated-disc-pendulum.png]
{title=Removing an off-centre disk and reattaching it on the opposite side changes both the pendulum inertia and its centre of mass}
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