A physical pendulum pivoted at a point executes angular oscillations. Its mass is m, has its centre of mass at distance r from the point of suspension. If its moment  of Inertia is  I, then its angular frequency is

For a body executing simple harmonic motion, Restoring torque acting on it after a  small displacement $\theta$, about an axis,
$torque \tau =-c\theta here c=couple acting; \theta is angular displacement$  
We know that this results in angular oscillations which can be also related by the equation  $torque=\tau =I\alpha ; here I=moment of inertia; \alpha =angular acceleration$
Equating the above two we get  $-c\theta =I\alpha$
Where for angular oscillations  $\alpha =-{\omega }^2\theta$
Hence $-c\theta =-I{\omega }^2\theta$ 
Simplifying this $c=I{\omega }^2$ 
Hence  $\omega =\sqrt{\frac{c}{I}}$
For angular oscillation $\omega =\sqrt{\frac{c}{I}}$ , where  $\omega =\frac{2\pi }{T}$
Restoring torque acting on a rod after a small displacement  $\theta$, about an axis  passing through point of contact O with the curved path,
$\tau =-Mgr\sin \theta =-r\times F=force\times perpendicular dis\tan ce$  , for small angles   $\sin \theta \approx \theta$
Therefore,  $\tau =-c\theta =-Mgr\theta$
Hence  $c=Mgr$. Substituting this is (1) we get 
$\omega =\sqrt{\frac{Mgr}{I}}$ .