Real bodies which can oscillate about an axis through a point of suspension are physical pendulums. These bodies are real bodies. In simple pendulum, a rigid bob was hanged by an imaginary massless thread, but in physical pendulum the entire body oscillates.
We also consider the rigid nature of physical bodies i.e. the moment of inertia about a fixed axis is the same throughout the oscillation. No body is rigid in nature which means every object is deformable but here we consider that the body is nearly rigid.
We have a body of mass \(m\) suspended by a pivot as shown in Figure 2. When the body is in equilibrium condition, the centre of gravity of the body is directly below the pivot point at a distance \(L\) from the pivot point.
When the body is slightly displaced from its equilibrium position, the centre of gravity of the body follows an arc of a circle of radius \(L\). When the pendulum (body) is not in its equilibrium position i.e. when the body is displaced, its weight has two components; one is parallel and another is perpendicular.
The tangential or the perpendicular component \({w_ \bot } = w\sin \beta = mg\sin \beta \) of the weight of the pendulum provides a restoring torque on the pendulum.
The torque provided by the parallel component \({w_\parallel } = w\cos \beta = mg\cos \beta \) is zero and hence does not contribute to the rotating effect. The restoring torque tends to keep the pendulum in its equilibrium position. Therefore, the restoring torque provided by the perpendicular component is
\[\tau = - (mg\sin \beta )L = - mgL\sin \beta \tag{7} \label{7}\]
In Eq. \eqref{7} the restoring torque is proportional to \(\sin \beta\) not to \(\beta\). So the oscillation of this pendulum can not be simple harmonic, however, if the angle \(\beta\) is small enough \(\sin \beta\) is approximately equal to \(\beta\) and hence we can replace \(\sin \beta\) in Eq. \eqref{7} by simply the angle \(\beta\) provided that \(\beta\) is small enough. And, therefore,
\[\tau = - mgL\beta \tag{8} \label{8}\]
Note that you should measure angles in radians not in degrees. If the angle is measured in degrees, you can easily convert it to radians. If \(I\) is the moment of inertia of the pendulum about the pivot, the rotational analogue of Newton's second law tells us that,
\[\tau = I\alpha \tag{9} \label{9}\]
Hence, using Eq. \eqref{8} and Eq. \eqref{9},
\[\begin{align*} I\alpha &= - mgL\beta \\ {\rm{or,}}\quad \alpha &= \frac{{{d^2}\beta }}{{d{t^2}}} = - \frac{{mgL}}{I}\beta \tag{10} \label{10} \end{align*}\]
The Eq. \eqref{10} shows that the angular acceleration is directly proportional to the angular displacement and hence the oscillation of the pendulum is nearly simple harmonic motion provided that the angular displacement from the equilibrium position is small enough.
Now we compare Eq. \eqref{10} with Eq. \((2)\) from simple harmonic motion. The role of \(k/m\) is played by \(\frac{{mgL}}{I}\) in Eq. \eqref{10} and therefore we can replace \(k/m\) by \(\frac{{mgL}}{I}\):
\[\omega = \sqrt {\frac{{mgL}}{I}} \tag{11} \label{11}\]
Once you found the angular frequency you can easily find the time period \(T\) and frequency \(f\) of the oscillation as
\[T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{I}{{mgL}}} \tag{12} \label{12}\]
\[f = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{{mgL}}{I}} \tag{13} \label{13}\]
If the physical pendulum in Figure 2 is a simple pendulum consisting a bob of mass \(m\) at a distance \(L\) from the point of suspension, the moment of inertia of the bob is \(I = m{L^2}\) and putting the value of \(I = m{L^2}\) in Eq. \eqref{11} the angular frequency of a simple pendulum is given by,
\[\omega = \sqrt {\frac{{mgL}}{I}} = \sqrt {\frac{{mgL}}{{m{L^2}}}} = \sqrt {\frac{g}{L}} \]
which is the same as the angular frequency for the simple pendulum.
