A simple pendulum consists of a rigid bob of mass \(m\) hanged by a massless thread of length \(L\). The massless thread is an idealization and nearly valid in real situations if the mass of the thread can be said as negligibly small. The weight \(\vec w\) of the bob acts vertically downwards. See Figure 1.

Initially the bob remains stationary at its straight down equilibrium position. The forces acting on the bob are its weight and tension in the thread. If the bob is slightly displaced from its equilibrium position, the weight of the bob will have two components.

The perpendicular component \({w_ \bot } = w\sin \beta \) tends restore the bob towards its equilibrium position while the parallel component \({w_\parallel } = w\cos \beta \) is balanced by the tension in the thread.

So the only force that tends to restore the bob towards its equilibrium position is the perpendicular component of the weight of the bob. Therefore, the restoring force \(F\) on the bob is

\[F = -w\sin \beta = -mg\sin \beta \tag{1} \label{1}\]

The negative sign is put because the restoring force tends to restore the bob towards its equilibrium position which is opposite to the angular displacement.

The Eq. \eqref{1} shows that the restoring force is proportional to \(\sin \beta\) instead of \(\beta\). So the oscillation can not be simple harmonic. However, if the angle \(\beta\) is small enough, then \(\sin \beta\) is approximately equal to \(\beta\). And hence we can write Eq. \eqref{1} for small angle \(\beta\) as

\[F = - mg\beta \tag{2} \label{2}\]

The Eq. \eqref{2} shows that the oscillation of the simple pendulum is nearly simple harmonic for small angular displacements. When the bob is displaced from its equilibrium position, the bob follows a curved path of an arc of a circle of radius equal to the length \(L\) of the thread. If the angle \(\beta\) is measured in radians, it is the ratio of arc length \(s\) to the radius \(L\) and so \(\beta = s/L\). The Eq. \eqref{2} becomes

\[\begin{align*} {F} &= - \frac{{mg}}{L}s\\ {\rm{or,}}\quad a &= - \frac{g}{L}s \tag{3} \label{3} \end{align*} \]

From simple harmonic motion

\[{a} = - \frac{k}{m}s\]

Now comparing Eq. \eqref{3} with above equation, the role of \(k/m\) is played by \(g/L\) in Eq. \eqref{3}.

And hence you can find the angular frequency of the simple pendulum for small angular displacements by replacing \(k/m\) by \(g/L\),

\[\omega = \sqrt {\frac{g}{L}} \tag{4} \label{4}\]

Similarly the time period \(T\) of the oscillation and frequency \(f\) are:

\[T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{L}{g}} \tag{5} \label{5}\]

\[f = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{g}{L}} \tag{6} \label{6}\]

Note that the time period of a simple pendulum only depends in \(L\) and \(g\). Hence it can be determined by the value of the length of the thread for a given value of \(g\).

On the other hand the time period of the oscillation of a simple pendulum of a given length for small angular displacements can be measured to find the value of the acceleration due to gravity \(g\). In simple harmonic motion the time period of the oscillation is independent of the amplitude of the oscillation.