## Simple Pendulum

A simple pendulum consists of a rigid bob of mass $m$ hanged by a massless thread of length $L$. The massless thread is an idealized concept which we can consider in our discussions and quiet 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

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

Now comparing Eq. \eqref{3} with Eq. $(2)$ from simple harmonic motion (you may need to read the article), 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 $g/L$ (you may need to read article simple harmonic motion),

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

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.

## Physical Pendulum

Real bodies which can oscillate about an axis through the point of suspension are called physical pendulums. These bodies are real bodies. Here we discuss the oscillation of a physical body. 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 in our discussion of physical pendulum we consider that the body is nearly rigid.

Consider 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 you obtained for the simple pendulum in Eq. \eqref{4}.