Simple and Compound Pendulum

Simple Pendulum

A heavy point mass suspended by a weightless and inextensible string from a rigid support such that the bob can oscillate freely under the action of gravity in a vertical plane is called a simple pendulum. The displacement is very small as compared to the length of the pendulum.

Using Newton's second law of motion,\begin{align*} F=m\frac{\mathrm{d^2x} }{\mathrm{d} t^2}\end{align*}\begin{align*} m\frac{\mathrm{d^2x} }{\mathrm{d} t^2}= -mg\sin \theta\end{align*}\begin{align*} \frac{\mathrm{d^2x} }{\mathrm{d} t^2}= -g\sin \theta\end{align*} for small displacements, \(x=l\theta\)\begin{align*} \frac{\mathrm{d^2x} }{\mathrm{d} t^2}=l\frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}\end{align*} Then, combining the equations,\begin{align*} \frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}=\frac{-g}{l}\sin \theta\end{align*} But, we have, for small angle \(\theta\), \(\sin\theta\approx\theta\). So, we can write,\begin{align*} \frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}=-\tfrac{gl}{\theta}\end{align*} This equation indicates that the angular acceleration of the simple pendulum is directly proportional to its displacement but in the opposite direction. Thus the motion must be simple harmonic. So, comparing it with the standard equation of SHM \(\alpha=-\omega^2x\), we get,\begin{align*} \omega ^{2}=\frac{g}{l}\end{align*}\begin{align*} \omega =\sqrt{\frac{g}{l}}\end{align*} Thus, the time period is,\begin{align*} T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l}{g}}\end{align*}Let, m be the mass of the body executing SHM and P be the instanteneous position of the bob where the displacement is \(x\) such that \(\theta\) is very small. The forces acting on the pendulum are weight of the bob acting downwards and the tension acting along the string. The component \(mg\cos \theta \) of the weight balance the tensin and the component\(mg\sin \theta \) helps to bring the bob towards the equilibrium; it is the restoring force.

The heavy point mass and an inextensible string is an ideal concept. The motion of bob cannot be completely linear and it is affected by air resistance. The motion of the bob would lag the motion of the string and the time period can be correctly found out only for small displacements. Due to these limitations, the concept of a compound pendulum was developed.

Compound Pendulum

A rigid body capable of oscillating in a vertical plane about a horizontal axis passing through the body but not through its centre of gravity is called a compound pendulum or a physical or a rigid pendulum.

Center of gravity(G)is a point in the body at which the total weight of the body acts. The point of intersection of a verticle line passing through the centre of gravity and the horizontal axis is called the point of suspension(S). The distance between the point of suspension and the centre of gravity is called the length of the compound pendulum.

Consider a compound pendulum of mass m and length l. Let, \(\theta\) be the angular displacement as it oscillates about the point of suspension. The forces acting on the pendulum in this case are weight mg of the pendulum acting vertically downward at G and the reaction force \(R=mg\) acting vertically up at the point of suspension. The, The restoring torque is \(-mgl\sin\theta\). For small displacement \(\theta\), \(\sin\theta\approx\theta\). So, restoring torque is \(-mgl\theta\). If I be the moment of inertia of the body abut the horizontal axis passing through S and, \(\alpha=\frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}\) be the angular acceleration, then the deflecting couple is \(I\alpha=I\frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}\). Then, the equation of motion of compound pendulum is,\begin{align*} \frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}=\frac{-mgl\theta}{I}\end{align*} From parallel axis theorem of moment of inertia, \(I=I_{o}+ml^2\), where \(I_{o}\) is the moment of inertia of the pendulum about the axis passing through G and \(I\) is the moment of inertia of the pendulum parallel to the axis of \(I_{o}\). If \(k\) be the radius of gyration of the pendulum about the horizontal axis through G, we can write \(I=mk^2+ml^2\). Then, the equation of motion becomes,\begin{align*} \frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}=\frac{-mgl}{mk^2+ml^2}\theta\end{align*}\begin{align*} \frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}=\frac{-g\theta }{\frac{k^2}{l}+l}=-\frac{g}{\frac{k^2}{l}+l}\theta\end{align*} For a given pendulum, l, k and g are constants,\begin{align*} \therefore \frac{\mathrm{d^2\theta} }{\mathrm{d} t^2}\propto -\theta\end{align*} Hence, the motion of the compound pendulum is simple harmonic. Comparing it with the standard equation of SHM, we get,\begin{align*} \omega =\sqrt{\frac{g}{\frac{k^2}{l}+l}}\end{align*} And, the period can be calculated by,\begin{align*} T=2\pi \sqrt{\frac{\frac{k^2}{l}+l}{g}}\end{align*} Comparing it with the time period of simple pendulum we find that \(\frac{k^2}{l}+l\) is the length equivalent to that of simple pendulum. At k=l, the time period is minimum and the time period is maximum(infinity) if it is suspended at the CG.

A point on the lower side of G in opposite direction of S at a distance \(\frac{k^2}{l}\) from G is called the centre of oscillation. The point of suspension and the centre of oscillation are interchangeable. Thus, there are four points collinear with CG about which the period of oscillation is the same in a compound pendulum.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.