Two pendulums have the same dimensions (length \(L\)) and total mass (\(m\)). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

Two pendulum have the same length (\(L\)) and total mass is (\(m\))

The letter "\(T\)" stands for the period of time needed for the pendulum to complete one complete oscillation.

\(T = 2\pi \sqrt {\frac{L}{g}} \) …(i)

Where, \(L\) is the length of pendulum and \(g\) is the acceleration due to gravity

As it given that pendulum A is a very small ball swinging at the end of a uniform massless bar then it is consider as a simple pendulum. So the time period of pendulum of A is expressed as,

\({T_A} = 2\pi \sqrt {\frac{L}{g}} \) …(ii)

Pendulum B consists of rod and a ball attached to the rod. Total mass of the system is given \(m\). The mass of the rod is \(\frac{m}{2}\) and mass of the ball is \(\frac{m}{2}\). It is consider as compound pendulum

Then time period of pendulum is expressed as,

\({T_B} = 2\pi \sqrt {\frac{I}{{mgd}}} \) …(iii)

Where, \(I\) is the moment of inertia and \(d\) is the distance between center of gravity and the point of suspension

Moment of inertia of pendulum B is the sum of moment of inertia of the rod about the point of suspension and moment of inertia of the ball rotating about the point of suspension. Thus,

\(\begin{aligned}{}I = \left[ {\frac{1}{3}\left( {\frac{m}{2}} \right){L^2} + \left( {\frac{m}{2}} \right){L^2}} \right]\\ = \frac{2}{3}m{L^2}\end{aligned}\)

Distance \(d\) is calculated as,

\(\begin{aligned}{}d = \frac{{\left( {\frac{m}{2}} \right)\left( {\frac{L}{2}} \right) + \left( {\frac{m}{2}} \right)L}}{{\left( {\frac{m}{2} + \frac{m}{2}} \right)}}\\ = \frac{{3L}}{4}\end{aligned}\)

Substitute these values in equation (iii)

\(\begin{aligned}{}{T_B} = 2\pi \sqrt {\frac{{\frac{2}{3}m{L^2}}}{{mg\frac{{3L}}{4}}}} \\ = 2\pi \sqrt {\frac{{8L}}{{9g}}} \\ = \frac{{2\sqrt 2 }}{3}\left( {2\pi \sqrt {\frac{L}{g}} } \right)\end{aligned}\)

By using (ii)

\(\begin{aligned}{}{T_B} = \frac{{2\sqrt 2 }}{3}{T_A}\\{T_A} = 1.06{T_B}\end{aligned}\)

Hence, the pendulum A takes longer time for swing than the pendulum B