Question

A simple pendulum, consisting of a mass on a string of length L, is undergoing small oscillations with amplitude A.

Answer 1

Answer:

A. Period is halved

Explanation:

The period of a pendulum swing, T, is given in terms of mass as:

$T = 2\pi \sqrt{\frac{I}{mgL} }$

where I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity

h = Length of string

If the mass is increased by a factor of 4, that means:

M = 4m

(M = new mass)

The new period of the pendulum, $T_n$, will now be:

$T_n = 2\pi \sqrt{\frac{I}{MgL} }\\\\\\T_n = 2\pi \sqrt{\frac{I}{4mgL} }\\\\\\T_n = 2\pi \sqrt{\frac{1}{4} * \frac{I}{mgL} }\\\\\\T_n = \frac{2\pi}{2} \sqrt{\frac{I}{mgL} }\\\\\\T_n = \frac{1}{2} * 2\pi \sqrt{\frac{I}{mgL} }\\\\\\T_n = \frac{1}{2} * T$

Hence, the period is halved.