A simple pendulum of length | Physics

JEE MAIN 2019

11-01-2019 S2

Question

A simple pendulum of length 1 m is oscillating with an angular frequency $10 \mathrm{rad} / \mathrm{s}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 \mathrm{rad} / \mathrm{s}$ and an amplitude of $10^{-2} \mathrm{~m}$. The relative change in the angular frequency of the pendulum is best given by

Select the correct option:

$10^{-3} \mathrm{rad} / \mathrm{s}$

$10^{-1} \mathrm{rad} / \mathrm{s}$

$10^{-5} \mathrm{rad} / \mathrm{s}$

$1 \mathrm{rad} / \mathrm{s}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution