JEE PYQ: Oscillations Question 40
Question 40 - 2019 (11 Jan 2019 Shift 2)
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of $10^{-2}$ m. The relative change in the angular frequency of the pendulum is best given by:
(1) $10^{-3}$ rad/s
(2) 1 rad/s
(3) $10^{-1}$ rad/s
(4) $10^{-5}$ rad/s
Show Answer
Answer: (1)
Solution
$\frac{\Delta\omega}{\omega} = \frac{1}{2}\frac{\Delta g}{g}$. $\Delta\omega = 10^{-3}$ rad/s.
BETA
