A simple pendulum is oscillating with frequency ' F ' on the surface of the earth. It is taken to a depth R/3 below the surface of earth. ( R = radius of earth). The frequency of oscillation at depth R/3 is (a) 2F/3 (b) (F)/(√(1.5)) (c) F (d) F/3

Answer: (b). F ∝ √(g) for a simple pendulum At depth R/3: g' = g(1 - 1/3) = 2g/3 F' = F√(2/3) = (F)/(√(1.5)) So the answer is (b).

A simple pendulum is oscillating with frequency ' F ' on the surface… | MHT CET 2023 (12 May, Shift 2)

Q67MHT CET 2023 · 12 MAY · SHIFT 2Medium

Acceleration Due to Gravity

A simple pendulum is oscillating with frequency ' $F$ ' on the surface of the earth. It is taken to a depth $\frac{R}{3}$ below the surface of earth. ( $R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

Show answer and solution

Answer: (b)

$F \propto \sqrt{g}$ for a simple pendulum
At depth $\frac{R}{3}$ : $g' = g\left(1 - \frac{1}{3}\right) = \frac{2g}{3}$
$F' = F\sqrt{\frac{2}{3}} = \frac{F}{\sqrt{1.5}}$
So the answer is (b).

Report a mistake

Acceleration Due to Gravity question from MHT CET 2023 (12 May, Shift 2)

More questions on Acceleration Due to Gravity