A pendulum is oscillating with frequency ' n ' on the surface of the earth. It is taken to a depth R/2 below the surface of earth. New frequency of oscillation at depth R/2 is [ R is the radius of earth] (a) n/3 (b) n/√(2) (c) 2n (d) n/2

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

A pendulum is oscillating with frequency ' n ' on the surface of the… | MHT CET 2021 (21 Sep, Shift 1)

Q95MHT CET 2021 · 21 SEP · SHIFT 1Medium

Acceleration Due to Gravity

A pendulum is oscillating with frequency ' $n$ ' on the surface of the earth. It is taken to a depth $\frac{R}{2}$ below the surface of earth. New frequency of oscillation at depth $\frac{R}{2}$ is

[ $R$ is the radius of earth]

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Answer: (b)

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

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Acceleration Due to Gravity question from MHT CET 2021 (21 Sep, Shift 1)

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