A pendulum is oscillating with frequency ' n ' on the surface of earth. If it is taken to a depth R/4 below the surface of earth, new frequency of oscillation of depth R/4 is ( R = radius of earth) (a) (2)/(√(3) n) (b) (√(3) n)/(2) (c) (2n)/(√(3)) (d) n/4

Answer: (b). Frequency n = 1/2π√(g/L), so n ∝ √(g) At depth R/4: g' = g(1 - 1/4) = 3g/4 n' = n√(3/4) = (√(3) n)/(2) So the answer is (b).

A pendulum is oscillating with frequency ' n ' on the surface of… | MHT CET 2024 (4 May, Shift 2)

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Acceleration Due to Gravity

A pendulum is oscillating with frequency ' $n$ ' on the surface of earth. If it is taken to a depth $\frac{R}{4}$ below the surface of earth, new frequency of oscillation of depth $\frac{R}{4}$ is ( $R =$ radius of earth)

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Answer: (b)

Frequency $n = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$ , so $n \propto \sqrt{g}$
At depth $\frac{R}{4}$ : $g' = g\left(1 - \frac{1}{4}\right) = \frac{3g}{4}$
$n' = n\sqrt{\frac{3}{4}} = \frac{\sqrt{3}\,n}{2}$
So the answer is (b).

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