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 $\frac{\mathrm{R}}{2}$ is [ $\mathrm{R}$ is the radius of earth]

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Aicharya · Accepted Answer

Frequency of a simple pendulum is given by $\begin{aligned} & \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \ & 	herefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \ & \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}ight)=\mathrm{g}\left(1-\frac{1}{2}ight)=\frac{\mathrm{g}}{2} \ & 	herefore \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}} \end{aligned}$

A pendulum is oscillating with frequency ' $n$ ' on the surface of the earth. It is taken to a depth…

Solution