A pendulum is oscillating with frequency 'n' on the surface of earth. If it is taken to depth $\frac{\mathrm{R}}{2}$ below the surface of earth where $\mathrm{R}$ is radius of earth. New frequency of oscillations at depth $\frac{\mathrm{R}}{2}$ is

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Aicharya · Accepted Answer

$\begin{aligned} T=2 \pi \sqrt{\frac{l}{g}} \quad T^{\prime} &=2 \pi \sqrt{\frac{1}{g^{\prime}}} \ &=2 \pi \sqrt{\frac{1}{g / 2}} \ T^{\prime} &=\frac{\sqrt{2}}{T} \ 	herefore n^{\prime} &=\frac{n}{\sqrt{2}} \end{aligned}$ $n=1 / T$ $\begin{aligned} g^{\prime} &=g(1-d / R) \ &=g\left(1-\frac{R / 2}{R}ight) \ &=g / 2 \end{aligned}$

A pendulum is oscillating with frequency 'n' on the surface of earth. If it is taken to depth…

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