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

The frequency of the pendulum at the surface is given as
\(
\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{l}}
\)
At depth the formula for gravitational acceteration is \(\mathrm{g}_{\mathrm{eff}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
For \(d=\frac{R}{3}, \quad g\left(1-\frac{1}{3}\right)\)
The frequency at depth \(\mathrm{d}=\frac{\mathrm{R}}{3}\)
\(
\mathrm{f}_{\mathrm{d}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}\left(1-\frac{1}{3}\right)}{l}}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{~g}}{3 l}}
\)
Take the ratio of both frequencies
\(
\begin{aligned}
\frac{\mathrm{f}_{\mathrm{d}}}{\mathrm{f}} & =\sqrt{\frac{2}{3}} \\
\therefore \quad \mathrm{f}_{\mathrm{d}} & =\frac{\mathrm{F}}{\sqrt{1.5}} \quad \ldots(\because \mathrm{f}=\mathrm{F})
\end{aligned}
\)