A simple pendulum is oscillating with frequency ' $F$ ' on the surface of the earth. It is taken to a depth…
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
$\frac{2 \mathrm{~F}}{3}$
$\frac{\mathrm{F}}{\sqrt{1.5}}$
$\mathrm{F}$
$\frac{F}{3}$
Solution
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
$\begin{aligned}
\frac{\mathrm{f}_{\mathrm{d}}}{\mathrm{f}} & =\sqrt{\frac{2}{3}} \\
\therefore \quad \mathrm{f}_{\mathrm{d}} & =\frac{\mathrm{F}}{\sqrt{1.5}} \quad \ldots(\because \mathrm{f}=\mathrm{F})
\end{aligned}$