Physics - Gravitation Question with Solution | TestHub

The frequency of a simple pendulum is given by $n=\frac{1}{2\pi }\sqrt{\frac{g}{L}}$.

This implies $n\propto \sqrt{g}$.

The acceleration due to gravity at a depth $d$ below the surface of the earth is given by $g^{\prime }=g\left(1-\frac{d}{R}\right)$.

Substituting $d=\frac{R}{3}$, we get:

$g^{\prime }=g\left(1-\frac{R/3}{R}\right)=g\left(1-\frac{1}{3}\right)=\frac{2}{3}g$

The new frequency $n^{\prime }$ at this depth is:

$n^{\prime }=n\sqrt{\frac{g^{\prime }}{g}}=n\sqrt{\frac{2g/3}{g}}=\sqrt{\frac{2}{3}}n$

Answer: $\sqrt{\frac{2}{3}}\mathrm{n}$