Physics - SHM Question with Solution | TestHub

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 $d=\frac{R}{3},g\left(1-\frac{1}{3}\right)$ The frequency at depth $\mathrm{d}=\frac{\mathrm{R}}{3}$ $${\mathrm{f}}_{\mathrm{d}}=\frac{1}{2\pi }\sqrt{\frac{\mathrm{g}\left(1-\frac{1}{3}\right)}{l}}=\frac{1}{2\pi }\sqrt{\frac{2\mathrm{g}}{3l}}$$ Take the ratio of both frequencies $$\begin{array}{rl}\frac{{\mathrm{f}}_{\mathrm{d}}}{\mathrm{f}}& =\sqrt{\frac{2}{3}}\\ \therefore {\mathrm{f}}_{\mathrm{d}}& =\frac{\mathrm{F}}{\sqrt{1.5}}\dots (\because \mathrm{f}=\mathrm{F})\end{array}$$