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(B) The frequency of a simple pendulum is given by $f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.At the surface of the earth,$f = F = \frac{1}{2\pi} \sqrt{\

Vedclass · Accepted Answer

$\frac{F}{\sqrt{1.5}}$

Vedclass · Answer

(B) The frequency of a simple pendulum is given by $f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.At the surface of the earth,$f = F = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.At a depth $d$ below the surface,the effective acceleration due to gravity is $g_d = g(1 - \frac{d}{R})$.Given $d = \frac{R}{3}$,the effective gravity becomes $g_d = g(1 - \frac{R/3}{R}) = g(1 - \frac{1}{3}) = \frac{2g}{3}$.The frequency at depth $d$ is $f_d = \frac{1}{2\pi} \sqrt{\frac{g_d}{l}} = \frac{1}{2\pi} \sqrt{\frac{2g}{3l}}$.Comparing this with the frequency at the surface:$f_d = \sqrt{\frac{2}{3}} 	imes \left( \frac{1}{2\pi} \sqrt{\frac{g}{l}} ight) = \sqrt{\frac{2}{3}} F$.Since $\sqrt{\frac{2}{3}} = \sqrt{\frac{1}{1.5}} = \frac{1}{\sqrt{1.5}}$,the frequency is $\frac{F}{\sqrt{1.5}}$.

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

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