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(A) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$.Since $L$ is constant,$n \propto \sqrt{g}$.On the surface o

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$\sqrt{2/3} n$

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(A) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$.Since $L$ is constant,$n \propto \sqrt{g}$.On the surface of the Earth,$g_s = g$.At a depth $d$,the acceleration due to gravity is given by $g_d = g_s \left(1 - \frac{d}{R}\right)$.Given $d = R/3$,we have $g_d = g \left(1 - \frac{R/3}{R}\right) = g \left(1 - 1/3\right) = \frac{2}{3}g$.The new frequency $n'$ is given by $n' = n \sqrt{\frac{g_d}{g_s}}$.Substituting the values,$n' = n \sqrt{\frac{(2/3)g}{g}} = n \sqrt{2/3}$.

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = R/3$ below the surface of the Earth,what is the new frequency of oscillation? ($R$ is the radius of the Earth)

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