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(B) The frequency of the pendulum at the surface is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.At a depth $d$,the gravitational acceleration is

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$\frac{\sqrt{3}}{2} n$

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(B) The frequency of the pendulum at the surface is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.At a depth $d$,the gravitational acceleration is given by $g_{eff} = g \left(1 - \frac{d}{R}\right)$.For $d = \frac{R}{4}$,the effective acceleration is $g_{eff} = g \left(1 - \frac{R/4}{R}\right) = g \left(1 - \frac{1}{4}\right) = \frac{3}{4} g$.The frequency at depth $d$ is $f_d = \frac{1}{2 \pi} \sqrt{\frac{g_{eff}}{l}} = \frac{1}{2 \pi} \sqrt{\frac{3g}{4l}}$.Taking the ratio of the new frequency to the original frequency $n$:$\frac{f_d}{n} = \frac{\frac{1}{2 \pi} \sqrt{\frac{3g}{4l}}}{\frac{1}{2 \pi} \sqrt{\frac{g}{l}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.Therefore,the new frequency is $f_d = \frac{\sqrt{3}}{2} n$.

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the Earth)

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