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

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

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(A) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.At the surface of the Earth,$n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.At a depth $d$ below the surface,the acceleration due to gravity $g'$ is given by $g' = g(1 - \frac{d}{R})$.Given $d = \frac{R}{2}$,we have $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The new frequency $n'$ at depth $d$ is $n' = \frac{1}{2\pi} \sqrt{\frac{g'}{l}}$.Substituting $g' = \frac{g}{2}$,we get $n' = \frac{1}{2\pi} \sqrt{\frac{g/2}{l}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2\pi} \sqrt{\frac{g}{l}} \right)$.Therefore,$n' = \frac{n}{\sqrt{2}}$.

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

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