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(B) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{\ell}}$.Since $n \propto \sqrt{g}$,the ratio of the new frequenc

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

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(B) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{\ell}}$.Since $n \propto \sqrt{g}$,the ratio of the new frequency $n'$ to the original frequency $n$ is $\frac{n'}{n} = \sqrt{\frac{g'}{g}}$.The acceleration due to gravity at a depth $d$ is given by $g' = g(1 - \frac{d}{R})$.Substituting $d = \frac{R}{2}$,we get $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.Therefore,$\frac{n'}{n} = \sqrt{\frac{g/2}{g}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,the new frequency is $n' = \frac{n}{\sqrt{2}}$.

$A$ pendulum is oscillating with frequency $n$ on the surface of the earth. It is taken to a depth $d = \frac{R}{2}$ below the surface of the earth. What is the new frequency of oscillation at this depth? [$R$ is the radius of the earth]

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