Question

A pendulum is oscillating with frequency 'n' on the surface of the earth. It is taken to a depth $\frac{R}{2}$ below the surface of earth. New frequency of oscillation at depth $\frac{R}{2}$ is [R is the radius of earth]

• A. $n\sqrt{3}$
• B. $\frac{n}{2}$
• C. $2n$
• D. $\frac{n}{\sqrt{2}}$

Hint:

At a depth $d$, gravity is proportional to the remaining radius $R-d$. At $d=R/2$, gravity is halved, so the frequency (proportional to $\sqrt{g}$) is divided by $\sqrt{2}$.

Answer 1

The Correct Option is D

Step 1: Link frequency to gravity.
A pendulum's frequency is $n = \tfrac{1}{2\pi}\sqrt{\tfrac{g}{\ell}}$, so $n \propto \sqrt{g}$. If $g$ changes, frequency changes by the square root of the ratio.

Step 2: Find $g$ at the depth.
At depth $d$, $g' = g\left(1 - \tfrac{d}{R}\right)$. With $d = \tfrac{R}{2}$, $g' = g(1 - \tfrac12) = \tfrac{g}{2}$.

Step 3: Take the ratio.
\[ \frac{n'}{n} = \sqrt{\frac{g'}{g}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt2}. \]

Step 4: Write the new frequency.
\[ \boxed{n' = \tfrac{n}{\sqrt2}} \]