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: Understanding the Question:
The frequency of a simple pendulum depends on the acceleration due to gravity ($g$). We need to find how the frequency changes when the pendulum is moved to a depth $d = R/2$ below the Earth's surface.

Step 2: Key Formula or Approach:
1. Frequency $n = \frac{1}{2\pi} \sqrt{\frac{g}{\ell}} \implies n \propto \sqrt{g}$.
2. Acceleration due to gravity at depth $d$ is $g' = g \left( 1 - \frac{d}{R} \right)$.

Step 3: Detailed Explanation:
Let $n$ be the frequency at the surface and $n'$ be the frequency at depth $d = R/2$.
Then $\frac{n'}{n} = \sqrt{\frac{g'}{g}}$.
Calculate $g'$ at depth $d = R/2$:
$g' = g \left( 1 - \frac{R/2}{R} \right) = g \left( 1 - \frac{1}{2} \right) = \frac{g}{2}$.
Now substitute $g'$ into the ratio:
$\frac{n'}{n} = \sqrt{\frac{g/2}{g}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
$\therefore n' = \frac{n}{\sqrt{2}}$.

Step 4: Final Answer:
The new frequency is $\frac{n}{\sqrt{2}}$, which corresponds to option (D).