Question

A pendulum is oscillating with frequency $n$ on the surface of earth. It is taken to depth $\dfrac{R{2}$ below the surface of earth where $R$ is radius of earth. The new frequency of oscillations at depth $\dfrac{R}{2}$ is}

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

Hint:

Frequency of a pendulum is proportional to the square root of acceleration due to gravity.

Answer 1

The Correct Option is A

Step 1: Frequency of simple pendulum.
\[ n = \frac{1}{2\pi}\sqrt{\frac{g}{\ell}} \]
Step 2: Acceleration due to gravity at depth $d$.
At depth $d$ below earth’s surface:
\[ g_d = g\left(1 - \frac{d}{R}\right) \]
Step 3: Substituting $d = \dfrac{R{2}$.}
\[ g_d = g\left(1 - \frac{1}{2}\right) = \frac{g}{2} \]
Step 4: New frequency at depth.
\[ n_d = \frac{1}{2\pi}\sqrt{\frac{g_d}{\ell}} = \frac{1}{2\pi}\sqrt{\frac{g}{2\ell}} = \frac{n}{\sqrt{2}} \]
Step 5: Conclusion.
The new frequency is $\dfrac{n}{\sqrt{2}}$.