Question 306: Gravitation

Question

A pendulum is oscillating with frequency ' $n$ ' on the surface of earth. If it is taken to a depth $\frac{R}{4}$ below the surface of earth, new frequency of oscillation of depth $\frac{\mathrm{R}}{4}$ is ( $\mathrm{R}=$ radius of earth)

Accepted Answer

The frequency of the pendulum at the surface is given as $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{~g}}{l}}$
At depth the formula for gravitational acceleration is $\mathrm{g}_{\mathrm{cff}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}ight)$
For $\mathrm{d}=\frac{\mathrm{R}}{4}, \mathrm{~g}_{\mathrm{eff}}=\mathrm{g}\left(1-\frac{1}{4}ight)=\frac{3}{4} g$
The frequency at depth $\mathrm{d}=\frac{\mathrm{R}}{4}$
$$\mathrm{f}_{\mathrm{d}}=\frac{1}{2 \pi} \sqrt{\frac{\frac{3}{4} \mathrm{~g}}{l}}=\frac{1}{2 \pi} \sqrt{\frac{3 \mathrm{~g}}{4 l}}$$
Take the ratio of both frequencies
$$\begin{aligned}
& \frac{f_d}{f}=\frac{\sqrt{3}}{2} \
& f_d=\frac{\sqrt{3}}{2} f=\frac{\sqrt{3} n}{2} \quad \ldots(\because f=n)
\end{aligned}$$

Answer

$\frac{2}{\sqrt{3} n}$

Answer

$\frac{\sqrt{3} n}{2}$

Answer

$\frac{2 \mathrm{n}}{\sqrt{3}}$

Answer

$\frac{\mathrm{n}}{4}$

Gravitation

A pendulum is oscillating with frequency ' n ' on the surface of earth. If it is taken to a depth \frac{R}{4} below the surface of earth, new frequency of oscillation of depth \frac{\mathrm{R}}{4} is ( \mathrm{R}= radius of earth)

Choose the correct answer: