Question 362: Gravitation

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]

Accepted Answer

Frequency of a simple pendulum is given by
$$\begin{aligned}
\mathrm{n} & =\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \
	herefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}} & =\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \
\mathrm{g}^{\prime} & =\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}ight)=\mathrm{g}\left(1-\frac{1}{2}ight)=\frac{\mathrm{g}}{2} \
	herefore \quad & \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}}
\end{aligned}$$

Answer

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

Answer

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

Answer

$$\mathrm{2 n}$$

Answer

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

Gravitation

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]

Choose the correct answer: