Question 326: Gravitation

Question

A simple pendulum is oscillating with frequency '$$F$$' on the surface of the earth. It is taken to a depth $$\frac{\mathrm{R}}{3}$$ below the surface of earth. ( $$\mathrm{R}=$$ radius of earth). The frequency of oscillation at depth $$\mathrm{R} / 3$$ is

Accepted Answer

The frequency of the pendulum at the surface is given as
$$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{l}}}$$
At depth the formula for gravitational acceleration is $$g_{	ext {eff }}=g\left(1-\frac{d}{R}ight)$$
For $$\mathrm{d}=\frac{\mathrm{R}}{3}, \mathrm{~g}_{\mathrm{eff}}=\mathrm{g}\left(1-\frac{1}{3}ight)$$
The frequency at depth $$d=\frac{R}{3}$$
$$f_d=\frac{1}{2 \pi} \sqrt{\frac{g\left(1-\frac{1}{3}ight)}{l}}=\frac{1}{2 \pi} \sqrt{\frac{2 g}{3 l}}$$
Take the ratio of both frequencies
$$\begin{aligned}
\frac{f_d}{f} & =\sqrt{\frac{2}{3}} \
	herefore \quad f_d & =\frac{F}{\sqrt{1.5}} \quad \ldots .(\because f=F)
\end{aligned}$$

Answer

$$\frac{2 \mathrm{~F}}{3}$$

Answer

$$\frac{\mathrm{F}}{\sqrt{1.5}}$$

Answer

F

Answer

$$\frac{\mathrm{F}}{3}$$

Gravitation

A simple pendulum is oscillating with frequency '$F' on the surface of the earth. It is taken to a depth \frac{\mathrm{R}}{3} below the surface of earth. ( \mathrm{R}= radius of earth). The frequency of oscillation at depth \mathrm{R} / 3$ is

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