Question 310: 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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