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Q. What effect occurs on the frequency of a pendulum if it is taken from the earth's surface to deep into a mine?

Punjab PMETPunjab PMET 2006Oscillations

Solution:

On going below depth $h$ from the surface of earth. The value of $g^{\prime}$ below is given by $g'=g\left(1-\frac{h}{R_{e}}\right)$ hence $g$ decreases
Also,
Time period $(T)=\frac{1}{\text { frequency }(n)}$
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and $T=2 \pi \sqrt{\frac{1}{g}}$
where $l$ is length of pendulum.
$\therefore \frac{1}{n}=2 \pi \sqrt{\frac{l}{g\left(1-\frac{h}{R_{e}}\right)}}$
Since, $n \propto g$, and $g$ decreases therefore frequency also decreases.