Question

A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.

Answer 1

Length of the pendulum = 40 $cm=0.4m,$ let acceleration due to gravity be g at the depth of 1600 km.

$\therefore g=g(1-\frac{d}{r})$

$=9.8(1-\frac{1600}{6400})$

$=9.8(1-\frac{1}{4})$

$=9.8\times \left(\frac{3}{4}\right)=7.35m{s}^{-2}$

$\therefore$ Time period,

$T=2\pi \sqrt{\left(\frac{l}{g_5}\right)}$

$=2\pi \sqrt{\left(\frac{0.4}{7.35}\right)}$

$=2\pi \sqrt{0.054}$

$=2\pi \times 0.23$

$=2\times 3.14\times 0.23$

$=1.465\approx 1.47sec$