Question

Show that the period of oscillation of simple pendulum at depth h below earths surface is inversely proportional to $\sqrt{R-h}$ where R is the radius of earth. Find out the time period of a second pendulum at a depth R/2 from the earths surface?

Answer 1

At earths surface the value of time period is given by
T = 2$\pi$ $\sqrt{L/g}$
where L is the effective length of the simple pendulum and g is the acceleration GM/$R^2$ = 4/3 $\pi$RGp. At depth h if the period is $T_h$ and acceleration due to gravity is $g_h$,
Then $T_h$ = 2$\pi$ $\sqrt{L/g_h}$
$T_h$ = T$\sqrt{R}/$$\sqrt{R-h}$
Thus $T_h$ is inversely proportional to $\sqrt{R-h}$
Time period at h = R/2 is given by
$T_{R/2}$ = 2 $\sqrt{R/R-/2}$ = 2$\sqrt{2}$ = 2.8 sec