Show that the period of oscillation of simple pendulum at depth `h` below earth's 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 earth's surface ?

Explanation

At earth.s surface the value of time period is given by
`T = 2pi sqrt((L)/(g))`
where L is the effective length of the simple pendulum and g is the acceleration due to gravity and is equal to `(GM)/(R^(2)) = (4)/(3) pi RG rho`. At depth h if the period is `T_(h)` and acceleration due to gravity is `g_(h)`.
Then `T_(h) = 2pi sqrt((L)/(g_(h)))`
Hence `(T_(h))/(T) = sqrt((g)/(g_(h))) = sqrt((R )/(R - h))`
or `T_(h) = (T sqrtR)/(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- ""^(R)//""_(2))) = 2 sqrt2 = 2.8 sec`