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 ?

Answer

B

Explanation

At earth' s surface, the value of time period is given by
`T = 2pi sqrt ((L)/(g))` or `T prop (1)/sqrt (g)`
At a depth `h` below the surface,
`g' = g(1 - (h)/(R))`
`:. (T')/(T) = sqrt ((g)/(g'))`
` = sqrt ((1)/((1 - (h)/(R)))) = sqrt((R )/(R - h))`
`:. T' = T sqrt((R)/(R - h))`
or `T' prop (1)/(sqrt (R - h))` Hence proved
Further, `T_(R//2) = 2 sqrt ((R)/(R - R//2))`
`= 2sqrt(2)s`