A pendulum is oscillating with frequency ' n ' on the surface of earth. If it is taken to a depth R 4 below the surface of earth, new frequency of oscillation of depth R 4 is ( R= radius of earth)

Step-by-step solution

The frequency of the pendulum at the surface is given as f = 1 2 ~g l At depth the formula for gravitational acceleration is g _ eff = g (1- d R ) For d = R 4 , ~g _ eff = g (1- 1 4 )= 3 4 g The frequency at depth d = R 4 f _ d = 1 2 3 4 ~g l = 1 2 3 ~g 4 l Take the ratio of both frequencies aligned & f_d f = 3 2 & f_d= 3 2 f= 3 n 2 ( f=n) aligned