A second pendulum is taken into a mine of depth 640 m and oscillated. What is its time period ?

A second pendulum is taken into a mine of depth 640 m and oscillated.

1.	A second pendulum is taken into a mine of depth 640 m and oscillated. What is its time period ?
Answer» Solution :The time period of a SIMPLE pendulum is `T=2pisqrt((l)/(g))` When a second pendulum is taken into a mine, .g. CHANGES (l constant) The ratio of its time periods is `(T_(1))/(T_(2))= SQRT((g_(2))/(g_(1))` On the surface of the earth `g_(1)= g` and in the mine `g_(2)= g(1-(d)/(R)) (T_(1))/(T_(2))= sqrt((g(1-(d)/(R))/(g))` or`(T_(1))/(T_(2))= sqrt(1-(d)/(R))` Here `T_(1)=2s, d= 640m= 0.6400 KM` R= Radius of the earth = 6400 km, `T_(2)`= ? `(2)/(T_(2))= sqrt((1-(0.640)/(6400)))= sqrt(1-0.0001)` `:. T_(2)= (2)/(sqrt(1-0.0001))= 2.0001s`