A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1 = 4 s. T2 would be its time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth] :
- 3T1 = 2T2
- T1 = T2
- 2T1 = 3T2
- 2T1 = T2
Answer
(1) 3T1 = 2T2
Solution
T = 2 lg', g' = GM(R + h)2
T (R + h) (distance from the centre of the earth)
T1T2 = R + RR + 2R = 23
3T1 = 2T2
Answer: (1)