A simple pendulum has time period `T_(1)`/ The point of suspension is now moved upward according to the realtion `y = kt^(2)(k = 1 m//s^(2))` where `y` is vertical displacement, the time period now becomes `T_(2)`. The ratio of `((T_(1))/(T_(2)))^(2)` is : `(g = 10 m//s^(2))`

Options

A. `4//5`
B. `6//5`
C. `5//6`
D. `1`

Answer

B

Explanation

Acceleration of the point of suspension,
`a = (d^(2)y)/(dt^(2))=2k=2m//s^(2)rArr T = 2 pi sqrt((L)/(g_(eff)))`
`rArr" "T_(1)=2pi sqrt((L)/(10))and T_(2) = 2pi sqrt((L)/(12))`
`rArr " "(T_(1)^(2))/(T_(2)^(2))=(6)/(5)`