A simple pendulum has time period `T_(1)`. When the point of suspension moves vertically up according to the equation `y=kt^(2)` where `k=1m//s^(2)` and `'t'` is time then the time period of the pendulum is `T_(2)` then `(T_(1)//T_(2))^(2)` is

Explanation

`y =kt^(2) = 1//2 at^(2) rArr (1)/(2) a = k =1`
`rArr a = 2m//sec^(2)` (acceleration)
`T_(1) = 2pi sqrt((l)/(g))` and `T_(2) = 2pi sqrt((l)/(g+a))`
`(T_(1)^(2))/(T_(2)^(2)) = (g+a)/(g) = (10+2)/(10) = (6)/(5)`