A simple pendulum has time period `T_1`. The point of suspension is now moved upward according to the relation `y = Kt^2`. (`K= 1 m s^(-2)`) where y is the vertical displacement. The time period now becomes `T_2`
The ratio of `T_1^2/T_2^2`(Take `g = 10 m s^(-2)`)

Options

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

Answer

B

Explanation

`y=kt^(2)` `(dy)/(dt)=2kt`
implies `(d^(2)y)/(dt^(2))=2k=2m//s^(2)` …(i)
(because `k=1m//s^(2)`,given)
We know that `T=2pisqrt((l)/(g)` therefore `(T_(1)^(2))/(T_(2)^(2))=(g_(2))/(g_(1))` implies `(T_(1)^(2))/(T_(2)^(2))=(12)/(10)=(6)/(5)` [because `g_(1)=10m//s^(2)`,`g_(2)=g+2=12m//s^(2)`]