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. `(5)/(6)`
B. `(6)/(3)`
C. `(4)/(5)`
D. `(5)/(4)`

Answer

B

Explanation

`T_(1)=2pi sqrt((l)/(g)) and T_(2)=2pi sqrt((l)/(g'))`
`therefore (T_(1)^(2))/(T_(2)^(2))=(g')/(g)`
It is given that `y=Kt^(2)`
`therefore (dy)/(dt)=2K t and (d^(2)y)/(dt^(2))=2K=2xx1=2 m//s^(2)`
`therefore` The point of suspension is moving upwards with an acceleration `a= 2 m//s^(2)`.
`therefore` The effective acceleration acting on the pendulum will be `g'=g+a =10+2=12 m//s^(2)`
`therefore (T_(1)^(2))/(T_(2)^(2))=(g')/(g)=(12)/(10)=(6)/(5)`