A simple pendulum has time period (T_1). The point of suspension is now moved upward according to the relation `y = K t^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)` is `(g = 10 m//s^2)`.

Options

A. `2//3`
B. `5//6`
C. `6//5`
D. `3//2`

Answer

C

Explanation

`y = Kt^(2) rArr (d^(2)y)/(dt^(2) = a_(y) = 2K = 2 xx 1 = 2m//s^(2)(:' K = 1 m//s^(2))`
Now `T_(1) = 2pi sqrt((l)/(g))` and `T_(2) = 2pi sqrt((1)/((g + a_(y)))`
`Dividing (T_(1))/(T_(2)) = sqrt((g + a_(y))/(g)) rArr sqrt((6)/(5)) rArr (T_(1)^(2))/(T_(2)^(2)) = (6)/(5)`