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

Answer

A

Explanation

Given, `y=kt^(2),(dy)/(dt)=2kt`
and `(d^(2)t)/(dt^(2))=2k=2xx1=2ms^(-2)`
So, point of suspension is moving upwards with acceleration `a=2ms^(-2)`. The effective acceleration due to gravity on pendulum is
`g^(')=g+a=10+2-12ms^(-2)`
`T_(1)=2pisqrt((l)/(g))` and `T_(2)=2pisqrt((l)/(g^(')))`
`:. (T_(1)^(2))/(T_(2)^(2))=(g^('))/(g)=(12)/(10)=(6)/(5)`