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

D

Explanation

As `y=Kt^(2) therefore v=(dy)/(dt)=2Kt`
and `a=(d^(2)y)/(dt^(2))=2K=2xx1=2ms^(-2)(because K=1ms^(-2))`
Since the point of suspension of pendulum is moving upwards with acceleration, so effective acceleration due to gravity on the pendulum is
`g.=(g+a)=10+2=12ms^(-2)`
`therefore T_(1)=2pisqrt((l)/(g))andT_(2)=2pisqrt((l)/(g.))`
Thus, `(T_(1)^(2))/(T_(2)^(2))=(g.)/(g)=(12)/(10)=(6)/(5)`