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A simple pendulum has time period T(1)/ ...

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))`

A

`2//3`

B

`5//6`

C

`6//5`

D

`3//2`

Text Solution

Verified by Experts

The correct Answer is:
C
` y = kt^(2) rArr (d^(2) y)/( dt^(2)) = a_(y) = 2 k = 2 xx 1 = 2 m // s^(2)`
`T_(1) = 2 pi sqrt((l)/(g)) and T_(2) = 2 pi sqrt((l)/((g + a_(y)))`
`:. (T_(1))/( T_(2)) = sqrt(( g + a_(y))/(g)) = sqrt((6)/(5)) rArr (T_(1)^(2))/(T_(2)^(2)) = (6)/(5)`
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