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A simple pendulum has time period T1. Th...

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

A

`5//6`

B

`6//5`

C

1

D

`4//5`

Text Solution

Verified by Experts

The correct Answer is:
B
`y = kt^(2)`
`:. (dy)/(dt) = 2kt`
`rArr (d^(2)y)/(dt^(2)) = 2k = 2 m//s^(2)`…(i)
`(because k = 1 m//s^(2), given)`
we know that
`T = 2 pi sqrt((l)/(g))`
`:. (T_(1)^(2))/(T_(2)^(2)) =(g_(2))/(g_(1)) rArr (T_(1)^(2))/(T_(2)^(2)) = (12)/(10) =(6)/(5)`
`[because g_(1) =10 m//s^(2), g_(2) =g+2=12 m//s^(2)]`.
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