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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

`6/5`

B

`5/6`

C

`1`

D

`4/5`

Text Solution

Verified by Experts

The correct Answer is:
A
`y = Kt^(2) rArr (d^(2))/(dt^(2)) = 2K`
`rArr a_(y) = 2m//s^(2)` (as `K = 1 m//s^(2)`)
`T_(1) = 2pisqrt((l)/(g))` and `T_(2) = 2pisqrt((l)/(g + a_(y)))`
`:. (T_(1)^(2))/(T_(2)^(2)) = (g+a_(y))/(g) = (10+2)/(10) = 6/5`
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