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A simple pendulum has time period (T1). ...

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)`.

A

`(5)/(6)`

B

`(6)/(5)`

C

`1`

D

`(4)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
B
`T_(1) = 2pi sqrt((l)/(g)) = 2pi sqrt((l)/(10))`
upward accelertion `(d^(2)y)/(dt^(2)) = 2k = 2 xx 1 = 2 m//s^(2)`
`:.` Acceleration `w.r.t.` point of suspension `= 12 m//s^(2)`
`T_(2) = 2pisqrt((l)/(12)) :. (T_(1))/(T_(2)) = sqrt((12)/(10)) :. ((T_(1))/(T_(2)))^(2) = (6)/(5)`
`T_(1) = 2pi sqrt((l)/(g)) = 2pi sqrt((l)/(10))`
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Knowledge Check

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