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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
`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

  • 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
    `6//5`
    B
    `5//6`
    C
    1
    D
    `4//5`

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