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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

(a)`6/5`

B

(b)`5/6`

C

(c)`1`

D

(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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