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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
`y=kt^(2)` `(dy)/(dt)=2kt`
implies `(d^(2)y)/(dt^(2))=2k=2m//s^(2)` …(i)
(because `k=1m//s^(2)`,given)
We know that `T=2pisqrt((l)/(g)` therefore `(T_(1)^(2))/(T_(2)^(2))=(g_(2))/(g_(1))` implies `(T_(1)^(2))/(T_(2)^(2))=(12)/(10)=(6)/(5)` [because `g_(1)=10m//s^(2)`,`g_(2)=g+2=12m//s^(2)`]
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