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A simple pendulum has time period T(1) T...

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 vertical displacement, the time period now becomes `T_(2)`. The ratio of `((T_(1))/(T_(2)))^(2)` is : `(g = 10 m//s^(2))`

Text Solution

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`y-kt^(2)/1//2at^(2)` acceleration
`implies1/2a=k=1" "impliesa=2m//sec^(2)`
`T_(1)=2pisqrt(l/g)` and `T_(2)=2pisqrt(l/(g+a))`
`(T_(1)^(2))/(T_(2)^(2))=(g+a)/g=(10+2)/10=6/5`
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