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 ms^(-2))` where y is the vertical diplacement. The time period now becomes `T_(2)`. What is the ration `(T_(1)^(2))/(T_(2)^(2))` ? Given `g=10 ms^(-2)`

Explanation

In first case,
`T_(1)=2pi sqrt(mug) " "...(1)`
In second case, displacement,
`y= kt^(2)`
upward velocity
`v=(dy)/(dt)=2kt` ,
upward acceleration,
`a=(dv)/(dt)`
`=2k=2xx1=2 ms^(-1)`
`:.` Time period,
`T_(2)=2pi sqrt((l)/(g+a))" "...(2)`
Hence `(T_(1)^(2))/(T_(2)^(2))=(4pi^(2)l)/(g)xx((g+a))/(4pi^(2)l)=(g+a)/(g)`
`=(10+2)/(10)=(6)/(5)`