A simple pendulum has time period `T_(1)`. Thetime period is changed to `T_(2)` when the point of suspension is moved upward according to the relation `y=2t^(2)` where y is the vertical displacemeent. The ratio of `(T_(1)^(2))/(T_(2)^(2))` is…………….`(g=10ms^(-2))`

Answer

`1.4`

Explanation

As `y=pt^(2)`
`:.(dy)/(dt)=2pt`
`implies(d^(2)y)/(dt^(2))=2p=2xx(2)=4ms^(-2)`
As `T=2pisqrt(L/g)impliesT^(2)=4pi^(2)L/g`
`implies(T_(1)^(.))/(T_(2)^(2))=(g_(2))/(g_(1))=(g+a)/g=14/10=7/5=1.4`