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)`.

Explanation

In first case
`T_(1)2pisqrt((l)/(g'))`
in second case, displacement `y=kt^(2)`
upward velocity , `v=(dy)/(dt)=2kt.`
upward acceleration `a=(d^(2)y)/(dt^(2))=2k`
`k=2xx1ms^(-2)=2ms^(-2) " "[:'k=1ms^(-2)]`
`T_(2)=2pisqrt((l)/(g+a))=2pisqrt((l)/(g+a))`
hence` (T_(1)^(2))/(T_(2)^(2))=(4pi^(2)l)/(g)xx(g+2)/(4pi^(2)l)=(g+2)/(g)=(10+2)/(10)=6/5`