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)` …(i)
Inecond case, displacement,` y=kt^(2)`
upward velocity, `v=(dy)/(dt)=2kt`
upwardd acceleration , `a=(dv)/(dt)`
`=2k=2xx1=2ms^(-2)`
`:.` Time period, `T_(2)=2pisqrt((l)/(g+a))` …(ii)
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)`