A simple pendulum has time period ' T ₁ '. The point of suspension is now moved upward according to equation y = kt ^2 where k =1 ~m / s ^2 . If new time period is ' T ₂ ' then T ₁^2 T ₂^2 will be ( g =10 ~m / s ^2 )

Step-by-step solution

Time period of a simple pendulum: T = 2 L g , with length L and gravitational acceleration g . At rest: g = 10 ; m/s ^2 . Squaring the period gives T₁^2 = 4 ^2 L g . Point of suspension moving upward with y = kt^2 : k = 1 ; m/s ^2 . Acceleration is a = d^2y dt^2 = 2k = 2 ; m/s ^2 . Effective gravity becomes g_ eff = g + a = 10 + 2 = 12 ; m/s ^2 , so T₂^2 = 4 ^2 L g_ eff . Ratio of squared periods: T₁^2 T₂^2 = g_ eff g = 12 10 = 6 5 .