A simple pendulum has time period T_(1). The point of suspension is now moved upwards according to the relation =kt^(2),Ik=1m//sec^(2)) where y is the vertical displacement. The time period now becomes T_(2) then find the ratio of (T_(1)^(2))/(T_(2)^(2)) (g=10m//sec^(2))

A simple pendulum has time period T_(1). The point of suspension is no

1.	A simple pendulum has time period T_(1). The point of suspension is now moved upwards according to the relation =kt^(2),Ik=1m//sec^(2)) where y is the vertical displacement. The time period now becomes T_(2) then find the ratio of (T_(1)^(2))/(T_(2)^(2)) (g=10m//sec^(2))
Answer» SOLUTION :`y-kt^(2)/1//2at^(2)` ACCELERATION `implies1/2a=k=1""impliesa=2m//sec^(2)` `T_(1)=2pisqrt(l/G)` and `T_(2)=2pisqrt(l/(g+a))` `(T_(1)^(2))/(T_(2)^(2))=(g+a)/g=(10+2)/10=6/5`