A simple pendulum has time period `T_(1)`. When the point of suspension moves vertically up according to the equation `y=kt^(2)` where `k=1m//s^(2)` and `'t'` is time then the time period of the pendulum is `T_(2)` then `(T_(1)//T_(2))^(2)` is

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

A simple pendulum has time period T_(1) / The point of suspension is now moved upward according to the realtion y = kt^(2)(k = 1 m//s^(2)) where y is vertical displacement, the time period now becomes T_(2) . The ratio of ((T_(1))/(T_(2)))^(2) is : (g = 10 m//s^(2))

A simple pendulum has a time period T_(1) . When its point of suspension is moved vertically upwards according as y=kt^(2) , where y is vertical covered and k=1 ms^(-2) , its time period becomes T_(2) then (T_(1)^(2))/(T_(2)^(2)) is (g =10 ms^(-2))

A simple pendulum has time period T_(1) . The point of suspension is now moved upward according to the relation y= kt^(2)(k=1 ms^(-2)) where y is the vertical diplacement. The time period now becomes T_(2) . What is the ration (T_(1)^(2))/(T_(2)^(2)) ? Given g=10 ms^(-2)