Home
Class 11
PHYSICS
[" 5.A simple pendulum has time period "...

[" 5.A simple pendulum has time period "T_(1)" .The point of "],[" suspension is now moved upward according to the relation "],[y=kt^(2),(k=1m/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 "],[[" (Take,"g=10m/s^(2)),(2005,2M)],[" (a) "6/5," (b) "5/6]]

Promotional Banner

Similar Questions

Explore conceptually related problems

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)

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 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) .

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) .