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 no

1.	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)
Answer» Solution :In FIRST case, `T_(1)=2PI sqrt(mug) ""...(1)` In second case, displacement, `y= KT^(2)` upward velocity `v=(dy)/(dt)=2kt` , upward acceleration, `a=(dv)/(dt)` `=2k=2xx1=2 ms^(-1)` `:.` Time period, `T_(2)=2pi sqrt((l)/(g+a))""...(2)` 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)`