A simple pendulum has time period 'T_(1)'. The point of suspension is now moved upwards according to the relation y = kt^(2), (k = 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")

A simple pendulum has time period 'T_(1)'. The point of suspension is

1.	A simple pendulum has time period 'T_(1)'. The point of suspension is now moved upwards according to the relation y = kt^(2), (k = 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")
Answer» Solution :`y= kt^(2)=1//2at^(2)` acceleration `IMPLIES (1)/(2)a= k=1 implies a= 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)`