In the previous problem if `t` is the time period of rotation
(i) `t=2pisqrt((L)/(g))`
(ii) `t=2pisqrt((Lcostheta)/(g))`
(iii) `T=(4pi^(2)mL)/(t^(2))`
(iv) The ball is in equilibrium

The number of variables in the formula T=2pisqrt((l)/(g)) is _____

What are the variables in the formula T=2pisqrt((l)/(g)) ?

How many times of relative error in I is the relative error in T, when T = 2pisqrt(l//g)

The time period in case of a conical pendulum is given T=2pisqrt-

The acceleration due to gravity on the surface of earth is 9.8 ms^(-2) .Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)

The acceleration due to gravity on the surface of earth is 9.8 ms^(-2) .Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)

The time period of an oscillating body is given by T=2pisqrt((m)/(adg)) . What is the force equation for this body?

The equation T=2pisqrt(l//g) is valid only when