The acceleration due to gravity at a place is `pi^(2)m//s^(2)`. Then, the time period of a simple pendulum of length 1 m is

The periodic time of a simple pendulum of length 1 m and amlitude

The acceleration due to gravity on the surface of moon is 1.7 m s^(-2) . What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s^(-2) )

The acceleration due to gravity on the surface of moon is 1.7 m s^-2 . What isthe time period of a simple pendulum on the surface of moon if itstime period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 m s6-2 )

The acceleration due to gravity on the surface of moon is 1.7 ms^(-2) . What is the time period of a simple pendulum on the moon if its time period on the earth is 3.5 s ? (g on earth = 9.8 ms^(-2) )

Motion of a simple pendulum is an example for simple harmonic motion. The acceleration due to gravity on the surface of the moon is 1.7 m/s^2 .What is the time period of a simple pendulum on the moon,if its time period on the earth is 3.5 seconds ?

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)