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)`

Explanation

` 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)`