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) = 2pisqrt((L)/(g_(m)))`
`(T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e))`
`g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)= ( 3.5 ^(2))/( 8.4^(2)) xx 9. 8 = 1.7 ms^(-1)`