The acceleration dula to gravity on the surface of moon is `1.7ms^(-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 3.5s ?

Explanation

(a) A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.
(b) For the moon : `g_(m) = 1.7 ms^(-2) , T_(m)` = ?
For the Earth : `g_(e) = 9.8 ms^(-2), T_(e)` = 3.5 s
But `T_(e) = 2 pi sqrt((l)/(g_(e)))` and `T_(m) = 2pi sqrt((l)/(g_(m)))`
`(T_(m))/(T_(e)) = sqrt((g_(e))/(g_(m)))`
`T_(m) = sqrt((g_(e))/(g_(m))) xx T_(e) = sqrt((9.8)/(1.7)) xx 3.5`
`T_(m) =` 8.4 sec
(c) Yes. Because the working of a wrist watch does not depend on gravity at that place but depends on spring action.