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

`g_(m) = 1.7 ms^(-2)`
`T_(m) = ?`
`g_(e ) = 9.8 ms^(-2)`
`T_(e) = 3.5 s`
As `T_(e ) = 2pi sqrt((l)/(g_(e)))`
`T_(m) = 2pi sqrt((l)/(g_(m)))`
`(T_(e))/(T_(m)) = sqrt((g_(m))/(g_(e)))`
`T_(m) = T_(e)sqrt((g_(e))/(g_(m)))`
` = 3.5 sqrt((9.8)/(1.7)) = 8.4s`