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`)

Explanation

We are given :
Acceleration due to gravity on moon, `g_(m)=1.7" m s"^(-2)`
Time period on earth, `T_(e)=3.5` s.
We know that, `t prop(1)/(sqrt(g))`.
This gives `(T_(m))/(T_(e))=sqrt((g_(e))/(g_(m)))rArr(T_(m))/(3.5)=sqrt((9.8)/(1.7))rArrT_(m)=8.4s`.