The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.

Explanation

Here, `g_(m)=1.7ms^(-2), g_(e)=9.8ms^(-2), T_(m)=?, T_(e)=3.5s^(-1)`
As, `T_(e)=2pisqrt((l)/(g_(e)))and T_(m)=2pisqrt((l)/(g_(m)))`
`:. (T_(m))/(T_(e)) or T_(m)=T_(e)sqrt((g_(e))/(g_(m)))=3.5sqrt((9.8)/(1.7))=8.4s`