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

Let g and `g^(')` be the acceleration due to gravity on the surface of earth and moon respectively.
On earth, `T=2pisqrt((l)/(g))`
On moon `T^(')=2pisqrt((l)/(g^(')))`
`(T^('))/(T)=sqrt((g)/(g^('))) or T^(')=Tsqrt((g)/(g^(')))=3.5 sqrt((9.8)/(1.7))=8.4s`