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

On earth time period
`t=2pisqrt((l)/(g))`
`3.5=2pisqrt((l)/(9.8))" "...(1)`
`:.` Acceleration due to gravity of the moon's is `1.7ms^(-2)`
`T=2pisqrt((l)/(1.7))" "...(2)`
Dividing (2) by (1) ,
Dividing `(T)/(3.5)=2pisqrt((1)/(1.7)),//2pisqrt((l)/(9.8))`
`T=sqrt((9.8)/(1.7))xx3.5=8.4s`