Theacceleration dula to gravity on the surface of moon is1.7ms^(-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 ?

Theacceleration dula to gravity on the surface of moon is1.7ms^(-2) .

1.	Theacceleration dula to gravity on the surface of moon is1.7ms^(-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 ?
Answer» Solution :For the moon: `g_(m)=1.7ms^(-2),T_(m)=?` For the EARTH: `g_(E)=9.8ms^(-2),T_(e)=3.5s` But, `T_(e)=2pisqrt((l)/(g_(e)))andT_(m)=2pisqrt((l)/(g_(m)))` `(T_(m))/(T_(e))=SQRT((g_(e))/(g_(m)))` `T_(m)=sqrt((g_(e))/(g_(m)))xxT_(e)=sqrt((9.8)/(1.7))xx3.5` `T_(m)=8.4sec`