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The acceleration dula to gravity on the...

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 ?

Text Solution

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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`
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