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

Text Solution

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Let l be the length of the simple pendulum and `g_(e)` and `g_(m)` the acceleration due to gravity on the earth and on the moon respectively. Then the time periods of the pendulum on the earth and on moon are given by :
`T_(e)=2pisqrt((l)/(g_(e)))" and "T_(m)=2pisqrt((l)/(g_(m)))`
Dividing, we get
`(T_(m))/(T_(e))=sqrt((g_(e))/(g_(m)))`
`T_(m)=T_(e)sqrt((g_(e))/(g_(m)))`
`T_(m)=3.5xxsqrt((9.8)/(1.7))`
`=3.5xx2.4=8.38s`.
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