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Question
A second’s pendulum is taken on the surface of the moon where the acceleration due to gravity is l/6th of that of earth. Will the time period of pendulum remain the same or increase or decrease? Give a reason.
Solution
We know the time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity
i.e. T α `1/sqrt("g")`
As acceleration due to gravity on the surface of the moon decreases as compared to that of the earth
∵ gmoon < gearth
∴ Tmoon > Tearth
⇒ The time period of the second’s pendulum increases when it is taken to the surface of the moon.
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