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Question
Calculate the time period of a simple pendulum of length 1.44 m on the surface of the moon. The acceleration due to gravity on the surface of the moon is 1/6 the acceleration due to gravity on earth, [g = 9.8 ms−2]
Solution
Length of simple pendulum = l = 1.44 m
Time period (T) =?
Acceleration due to gravity on the surface of the moon
= g′ = `1/6xx"g"`
g = `9.8/6`
T = `2πsqrt(1/"g′")`
T = `2xx22/7xxsqrt((1.44xx6)/9.8)`
T = `14/7xx0.9389`
T = 5.90 s
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