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Question
The acceleration due to gravity on the surface of the moon is 1.7 m/s2. What is the time period of a simple pendulum on the surface of the moon if its time period on the surface of the earth is 3.5 s? (g on the surface of earth = 9.8 m/s2)
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Solution
Given: gm = 1.7 m/s2, gE = 9.8 m/s2, TE = 3.5 s
To find: Time period on the surface of the moon (Tm)
Formula: `"T" = 2pisqrt("L"/"g")`
Calculation:
From formula,
`"T"_"m"/"T"_"E" = sqrt("g"_"E"/"g"_"m")`
∴ Tm = `sqrt(9.8/1.7) xx 3.5`
= 8.40 s
The time period of a simple pendulum on the surface of the moon is 8.40 s.
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