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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?

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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

Sum
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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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Chapter 5: Oscillations - Short Answer I

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