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Question

The free-fall acceleration on the moon is 1.62m/s21.62 \mathrm { m } / \mathrm { s } ^ { 2 }. What is the length of a pendulum whose period on the moon matches the period of a 2.00-m-long pendulum on the earth?

Solution

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On the moon and on the earth, respectively, the periods of the pendulum are expressed as:

Tearth=2πLearthgearthandTmoon=2πLmoongmoonT_{earth}=2\pi\sqrt{\dfrac{L_{earth}}{g_{earth}}}\:\:and\:\: T_{moon}=2\pi\sqrt{\dfrac{L_{moon}}{g_{moon}}}

Since we have Tearth=TmoonT_{earth}=T_{moon}

2πLearthgearth=2πLmoongmoonLmoon=(gmoongearth)Learth=(1.62m/s29.8m/s2)(2.00m)=33.1cm2\pi\sqrt{\dfrac{L_{earth}}{g_{earth}}}=2\pi\sqrt{\dfrac{L_{moon}}{g_{moon}}}\Rightarrow L_{moon}=\left(\dfrac{g_{moon}}{g_{earth}}\right)L_{earth}=\left(\dfrac{1.62\:m/s^2}{9.8\:m/s^2}\right)(2.00\:m)=33.1\:cm

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