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Question

# The free-fall acceleration on the moon is $1.62 \mathrm { m } / \mathrm { s } ^ { 2 }$. 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:

$T_{earth}=2\pi\sqrt{\dfrac{L_{earth}}{g_{earth}}}\:\:and\:\: T_{moon}=2\pi\sqrt{\dfrac{L_{moon}}{g_{moon}}}$

Since we have $T_{earth}=T_{moon}$

$2\pi\sqrt{\dfrac{L_{earth}}{g_{earth}}}=2\pi\sqrt{\dfrac{L_{moon}}{g_{moon}}}\Rightarrow L_{moon}=\left(\dfrac{g_{moon}}{g_{earth}}\right)L_{earth}=\left(\dfrac{1.62\:m/s^2}{9.8\:m/s^2}\right)(2.00\:m)=33.1\:cm$

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