The free-fall acceleration on the moon is 1.62 m/s2. What is the length of a pendulum whose period on the moon matches the period of a 1.70-m-long pendulum on the earth?

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Answer 1 · 2017-11-20T07:32:15

The length of the pendulum is #=0.281m#

The period of a pendulum is

#T=2pisqrt(l/g)#

On earth, the length of the pendulum is

#l=1.7m#

The acceleration due to gravity on earth is #g=9.8ms^-2#

The period is

#T=2pisqrt(1.70/9.8)#

On the moon, the acceleration due to gravity is #g_1=1.62ms^-2#

Let the length of the pendulum on the moon be #l_1#

Therefore,

#2pisqrt(l_1/g_1)=2pisqrt(1.7/9.8)#

#l_1/(g_1) =1.7/9.8#

#l_1= 1.7/9.8*1.62=0.281m#