The free-fall acceleration on the moon is 1.62 m/s2 . What is the length of a...
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? Lmoon= _____ m
Homework Answers
formua for time period of simple pendulum is given
by
T = 2pi sqrt(L/g)
for length 1.7 m
time period on earth is
T = 2* 3.14* sqrt(1.70/9.81)
T = 2.614 secs
now for this period on moon, the length Corresponds to
T = 2pi * sqrt(L/g)
sqrt(L/g) = 2.614/(2pi)
L/g = 0.416^2
L = 0.173056 * 1.62
L = 0.28 m or 28 cm --------<<<<<<<<<<Answer
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