(a) Your pendulum clock which advances 1.0 s for every complete oscillation of t
ID: 1475872 • Letter: #
Question
(a) Your pendulum clock which advances 1.0 s for every complete oscillation of the pendulum is running perfectly on Earth at a site where the magnitude of the acceleration due to gravity is 9.80 m/s2. You send the clock to a location on the Moon where the magnitude of the acceleration due to gravity is 1.65 m/s2. (b) If the clock is started at 12:00 noon, what will it read after a time of 11.9 h has elapsed on Earth? Enter the time to the nearest minute. PM I put 4:53 PM as answer. Program says hour is right but minute is wrong. Thank you in advance.
Explanation / Answer
The Time period of a pendulum is proportional to sqrt(1/g).
Let T be the period of the pendulum on earth, and t be the period on the moon.
Then, T/t = sqrt(1.65 / 9.8) = 0.409
Hence in 11.9 hours, the clock on moon will advance 11.9*0.409 = 4.8671hours which is equal to 4 hours and 52.026. Hence answer is 4:52 P.M
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