A pendulum clock that works perfectly on Earth is taken to the Moon. (a)Does it run fast or slow there? (b) if the clock is started at 12:00:00 am, what will it read after one Earth day(24hrs)? Assume that the free-fall acceleration on the Moon is 1.63 m/s^2

I tried to do this problem and i still cant figure it out..please would be great is the work was shown ..thanks in advance

"it would be great if the work was shown .."
It sure would.
The period of a pendulm is
P = 2 pi sqrt (L/g)
L is the length of the pendulum and g is the acceleration of gravity.
On the moon, g is 1.63/9.81 = 0.166 as high as it is on earth. Ask yourself How long it would take to register 24 hours on the moon. Pendulum clocks work by counting the number of swings.

And please show your work if you need further help.

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asked by steph
  1. 9g of the moon is .166*g on earth.
    T pendulum is 2*pi(L/g)^.5
    Because g of the moon is .166*g on earth, T moon is 2*pi(L/.166g)^.5
    This goes to T moon= (1/.166)^.5*T on earth.
    T on earth for 24 hours is 24 hours. Convert that to seconds and multiply by (1/.166)^.5
    Your answer will be the seconds that pass on the moon. From there you can find hours, minutes, and second elapsed.

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    posted by Fig
  2. There should be no 9 at the beginning of my previous post. Sorry.

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    posted by Fig

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