Reading of a pendulum clock from the earth to the moon

1. Jan 28, 2008

missashley

A certain pendulum clock that works perfectly on Earth is taken to the moon, where g = 1.62 m/s^2. Acceleration of gravity is 9.81 m/s^2 on Earth. THe clock is started at 12:00:00 AM and runs for 22 h.

What will be the reading for the hours on the moon? answer in h

T = 1 second
g = 9.81
T = 2pi * square root of length/g or rearrange into g (T/2pi)^2 = l
l = 0.248

g = 1.63
T = 2pi * square root of l/1.63
T = 2.453

22h * 3600s = 79200 sec
79200/2.453 = 32286.99552
32286.99552/3600 = 8.9686 hours

i also tried

22*3600 = 79200s
79200 = 2pi * square root of l/9.81
l = 1557100708
T = 2pi *square root of 1557100708/1.63
T = 194197.8499s
194197.8499s / 3600 = 53.9438472 hr

Last edited: Jan 28, 2008
2. Jan 28, 2008

Staff: Mentor

(1) Does the clock run slower or faster on the moon?
(2) What's the ratio of the moon period to the earth period?

(One of your answers is correct, but you did more work than needed.)

3. Jan 28, 2008

missashley

i tried those answers and it says both were wrong

4. Jan 29, 2008

Staff: Mentor

It's not clear what format the answer must have. Sounds like X hours, like you have. But it could also mean the time, as in 5:17:34 PM.

In any case, the ratio of periods is:

$$T_m/T_e = \sqrt{\ell / g_m} / \sqrt{\ell / g_e} = \sqrt{g_e / g_m}$$

Thus $T_m = 2.46 T_e$, so the clock on the moon is 2.46 times slower than the clock on earth.