Reading of a pendulum clock from the earth to the moon

[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

 

[Doc Al]

Answer these questions:
(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.)

 

[missashley]

i tried those answers and it says both were wrong

 

[Doc Al]

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

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.