Time When Pendulum Clock is Put on Moon?

[petern]

A pendulum clock that works perfectly on Earth is taken to the moon where g = 1.63 m/s^2. If the clock is started at 12:00 A.M., what will it read on the moon 24 Earth hours?

The equation you would probably use is T = 2(pi)(sq. root of m/k)

Can someone get me started?

 

[rock.freak667]

The pendulum clock works on SHM...how is angular velocity affected by the value of 'g' with a pendulum?

Use T=2\pi\sqrt{\frac{l}{g}}are you allowed to use the value of g on earth? if so consider how you can express the acc.due to gravity on the moon in terms of 'g' on earth

 

[petern]

I don't really understand but I know that since gravity is less on the moon, the pendulum doesn't swing farther out than on Earth and therefore the clock is slower than on earth.

 

[petern]

I would find the ratio so on Earth but plugging g on Earth and on the moon into T = 2(pi)(sq. root of m/k). The ratio I got for moon to Earth was .783 to .319 or 1 to 2.45455.

So 24 h x 2.45455 = 58.9092 h. So the answer would be 10:55 A.M. on the third day. Did I do this problem correctly?

 

[DaveC426913]

Is it possible that this is a trick question? Doesn't a pendulum clock tell time on Earth by way of the Earth's rotation? Wouldn't a pendulum clock on the Moon depend on the Moon's rotation? And isn't the swing speed completely irrelevant?

 

[petern]

I don't think it is. My physics class isn't that advanced and we usually never have any trick questions. Thanks for the input anyways. Do you think I'm doing it right?

 

[rock.freak667]

well the way I intended to do it was the method with finding the ratio...so i think it would be correct...not too sure if you have to take into account the rotation of the moon and such

 

[petern]

I think I did it right because my teacher has gone over anything related to the rotation so thanks! Problem solved!