# Time When Pendulum Clock is Put on Moon?

1. Nov 25, 2007

### 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?

Last edited: Nov 25, 2007
2. Nov 25, 2007

### 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

Last edited: Nov 25, 2007
3. Nov 25, 2007

### 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.

4. Nov 25, 2007

### 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?

5. Nov 25, 2007

### 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?

6. Nov 25, 2007

### 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?

7. Nov 25, 2007

### 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

8. Nov 25, 2007

### petern

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