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- Thread starter whozum
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FredGarvin

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HallsofIvy

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By the way, a vacuum would not be enough for a "perpetual motion" clock- there would still be friction in the gears and other operating parts. I have a grandfather clock sitting about 10 feet from me right now that is rubbing somewhere and I can't find it! I wonder if a good hard kick would help?

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From there, is it not possible to create a detective system (probably optic) to sense the completion of a period and link that to a clock?

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chroot

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In the absence of any kind of friction, it would never lose energy, so yes, in theory. Of course, no real pendulum can ever be totally free of all friction, so no, in practice.whozum said:Slightly deviating: Provided a simple pendulum in vacuum, would it swing forever? Close to?

- Warren

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My question is in a practical sense, are those small enough to neglect for a certain time (that is reasonable for a clock of such delicate construction) before recalibration? I'm talking at least a week or so. We can attribute the error in the correct time to the energy lost in friction.

But even if some energy is lost, is the period of a pendulum not the same regardless of the displacement from equilibrium? If not, perhaps we could adjust our detector to compensate for such effects?

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chroot

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Precisely.whozum said:Im saying putting it in a vacuum, as close a vacuum as possible. There will be an extremely small bit of friction from possibly stray air molecules, and some from the junction of string and ceiling.

Sure, you can spend some engineering effort to design a pendulum support that will allow the pendulum to lose only a specified amount of its energy in a specified time. For example, you could design a pendulum that would only lose 1% of its energy to friction in a week, but that design will be more difficult than one that loses 10% of its energy in a week. There is, however, no practical way to design a pendulum that never loses any energy.My question is in a practical sense, are those small enough to neglect for a certain time (that is reasonable for a clock of such delicate construction) before recalibration? I'm talking at least a week or so. We can attribute the error in the correct time to the energy lost in friction.

The period does not stayBut even if some energy is lost, is the period of a pendulum not the same regardless of the displacement from equilibrium? If not, perhaps we could adjust our detector to compensate for such effects?

[tex]\sin \theta \approx \theta \,\, \textrm{for small angles}[/tex]

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brewnog

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Whozum, air resistance is not really a factor in determining the accuracy of a mechanical clock. It would be pretty pointless to use a string (rather than a bar of some kind) as a pendulum, but even so, mechanical resistances are largely the cause of energy losses, traditionally accounted for by using a falling weight.

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For an ideal mathematical pendulum (meaning no friction),it can be shown that the period of oscillations,though depending of amplitude for big angles,is independent of time.That means the oscillations are tautochrone...

One of favorite formulas is this

[tex] T_{pendulum} (\alpha,l) =4\sqrt{\frac{l}{g}} \int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1-\kappa^{2} \sin^{2}\varphi}} [/tex]

,where

[tex] \kappa^{2} = \sin^{2}\frac{\alpha}{2} [/tex],where [itex] \alpha [/itex] is the angular amplitude of oscillation.

Note that the modulus of the complete elliptic integral of the first kind is always less than unity which can be used for a series expansion and a proof of the intimate connection between elliptic integrals and Gauss hypergeometric functions.

Incidentally

[tex] T_{pendulum} (\alpha,l) =2\pi \sqrt{\frac{l}{g}} \ _{2}F_{1}\left(\frac{1}{2};\frac{1}{2};1;\kappa^{2}\right) [/tex]

Daniel.

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Bibliography:

Mocioaca,Ghe.,*Functii speciale*,Ed.Tehnica,Bucuresti,1983.

One of favorite formulas is this

[tex] T_{pendulum} (\alpha,l) =4\sqrt{\frac{l}{g}} \int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1-\kappa^{2} \sin^{2}\varphi}} [/tex]

,where

[tex] \kappa^{2} = \sin^{2}\frac{\alpha}{2} [/tex],where [itex] \alpha [/itex] is the angular amplitude of oscillation.

Note that the modulus of the complete elliptic integral of the first kind is always less than unity which can be used for a series expansion and a proof of the intimate connection between elliptic integrals and Gauss hypergeometric functions.

Incidentally

[tex] T_{pendulum} (\alpha,l) =2\pi \sqrt{\frac{l}{g}} \ _{2}F_{1}\left(\frac{1}{2};\frac{1}{2};1;\kappa^{2}\right) [/tex]

Daniel.

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Bibliography:

Mocioaca,Ghe.,

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FredGarvin

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Here you go Brewnog...The anniversary clock has a rotating pendulum. Those are the ones I was referring to.

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brewnog

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Cheers for that Mr G, I like the cut of your jib. But not a grandfather clock...FredGarvin said:Here you go Brewnog...The anniversary clock has a rotating pendulum. Those aer the ones I was referring to.

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