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Let a wire be shaped according to some even function $y=f(x)$, with $f'(0)=0$ and $f''(0)>0$, and let a bead of negligible size slide frictionlessly on the wire. Let the bead oscillate under the influence of gravity about $x=0$ with amplitude $A$ (i.e., between $x=-A$ and $x=+A$) and frequency $\omega(A)$. Clearly $\omega$ is nearly constant for small $A$; it differs from the frequency $\omega_o$ of simple harmonic motion by at most $O(A^2)$. By choosing $f$ to be a fourth-order polynomial, we could presumably adjust the wire's shape so as to eliminate the errors of order $A^2$ and make $\omega(A)$ constant up to $O(A^4)$. Possibly we could continue this process of approximation and make all the derivatives $d^n\omega/dA^n$ vanish up to some finite $n$, or maybe for all $n$.

If the derivatives can be made to vanish for all $n$, then I think $\omega$ would have to be nonanalytic at $x=0$. It seems impossible that there is any $f$ such that isochrony holds for arbitrarily large $A$. No matter how steep you make the sides, the bead can't do any better than accelerating downward with acceleration $g$. Therefore I think the best $f$ you can find is probably one that blows up to infinity at $|x|$ equal to some $x_{max}$. On dimensional grounds, we would have to have $x_{max}=cL$, where $c$ is a unitless constant and $L=g/\omega_0^2$.

So my multipart question is: (1) Is there a function $f$ that gives $d^n\omega/dA^n=0$ for all $n$? If so, ... (2) How is $f$ characterized, and what is $c$? (3) Is $\omega(A)$ analytic at $x=0$, and if so, what is its radius of convergence to its Taylor series in units of $L$?

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To (1), there is such a function. I read about it lang ago, Somewhere in connection with my hobby of pendulum (master) clocks. But I do not remember where. :=( –  Georg Aug 5 '11 at 19:09
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up vote 6 down vote accepted

Here we are: such a pendulum was developed centurys ago:

http://myreckonings.com/wordpress/2007/11/19/the-not-so-simple-pendulum/

The trajectory is a cycloid, a possible realisation is a kind of guide in form of the evolent of that cycloid close to the fixation point of the pendulum.

AfaIr, a similar cycloid is the fastest trajectory for a body (on a rail) in free fall.

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Cool, thanks for the answer! It's amazing that the tautochrone, the brachistochrone, and their evolutes are all cycloids! –  Ben Crowell Aug 5 '11 at 19:34
    
@Ben, In fact You gave the hint in the headline of the question: "isochronous". After some thinking somthing like isochronous pendulum resurfaced in my mind. –  Georg Aug 5 '11 at 19:46
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