Is there a formula relating wavelength to amplitude for a sinusoidal curve of fixed line length I have a piece of string of fixed length 'x'.  It is possible to lay it down in a set of sinusoidal waves of varying amplitude.  The resultant wavelength will be a function of the amplitude.  The maximum amplitude will be x/4, the minimum will be zero.  For a given amplitude, is there a formula that defines the wavelength ?
 A: Treat the sine function as the parametric equation $x=t,y=a\sin kt$; this has amplitude a and wavelength $\frac{2\pi}k$. The arc length of one whole cycle is given by
$$L=\int_0^{2\pi/k}\sqrt{x'(t)^2+y'(t)^2}\ dt$$
$$=\int_0^{2\pi/k}\sqrt{1+(ka\cos kt)^2}\ dt$$
Make the substitution $u=kt$, with $\frac{du}{dt}=k$:
$$=\frac1k\int_0^{2\pi}\sqrt{1+(ka)^2\cos^2(u)}\ du$$
Because of the symmetries of the squared cosine function, where $\cos^2 x=\cos^2(\pi+x)=\cos^2(\pi-x)$, the integral can be divided by four:
$$=\frac4k\int_0^{\pi/2}\sqrt{1+(ka)^2\cos^2u}\ du$$
$$=\frac4k\int_0^{\pi/2}\sqrt{1+(ka)^2-(ka)^2\sin^2u}\ du$$
$$=\frac4k\sqrt{1+(ka)^2}\int_0^{\pi/2}\sqrt{1-\frac{(ka)^2}{1+(ka)^2}\sin^2u}\ du$$
This last integral is in the form of the complete elliptic integral of the second kind, and yields our final result:
$$L=\frac4k\sqrt{1+(ka)^2}E\left(\sqrt{\frac{(ka)^2}{1+(ka)^2}}\right)$$
The question now asks for $\frac{2\pi}k$ given L and a, but the relation derived above has no closed-form solution for k; iterative numerical methods must be used. For example, with $a=1$ and $L=8$, plonking the relation into Wolfram Alpha yields $k=0.936355\dots$, whence the wavelength $\frac{2\pi}k=6.710259\dots$
