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 ?
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Hello and welcome to math.stackexchange! This is an interesting question, thank you. Where is it coming from? – Hans Engler Aug 17 '16 at 13:10
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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 $\frac{2\pi}k=6.710259\dots$
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