0

I'm trying to figure out how to calculate the wavelength of a sine function from a specified arc length and peak-to-peak amplitude.

I'm using this formula to generate a sine wave mould for a mechanical press, which needs to line up with other parts. Think of a sheet of graph paper being bent into the shape of a sine wave. How would I ensure the lines on the paper match up with the peaks and valleys of the sine wave?

I started with the basic formula for a sine wave: $$y=\frac A2 \sin \left( \frac {2\pi x}{\lambda} \right) $$

Then took the derivative: $$\frac {d}{dx} = \frac {\pi A \cos \left( \frac {2\pi x}{\lambda} \right) }{\lambda}$$

I applied this to the standard arc length formula: $$\int_0^\lambda \sqrt {1+ \left( \frac{d}{dx} \right) ^2} dx$$

$$\int_0^\lambda \sqrt {1+ \left( \frac{\pi A \cos \left( \frac{2 \pi X}{\lambda} \right) } {\lambda} \right) ^2} dx$$

Trying to solve this integral leads to things like elliptic integrals. Unfortunately, this is where my math skills end. How do I proceed from here?

SchrodingersCat
  • 24,924
Rachie
  • 1
  • Why is your integral from o to $\lambda$? Assume $x_1$ and $x_2$, have the integral for this range to get the length. You have two other equations for the amplitudes. I wonder if there is a unique solution. – Moti Jan 26 '16 at 21:10
  • I chose $0$ to $\theta$ because I'm interested in the arc length of a single period. I'm not sure what you mean by $x_1$ and $x_2$. – Rachie Jan 26 '16 at 22:42
  • There's no way to avoid elliptic functions here, but http://tinyurl.com/hcldrcf may help –  Jan 27 '16 at 02:53
  • A related question. – Lucian Jan 27 '16 at 03:04

0 Answers0