Why not any other shape to represent a single frequency (like triangular waves)?

I haven't been able to find a glorious "easy to understand answer", but I did find that sinusoidal displacement of a particle produces a sinusoidal velocity and acceleration. Could this explain why this is the ideal or natural shape of a motion and waves in nature, as well as it's use in analysis techniques like Fourier transforms?

  • $\begingroup$ The answer is much simpler. When you look at a wheel that rotates at a constant speed, a coordinate of any point on the wheel changes in time as a sine. So frequency is a projection or rotation. You can rotate a wheel only with one constant speed at a time. Therefore a sine represents only one frequency. $\endgroup$ – safesphere Oct 17 '17 at 4:19
  • $\begingroup$ @safesphere And yet, in the dispersive systems you discuss below a sinusoidal driving influence will result in a mixture of frequencies arriving at the detector at different times. And the question is about waves after all, not about circular motion. (For all that the 'reference circle' is a useful way to teach harmonic waves without calculus; but then in non-linear system harmonic waves are—as you said—not special). $\endgroup$ – dmckee Oct 17 '17 at 4:25

Sinusoids (sines and cosines) are the eigenfunctions of the wave equation. That is if you look for a set of functions $f_{i,\omega}$ for which it is true that $$ \frac{\partial^2 f_{i,\omega}}{\partial x} \pm \frac{1}{c^2} \frac{\partial^2 f_{i,\omega}}{\partial t} = \lambda f_{i,\omega} \;, $$ for some real number $\lambda$, then all the answers would be some combinations of $\sin(kx \mp \omega t)$ and $\cos (kx \mp \omega t)$.1

This is a pretty special property. For any other periodic functions (such as triangle waves or square waves) there are no qualifying solutions at all. However, you can write all those non-qualifying solutions as a sum of the ones that do qualify. Letting $F$ be a stand in for any non-qualifying periodic solution, then $$ F = \sum_i \int \mathrm{d}\omega\, c_{i,\omega} \, f_{i,\omega} \;, $$ for some selected set of coefficient $c_{i,\omega}$s.2

So, long and mathematical story short: sinusoidal waves have the special properties of having unique frequency and a being usable to compose all periodic functions.

1 I included the $i$ subscript on $f_{i,\omega}$ so that we could distinguish $\sin$ and $\cos$. It turns out that $\lambda = k^2 - \frac{\omega^2}{c^2}$, and that by choosing $k$ (for a given $\omega$) so that $\lambda = 0$ you force the wave speed to be $c$ and have a description of a physical wave that matches the conditions of your apparatus.

2 There is even a way to find the correct $c_{i,\omega}$s from $F$, but writing it down here wouldn't actually explain any more about why the sinusoids are special.

  • $\begingroup$ This equation is only true for a linear wave and thus hardly answers the question. The special property of its solution is due to the linearity of the equation, not to the uniqueness of the sine function. You can do a Fourier-like transform using different periodic functions, which are equally capable to compose all periodic functions. This answer doesn't explain why a sine represent a single frequency. $\endgroup$ – safesphere Oct 17 '17 at 4:12
  • $\begingroup$ @safesphere This is certainly specific to the plain vanilla "wave equation" but the line of argument extends to other wave equations as well: the eigenfunctions will still be special. And indeed they will form the basis of the 'Fourier-like transform' that you allude to up to and including representing the dispersive (or non-dispersive) behavior of compound solutions in those cases as well. An introductory course in Schrödinger mechanics will put students through the paces on this for one non-linear wave equation. $\endgroup$ – dmckee Oct 17 '17 at 4:20
  • $\begingroup$ None of this explains why sine waves are called "pure" tones. IMO that is question about human psycho-acoustics, and has nothing to do with physics or mathematics. For example humans can hear a single tone with apparent frequency $f$ when the actual periodic waveform has no Fourier component at all with frequency $f$! $\endgroup$ – alephzero Oct 17 '17 at 4:39
  • $\begingroup$ @alephzero I certainly haven't addressed the matter of human sensory response here, but our inner ears do read the incoming signal in some approximation of frequency space. It's just that there are edge cases that mess up the interpretation of the data. (And I've seen the demo and it is weird. As is the one where you can't hear the 30 Hz tone until a 90 Hz signal is mixed in.) Feel free to write up the sensory response end: I'd be out of my depth. $\endgroup$ – dmckee Oct 17 '17 at 4:46
  • $\begingroup$ @alephzero For my question, I was looking for a maths/physics explanation of the "perfect waveform" and less of the human interpretation of it's sound. dmckee answered my question perfectly (though I wasn't expecting such beautiful maths in a physics forum). Your contribution on the human psycho-acoustics bit is quite interesting though, so I'll be researching that later :D $\endgroup$ – Sonny6155 Oct 17 '17 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.