3
$\begingroup$

The propagation of high frequencies sound waves is more directional (specular), and they don't diffract as much as low frequencies. Low-frequencies diffract and thus propagate in a more omni-spherical fashion. This, to my knowledge, applies to all waves, not only sound ones.

Now trying to imagine the air particles hitting one another at low/high frequencies doesn't really unlock why this happens. I am aware that deep understanding of wave theory will provide the answer.

But is there a simple, visual explanation to why this happens? And if not, is there at least some simple mathematical equation that can provide the answer?

$\endgroup$

2 Answers 2

3
$\begingroup$

Diffraction angle on the body of size $L$ is about $\lambda/L$ thus it is bigger for LF waves.

Another way to think about it is to remember that $\lambda \to 0$ case is a geometric optic with no diffraction at all, and the bigger the wavelength is, the further we away from the geometric optic to meet diffraction, etc.

PS. I am not sure the picture of particles hitting each other plays well here because of the ideal acoustic targets cases with wavelength much bigger than mean free path in order to arrive to the dissipation-free wave equation we use to describe the diffraction.

$\endgroup$
2
  • $\begingroup$ This is a fine answer, but would be nice to see an explanation as to why it happens? Also, $L$ implies that this works with bodies, but what about the propagation in air with no obstacles? $\endgroup$
    – Izhaki
    Commented Feb 19, 2013 at 9:58
  • $\begingroup$ I guess we will always have at least a size of the source (with corresponding angles). To show that the source is important, one can take HF case, increase the full system (including wavelengths), and obtain exactly the same angles, but the bigger scales. There will be no difference in 'sphericality'. If HF case is less spherical, than LF, go upstream to the source to discover different $\lambda/L$ ratio; and we will always have an implicit scale we compare with. Even HF/LF means small/big $\lambda$, but small with respect to what? With respect to the (unstated) $L$. $\endgroup$ Commented Feb 24, 2013 at 6:16
0
$\begingroup$

OK, I believe that a simple explanation can be based on Huygen's Principle (the animation on this page was particularly helpful).

If we take the simple case of 2D propagation, the object resonating can be looked at as a line. As the line moves back and forth, wavelets are created along its length. The wavefronts moving forward reinforce to create new wavefronts; but the wavefronts moving sidewise cancel out. So based on this:

Waves do travel in all directions regardless of the frequency, it's just that side motion for higher frequencies results in phase cancelations (and thus reduced pressure changes).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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