Can sound waves be cancelled out with a perfect mirror? Can this be applied to active noise cancelling headphones to make it perfect?
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$\begingroup$ What does perfect mirror mean in this context? $\endgroup$– nicoguaroCommented Sep 2, 2021 at 15:58
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$\begingroup$ @nicoguaro I think what the OP means is if the sound wave bounces back from a surface perfectly, shifting it's phase by 180 deg and superposition would then give zero. $\endgroup$– NeuroEngCommented Sep 2, 2021 at 16:03
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1$\begingroup$ Oh, I see. In that case, it would work for waves that impinge perpendicularly to the surface but that's it. $\endgroup$– nicoguaroCommented Sep 2, 2021 at 16:06
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1$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented Sep 2, 2021 at 16:09
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1 Answer
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Yes they can.... at only a finite set of frequencies ($\omega=ck$) such that the two paths to the microphone (or ear) are 180 degrees ($\pi$ radians) out of phase.
If the path length difference (plus the $\pi$ upon reflection) add to $\pi$ (modulo $2\pi$):
$$k\Delta L + \pi = \pi\,\ {\rm mod} \, 2\pi$$
or
$$ k = \frac{2\pi n}{\Delta L}\ \ \ \ n\in(0,1,2,\cdots) $$
where $k=\lambda/2\pi$ is the wave number.
Hence:
$$ \lambda = \frac{\Delta L}n $$
Now if
$$ \lambda = \frac{\Delta L}{n+\frac 1 2} $$
the amplitude should be doubled.
(Note: I assumed "perfect mirror" means the reflected amplitude exactly equals the input amplitude...but is that even possible? A parabolic mirror will amplify the reflect sound, a flat mirror could disperse it. There is a lot to consider.)
