# Is there a point of no hearing if two speakers are emitting the same sounds - Wave Superposition

I'm talking bout this phenomenon (The third one):

I would like to know if with sound waves, happen the same as with (water) waves.

So, If two speakers were emitting the same sound, would there technically be a point where you wouldn't hear. (I guess this would only happen if you heard with only one ear).

Is this true for sound wave too?

Does this happen in real life, say, in car?

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Since you've gotten one 'yes' answer, let me give you a 'no' answer. :-) More precisely, there are two different questions here: Brian's 'yes' answer handles the question 'for any sound wave, is there some other wave that will cancel that wave out?' The question you asked, on the other hand, was closer to 'for any sound wave, is there some shift of that wave that will cancel the original out' - in other words, does every wave have a set of peaks and troughs that can be shifted in time to cancel it out? In general, the answer is no: there are wave functions for which no shift can possibly cancel out the function.

What does it mean to be self-cancellable? It means there's some shift of your wavefunction $f(t)$ - say, by $t_0$ - such that the sum of the two equals zero: $f(t)+f(t-t_0) = 0$. But then, this means that $f(t) = -f(t-t_0)$, and $f(t-t_0) = -f(t-t0-t0) = -f(t-2t_0)$ - so $f(t) = -f(t-t_0) = -(-f(t-2t_0)) = f(t-2t_0)$ - in other words, $f(t)$ is periodic with period $2t_0$. So since every self-cancellable wavefunction is periodic, no non-periodic function can be self-cancellable; as a simple example, any glissando (rising tone) won't be self-cancellable.

On the other hand, being periodic isn't sufficient to be self-cancellable, either. The simplest example is probably the Sawtooth wave: $f(t) = t - \lfloor t\rfloor$. Note that this function has derivative (i.e. slope) 1 everywhere its derivative is defined, and so every shift of it (by, say, $t_0$) also has derivative 1 everywhere its derivative is defined. But then $f(t) + f(t-t_0)$ has derivative 2 everywhere it's defined, whereas to cancel out they would need to add to the zero function - whose derivative is also zero.

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Very interesting answer, but I'm having some trouble following (No expert). What I did not get is why some functions can't be self cancelable (couldn't quite follow the equations). Shouldn't each wave have it's "anti-wave" even if it's a rising tone, by any chance do you have any visual representation of this phenomenon? (helps too see sometimes). Thank you very much for your answer! – Trufa Feb 3 '11 at 0:56
Unfortunately, I'm lousy at drawing pictures. :-) You're right that each wave has an 'anti-wave' - some other wave that you could add to it to come up with cancellation; that, just like Brian said, is how noise-cancelling headphones worked. But what you asked was 'for two speakers playing the same sound, is there a point where you wouldn't hear?' If two speakers are playing the same sound, then they're going to send the exact same wavefunction out; the only thing that changing your position can do then is to shift the time difference between them. – Steven Stadnicki Feb 3 '11 at 1:00
For instance, if you're 10m from one speaker and 20m from another, then you would have a delay of about .03 seconds between the sound from one speaker and the sound from the other. That's why I talk about self-cancellation specifically. Imagine taking the graph of the sawtooth and sliding it left and right along itself to find some way of 'cancelling' the notches; you can see that this is impossible, whereas doing it for (for instance) a square wave or a sine wave is easy. – Steven Stadnicki Feb 3 '11 at 1:03
@Steven (sorry for the delay I got no notification) Ok I think I "got it" now with the moving around of the graph, (I'm and image kind of graph). So then in a car (emitting the same sounds) is a definite no go. If the waves are no inverse, there is no way they could cancel each other, this makes sense! is this true? But the this leads me to this question, what does an "inverse" sound, sound like, I mean I get it is obviously not playing the tape backwards :) what would you actually have to do to, let's say a song to cancel the original song? – Trufa Feb 3 '11 at 1:27
Trufa: it turns out that it sounds almost exactly like the original sound! The short reason for this is that your ears/brain can't distinguish the 'shape' of a waveform, only how much energy is in different frequency bands (roughly, think of it as how hard the individual notes are being hit in a piano chord), and 'flipping' the wave upside-down doesn't change its energy profile. – Steven Stadnicki Feb 3 '11 at 2:00

Yes. Take a look at this: Noise Cancelling Headphones

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en.wikipedia.org/wiki/Active_noise_control Some stores use this to cut down on the noise from their ventilation systems. – Brian Feb 3 '11 at 0:14
Thank you and interesting references! – Trufa Feb 3 '11 at 0:28
But does this really answer the question? The headphones answer the question, given a noise source, is there another noise that if we emit it at known distance, then it will cancell the given noise. That answer is clearly yes, essentially by the third diagram in the question. But the original question is: given two sources emitting the same noise, is there a position where they are exactly cancelled? – JDH Feb 3 '11 at 0:36
@JDH en.wikipedia.org/wiki/Interference_(wave_propagation) This might answer that question a little better. It works with any type of wave (interference patterns were a main argument for light being a wave, although there are other experiments were it acts like a particle.) – Brian Feb 3 '11 at 0:38
Ok, sorry that all I'm doing is linking to wikipedia articles, but here is another one you might find interesting: en.wikipedia.org/wiki/Interferometry – Brian Feb 3 '11 at 0:42