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?
Thanks in advance!
 A: 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.
A: Yes.  Take a look at this: Noise Cancelling Headphones
