# How do sound waves from different distances not mess up the pitch?

This might be a too begginer question for this site, but, if two sounds are produced with same pitch from different places, why do they sound like the same? Why do they "sync" their vallies and peaks with each other? Intuitively, if for example one sine wave was sent exactly half of the frequency distance from the other, their sum would be messed up, and not a pure sine wave anymore, but this doesn't seem to happen. Why?

• This is actually a really interesting question, some possibilities to take into account involve that the wavelength of sound is a macroscopic distance comparable to the cones used in the speakers and those speakers cannot be placed perfectly symmetrically, plus reflections from nearby floors and walls, plus the fact that the $1/r^2$ power dispersion means that they don't cancel perfectly at nodes, plus the fact that the sounds you hear in reality are not pure sine waves but have overtones, plus the finite size of a head. Images illustrating these would be amazing. Oct 10, 2018 at 22:20
• The thing you described actually does happen! This is what noise cancelling headphones take advantage of. They produce a sound wave that is shifted in distance just right to cancel out the outside sound wave. Oct 11, 2018 at 4:29
• It is the case that the resulting wave is still a sine wave, it just might have a different amplitude. Oct 11, 2018 at 4:31

Long story short, maths happens. Mathematically if you add two sine waves together with different amplitudes and a phase difference, the result is still a sine wave.

You can calculate the unified wave with a bit of trig and algebra:

$$A\sin(\omega t) + B\sin(\omega t + \phi) \equiv C \sin(\omega t + \theta)$$

Expanding both sides with the addition identity gives

$$\left[A + B\cos\phi \right]\sin(\omega t) + \left[ B\sin\theta\right]\cos(\omega t) \equiv C\cos\theta\sin(\omega t) + C\sin\theta\cos(\omega t),$$ and collecting terms gives you

$$C\cos\theta = A + B\cos\phi$$ $$C\sin\theta = B\sin\phi$$

You can solve these two to get:

$$C^2 = A^2 + B^2 + 2AB\cos\phi$$ $$\tan\theta = \frac{B\sin\phi}{A + B\cos\phi}$$

If you don't want to follow the maths it basically comes down to this- when you add together two sine waves with the same frequency, the resultant wave has a different amplitude and phase but the same frequency.

If two sources of traveling waves (i.e, your average sound wave) of same frequency intersect in opposite directions, they form a standing wave. In a standing wave, there are points in space at which the increase in one with time is exactly matched by the decrease of the other. At these points, the sound does decrease, but it does not reach zero practically due to multiple reflections. At a point half a wavelength away, they add up and create double the sound.
If they are both in the same direction, they just add up and are still a traveling sine wave, the exact math depends on phase differences and such