Superposition of waves: why do they not cancel?

I am not sure what I am doing wrong here or if I have made some wrong assumptions, however my intuition is contrasting with my mathematical result.

I am considering a case where there are two speakers opposite each other, separated by a distance $R$. They are in phase and produce waves of the same amplitude $a_0$. So they produce waves of the form $a_0 \cos (\omega t)$.

If I place a detector between them and a variable distance r away from one of the speakers, I get that detected waves from one speaker are of the form $$a_0 \cos \left(\omega t -\frac {\omega r}{c} \right)$$ and from the other is $$a_0 \cos \left(\omega t -\frac {\omega (R-r)}{c} \right) \, .$$

Now this gives me a standing wave when I consider the case of R being fixed and varying r only, and I considered where the phase diffeence between the waves detected was $2n\pi$ for constructive interference and $(2n+1)\pi$ for destructive. This gave me the expected result, so my expressions for the waves seem to work here.

However I am stuck when I considered varying the distance between the speakers $R$. Intuitively, I would think that if the distance between the speakers is $(n+1/2)\lambda$ everywhere, then the waves at each point between the speakers would be in antiphase and would cancel. However when I looked at this mathematically, I thought: is there an $R$ for which

$$-\frac {\omega r}{c})=\frac {\omega (R-r)}{c})+(2n+1)\pi$$ independent of $r$. Thinking about it now, this is actually the exact same equation I used to find where the nodes are in the standing wave, and that was at particular values of $r$. It seems from this equation that there is no distance between the speakers for which the waves are in anti-phase everywhere.

This goes against my intuition. I'm not sure which is right.

• I fixed some of the math typesetting here but there's still an obvious error in the third equation. Please fix it. Also, there's no actual question in this post. Please as a specific, well defined question :) Commented Dec 26, 2016 at 0:57

The maths is $$R=(n+1/2)λ$$
Trying to resolve it by equating phase angles is not necessary. focus on the total distance including that odd half wavelength to give a $\pi$ phase shift.