We know a standing wave is defined by $D(x,t)=2a \sin kx\cos wt$. Intuitively, all particles within the same "loop" of a standing wave are vibrating in phase; all particles within 2 adjacent "loops" are vibrating in opposite phase. However, is there a mathematical proof of this?
Below is my attempt:
For a progressive wave $D(x,t)=A \sin (kx-wt+\Phi_0)$, the phase is $kx-wt+\phi_0$, which makes the phase difference $\Delta\Phi = (kx_2-wt+\Phi_0) - (kx_1-wt+\Phi_0) = k\Delta x$. Then if $\Delta\Phi = 2\pi$, the two particles are vibrating in phase; if $\Delta\Phi = \pi$, two particles are vibrating out of phase.
But using the same logic for standing waves, it seems the phase for them would be $wt$ thus phase difference $\Delta\Phi = wt - wt = 0$. This makes sense for particles in the same loop, but does not take into account particles in adjacent loops.