# How to derive the phase difference of a standing wave?

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.

The phase difference you are trying to calculate is the phase difference between different points in space $x$ at the same time $t$. In other words you are choosing some constant time $t$ then calculating how the phase $\Phi$ varies with $x$.

In your example of the travelling wave:

$$D(x,t)=A \sin (kx-\omega t+\Phi_0)$$

your method works because you take two different values of $x_1$ and $x_2$ at the same time $t$ so when you calculate:

$$\Delta\Phi = (kx_2-\omega t+\Phi_0) - (kx_1-\omega t+\Phi_0)$$

the $\omega t$ terms are constant and cancel out.

This works in exactly the same way for the standing wave:

$$D(x,t)=2a \sin kx\cos \omega t$$

If we take constant $t$ then $\cos \omega t$ is constant and we can write our snapshot in time as:

$$D(x) = A\sin kx$$

where $A$ is a constant given by $A = 2a\cos\omega t$. And just as for the travelling wave we get:

$$\Delta\Phi = k(x_2 - x_1)$$

• Thank you, this clears things up! May I ask another (a little bit irrelevant) question though, how can I rewrite the formulas for $\Delta\Phi=2\pi$ (in phase) and $\Delta\Phi=\pi$ (out of phase) so that they would work for $\Delta\Phi > 2\pi$, say, $\Delta\Phi = 4\pi$ which should be considered in phase? – TigerHix Jun 3 '17 at 10:48
• @TigerHix: the phase is given by your equation $\Delta\Phi = k(x_2 - x_1)$ and it can have any value from (in principle) $-\infty$ to $+\infty$. Points are in phase if $\Delta\Phi$ is a multiple of $2\pi$, and they are in antiphase if $\Delta\Phi$ is a multiple of $2\pi$ plus $\pi$. – John Rennie Jun 3 '17 at 10:57