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.


1 Answer 1


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:

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

  • $\begingroup$ 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? $\endgroup$
    – TigerHix
    Commented Jun 3, 2017 at 10:48
  • $\begingroup$ @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 $\pi$. $\endgroup$ Commented Jun 3, 2017 at 10:57

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