So I understand that you get standing waves if there are waves of the same amplitude and wavelength traveling in opposite directions. But what happens if the wave traveling in an opposite direction has a phase shift, do you still get a standing wave, and if so how does it affect the overall standing wave if at all. For example, if you have one wave with equation of y(x,t)=A*sin (kx+wt) and the other wave was y(x,t)=Asin(kx-wt+φ) in the opposite direction. (I guess you might do this by reflecting a microwave off some metal surface at a spot where it is not naturally a node or antinode based off the frequency).

  • $\begingroup$ Well what do you think? What have you tried so far? Will the peaks and valleys still line up at some point in time? If so, where will this happen compared to no phase shift? Will this still allow for nodes at the places you need the nodes to be? $\endgroup$ Commented Mar 14, 2019 at 3:37
  • $\begingroup$ I would think that peaks and valleys may not line up if the phase shift is not some multiple of pi, but that there could still be points of destructive and constructive interference that would equate to nodes and antinodes $\endgroup$
    – Student
    Commented Mar 14, 2019 at 3:44
  • $\begingroup$ Nodes and antinodes are created when peaks of one wave line up with valleys of the other and when peaks/valleys of both waves line up respectively. So your comment contradicts itself. If you can't envision this scenario I would suggest simulating it. Just plot the superposition as a function of $x$ and let time run with various phase differences. $\endgroup$ Commented Mar 14, 2019 at 3:51
  • $\begingroup$ can you not have a node where the one wave is only displaced up A/2 and the other is displaced down A/2, thereby making a node there even though it wasn't where a peak met a trough $\endgroup$
    – Student
    Commented Mar 14, 2019 at 3:55
  • $\begingroup$ Good point. I should have just talked about the antinodes. $\endgroup$ Commented Mar 14, 2019 at 3:57

2 Answers 2


Looking at graphs of this like I mention in the comments it's always helpful. You can also look at the math.

It's easier to work with complex exponentials and then take the imaginary part at the end. So we start with $$y=Ae^{i(kx-\omega t+\phi)}+Ae^{i(kx+\omega t)}=Ae^{ikx}\left(e^{-i(\omega t-\phi)}+e^{i\omega t}\right)$$

We can then do a fun trick where we factor out $e^{i\phi/2}$ from what is inside the parentheses: $$y=Ae^{i(kx+\phi/2)}\left(e^{-i(\omega t-\phi/2)}+e^{i(\omega t-\phi/2)}\right)$$

This is nice because now the stuff inside the parentheses is just $2\cos\left(\omega t-\frac\phi2\right)$. Using Euler's equation as well gives us $$y=2A\cos\left(\omega t-\frac\phi2\right)\left(\cos\left(kx+\frac\phi2\right)+i\sin\left(kx+\frac\phi2\right)\right)$$

We can finally then take the imaginary part like I mentioned at the beginning $$\text{Im}(y)=2A\cos\left(\omega t-\frac\phi2\right)\sin\left(kx+\frac\phi2\right)$$

This is our standing wave for phase shift of $\phi$ applied to the right traveling wave. Note that this is still a standing wave with the same amplitude, wavelength, frequency as the "original standing wave", as can be verified for when $\phi=0$.

However, you will notice that the location of our nodes (and hence antinodes) depend on the phase. We can see this by finding the points in space $x_n$ where our function is $0$ for all times. This is then when $$\sin\left(kx_n+\frac\phi2\right)=0$$ or $$x_n=\frac{\pi n-\phi/2}{k}$$ for any integer $n$.

So if you want your nodes to be at specific locations, then you can't choose any phase difference you want. For example, if you always want nodes at $x=0$, then you would always need $\phi=2\pi n$, which doesn't change the left traveling wave at all compared to the "original wave" of $\phi=0$.

If you want a node at some general location $x=x'$, then the phase shift can only be $$\phi=2(\pi n-kx')$$ And I should also point out that specifying where you want one node automatically determines where all nodes and antinodes will exist assuming a fixed wave number $k$


Sometimes the use of phasors reduces the use of Mathematics but increases the use of Physics?

Two waves travelling in opposite directions each have an amplitude of amplitude $A$ at at position $x=0$ and out of phase by $\phi$ can be represented by $L(x,t) = A \sin \left (\omega t + \frac \phi 2 +kx\right)$ and $R(x,t) = A \sin \left (\omega t - \frac \phi 2 - kx\right )$.
I have changed your wave equations slightly to make my derivation easier to understand.

At $x=0$ the phasor representation of the combination of these two waves is shown below.

enter image description here

The amplitude of the addition of the two waves is $X(0)= 2A\cos \frac \phi 2$ and so the displacement at position $x=0$ is given by $X(0,t) = 2A\cos \frac \phi 2\,\sin \omega t$

At position $x$ the phasor diagram changes to the following with the $kx$ representing the phase lead/lag relative to the $x=0$ position of the two waves.

enter image description here

The amplitude of the addition of the two waves is $X(x)= 2A\cos \left (\frac \phi 2+2kx \right )$ and so the displacement at position $x$ is given by $X(x,t) = 2A\cos \left (\frac \phi 2+2kx\right )\,\sin \omega t$

Looking at the second phasor diagram (or the equation) when $L(x)$ and $R(x)$ differ in phase by $\pi$, ie are in antiphase, the displacement is always zero so the condition for a node at position $x$, if the phase difference between the waves is $\phi$ at position $x=0$, is $\phi +2kx = \pi \Rightarrow x = \dfrac {\pi- \phi }{2k}$

As a check if $\phi=0$, ie the waves are in phase at $x=0$ and it is an antinode, the first node occurs at $x = \dfrac {\pi- 0 }{2k} \Rightarrow x = \frac \lambda 4$ as $k = \frac{2\pi}{\lambda}$.

In general $x = \dfrac {n\pi- \phi }{2k}$ where $n$ is an odd integer.

I have just found this animation which may be of interest?


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