I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Related question and discussions here $\endgroup$– BioPhysicistCommented Mar 14, 2019 at 6:29
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
7
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In the following animation$^\dagger$ the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
$^\dagger$ Frames for the animations were made with Mathematica code available here, then combined into GIFs with ImageMagick's convert command.
-
1$\begingroup$ S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to. $\endgroup$ Commented Mar 14, 2019 at 6:28
-
1$\begingroup$ The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation. $\endgroup$ Commented Mar 14, 2019 at 13:00
-
2$\begingroup$ @CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave? $\endgroup$– FarcherCommented Mar 14, 2019 at 13:22
-
$\begingroup$ @Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently. $\endgroup$ Commented Mar 14, 2019 at 15:02
-
$\begingroup$ All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission. $\endgroup$– tpg2114Commented Mar 14, 2019 at 20:28
$\begingroup$
$\endgroup$
4
You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.
-
7$\begingroup$ +1 for the physicist's old trick of deducing general results from what happens in between special cases. $\endgroup$– J.G.Commented Mar 14, 2019 at 11:53
-
2$\begingroup$ Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more. $\endgroup$ Commented Mar 14, 2019 at 14:26
-
4$\begingroup$ @DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not. $\endgroup$– YakkCommented Mar 14, 2019 at 19:55
-
$\begingroup$ @Yakk That seems to fill the gap -- thanks. $\endgroup$ Commented Mar 14, 2019 at 20:01