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When we have 1D standing waves, we can write them as the sum of two propagating wave in opposite directions that give the formula $\sin kx\cos\omega t$.

When I try to do this for 2D waves (I mean 2D by the fact there are $k_x$ and $k_y$), I don't have the right expression: $$\exp\left(i\left(k_x x + k_y y - \omega t\right)\right)+\exp\left(i\left(k_x x + k_y y + \omega t\right)\right)=\exp\left(i\left(k_x x + k_y y\right)\right)2\cos(\omega t).$$

If I take the imaginary part, I will have $\sin(k_x x + k_y y)$ and not $\sin k_x x \sin k_y y$.

Am I wrong somewhere or is it not true that 2D stationary waves are the sum of two propagating waves in opposite directions?

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  • $\begingroup$ If you expand $\sin(a+b)$, you'll get sum of two terms of the form you look for. $\endgroup$
    – Ruslan
    Commented Apr 6, 2016 at 15:16
  • $\begingroup$ Yeah but if I expand it I will get a sum indeed. But a 2D standing wave is supposed to be $sin(k_x.x)sin(k_y.y)sin(wt)$ and not something like $sin(k_x.x)cos(k_y.y)sin(wt)+cos(k_x.x)sin(k_y.y)sin(wt)$ right ? What I wrote would be a sum of stationnary waves and not a stationnary wave ? $\endgroup$
    – StarBucK
    Commented Apr 6, 2016 at 15:20
  • $\begingroup$ For the future please note that in $\LaTeX$, standard functions like $\sin$ and $\exp$ should be written as $\sin$ and $\exp$ (note the backslash). Also to make the equation displayed (as opposed to inline), use double dollars like $$this$$. $\endgroup$
    – Ruslan
    Commented Apr 6, 2016 at 16:17
  • $\begingroup$ Ok I take note of this $\endgroup$
    – StarBucK
    Commented Apr 6, 2016 at 16:24

1 Answer 1

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So, you summed two waves and got $[\sin(k_xx)\cos(k_yy)+\cos(k_xx)\sin(k_yy)]\cos(\omega t)$. Now, add two more with wavevector $(k_x,-k_y)$ and you will obtain $2\sin(k_xx)\cos(k_yy)\cos(\omega t)$, the standing wave you're looking for.

However, the function $\sin(k_xx+k_yy)\cos(\omega t)$ already is a standing wave, since the spatial and the temporal oscillations are independent: $f(x,y)g(t)$. Here nodes and anti-nodes consist of lines perpendicular to $\vec{k}$.

Experimentally, a standing wave results from the superposition of incident and reflected waves. In one dimension you need one wall (one reflection). In two dimensions you generally need two non-parallel walls; the second wall will reflect the incident wave and the wave reflected on the first wall, hence the need of four travelling waves to make a standing one. But, if you place a single wall perpendicularly to $\vec{k}$, you obtain (in 2D- or 3D-space) a one-dimensional standing wave, it just has a peculiar orientation. Experimentally this can be approximately obtained by placing a loudspeaker facing a wall.

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  • $\begingroup$ Thank you ! So if I summarise in 1D we only need one reflexion (so 2 propagative waves in opposite directions) whereas in 2D we need 3 propagative waves (kx,ky,+wt)+(kx,ky,-wt)+(kx,-ky,+wt) $\endgroup$
    – StarBucK
    Commented Apr 6, 2016 at 15:53
  • $\begingroup$ 2+2=4. The second wall reflects both the incident wave and the wave reflected on the first wall. Also, I'll edit my answer because the function you obtained already is a standing wave. $\endgroup$
    – L. Levrel
    Commented Apr 6, 2016 at 17:42

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