Wave reflection

Question

I have a simple question on wave reflection.

I know that if I have a progressive monochromatic EM wave and a mirror, the reflected wave will be opposite in phase on the mirror to assure a total E field equal to 0 on the mirror.

But when I think about summation of two progressive waves (in opposite directions), I have :

$$\textrm{Im} \left(e^{i(kx-\omega t)}+e^{i(kx+\omega t)}\right)=2\sin(kx)\cos(\omega t)$$

Cool, I have a stationary wave.

If I study these two propagative functions and I want to see where they cancel (the place where my mirror would be), I have : $kx-\omega t=kx+\omega t+\pi$

And it gives : $2\omega t=\pi$.

So I don't understand, when I study this equation I see that it is not possible to have a mirror because if there would have a mirror, I would have the equation $2\omega t=\pi$.

Could anyone help me ?

Answer 1

Your expression tells us that the standing wave amplitude is zero twice per period. But the waves also cancel where $\sin(kx) = 0$ - that is where the nodes of the standing wave are.

Answer 2

You only need the imaginary part of $e^{i(kx-\omega t)}+e^{i(kx+\omega t)}$ to vanish to get a node of the wave. In particular, you have $$ \Im( e^{i(kx-\omega t)}+e^{i(kx+\omega t)}) = 2 \Im (e^{ikx} \cos \omega t). $$ If the quantity inside parentheses is purely real, you will also get the wave to vanish. This occurs when $kx = n \pi$, which (in turn) implies $\sin (kx) = 0$.