Prof that it is not possible to get a standing wave.

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Homework Statement


As in the Title starting from two planewaves with the same amplitude, but different frequency.

Homework Equations


Starting from
Ae^{i(kx +\omega_1 t)} + Ae^{i(kx +\omega_2 )}

The Attempt at a Solution



I got as far as

Ae^{i(kx +\omega_1 t)} + Ae^{i(kx +\omega_2 )}
= 2Ae^{ikx} e^{i \frac{\omega_2 + \omega_1}{2} t} \cos \frac{\omega_2 - \omega_1}2 t

taking the real part

= 2A \left( \cos kx \cos{ \frac{\omega_2 + \omega_1}{2} }t - \sin kx \sin { \frac{\omega_2 + \omega_1}{2} } \right) \cos{ \frac{\omega_2 - \omega_1}{2}}

Now how do i argue to conclude that this is definitley not a standing wave? I know that position and time has to be decoupled, but how can I conclude that they are not?

Can this expression be simplified further?

is it enough to say that this is a product of two factors one which depends only on time and one which does depend on both space and time and therefore can not be decoupled?
 
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I would start with an expression that reduces to a standing wave when the frequencies are the same. Does your starting expression do that?
 
Yes it does when \omega_1 = - \omega_2
 
By your logic, if ω1 = 100 Hz, then ω2 = -100 Hz. Is 100 Hz equal to -100 Hz? Do it right and put the minus sign where it belongs. Wave frequencies are always positive.
 
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