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In an introductory electromagnetism text I am using the following is stated regarding waves of a string:

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Question: Why is it a given that the reflected wave has the same wave number and angular frequency as the incident wave and the transmitted wave has the same angular frequency as both the incident and reflected wave?

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  • $\begingroup$ What's the other half of the final sentence in your extract? Does it tell you. $\endgroup$
    – JMLCarter
    Commented Jan 23, 2017 at 14:33
  • $\begingroup$ @JMLCarter "the person at $z = - \infty$, who is shaking the string in the first place." $\endgroup$
    – user100411
    Commented Jan 23, 2017 at 14:35
  • $\begingroup$ so basically all frequencies are the same. There's no reason to suggest a change in frequency. Imagine at any point in the path the string can be cut and the half back to the source can be replaced by another guy shaking what's left up and down. For this to have no effect on the rest of the string they will have to shake at the same rate. $\endgroup$
    – JMLCarter
    Commented Jan 23, 2017 at 14:45
  • $\begingroup$ Now in the surface of a medium and with light waves, the guy shaking the string is a charge carrier in the medium. $\endgroup$
    – JMLCarter
    Commented Jan 23, 2017 at 14:46
  • $\begingroup$ @John Doe Just being curious, did you find any interest in my answer below? $\endgroup$
    – user130529
    Commented Jan 26, 2017 at 7:51

1 Answer 1

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Regarding frequency: this is a general property of convolution operators: for any given system that can be put under the form output $=$ input $*$ impulse ($*$ being the convolution operator of the two functions input and output), if the input is a function containing a single frequency $f$, then the output is another function containing the same single frequency $f$ $^{(1)}$.

So the question becomes here: are the two systems transforming the incident wave into reflected and transmitted wave convolution operators?

The answer is yes, and follows from the filter theorem: if a system is linear and time invariant $^{(2)}$, then it can be represented by a convolution operator, that is, there exists a function $h$ such that the output $y(t)$ takes the form $$y(t) = (x*h)(t)$$ where $x(t)$ is the input.

In the case of the string, the system is linear (as far as you are solving the usual wave equation) and time invariant (property of linear partial differential equations).

(1) Moreover, in that single frequency case, the output is equal to the input multiplied by the transfer function, which is the Fourier transform of the impulse.

(2) With the notation above, time invariant means that time delay commutes with the system, that is, for all $\tau$, the output corresponding the $x(t-\tau)$ is exactly $y(t-\tau)$ for all $t$.

Regarding the wave number: it is a simple consequence of the constancy (with respect to frequency) of the speed of traveling waves on the string, valid for the ordinary second order linear wave equation.

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