Why must the term $kx-\omega t$ be constant? I mean from an intuitive point of view. I can mathematically prove it, however I don't seem to understand it very well.
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2Can your little further explain what you are asking? – Young Kindaichi Oct 24 '20 at 11:17
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It doesn't have to be constant unless you are interested in the behavior of a point of constant phase. For example, if you want to know the phase velocity of a wave, you would follow a point of constant phase ... say, a peak or a zero crossing (or any other [constant] value of phase). – garyp Oct 24 '20 at 13:19
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This is the phase $\phi=\left(kx-\omega t\right)$ of a wave. If you follow a point moving so that the phase on it to be constant, that is $\phi=\left(kx-\omega t\right)=\texttt{constant}$, then $\mathrm d\phi=\left(k\mathrm dx-\omega \mathrm d t\right)=0$ so the wave is propagating with speed $\upsilon_{ph}=\mathrm dx/\mathrm d t=\omega/k$.
Related : Significance of wave number.
Frobenius
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