Wave explanation and derivation

Question

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.

Answer 1

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 $\upsilon_{ph}=\mathrm dx/\mathrm d t=\omega/k$.

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