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I'm trying to find a relation between the phase velocity, $v_p=$$\omega \over k$ and $\delta\omega$ and $\delta k$, of a wave formed by adding the two waves:

$$y_1=\cos(k_1 x-\omega_1 t)$$ and $$y_2=\cos(k_2 x-\omega_2 t)$$

which gives

$$y=2\cos\left(\frac{\delta kx}2-\frac{\delta\omega t}2\right)\cos\left(\bar kx-\bar\omega t\right)$$

where $\bar\omega=(\omega_1+\omega_2)/2$ , $\bar k=(k_1+k_2)/2$ , $\delta k=k_1-k_2$ and $\delta \omega=\omega_1-\omega_2$.

I know that the group velocity is given by $d\omega\over dk$ and wanted to know how phase velocity changes as the values of $\delta\omega$ and $\delta k$ increase while remaining equal to each other.

For example, what is the difference between $v_p$ for $\delta\omega=\delta k=0.02$ and $\delta\omega=\delta k=0.06$?

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    $\begingroup$ What phase velocity? Each one of the two components have their phase velocity. $\endgroup$
    – nasu
    Nov 13, 2021 at 19:09
  • $\begingroup$ @nasu phase velocity of the fast oscillating component given by $cos(\bar k x-\bar \omega t)$ $\endgroup$ Nov 14, 2021 at 5:24

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The phase velocity of the first wave is $$ \frac{\omega_1}{k_1}. $$ The phase velocity of the second wave is $$ \frac{\omega_2}{k_2}. $$

In the expression $$y=2\cos\left(\frac{\delta k\, x}2-\frac{\delta\omega\, t}2\right)\cos\left(\bar kx-\bar\omega t\right)$$ you have an oscillation whose phase is $(\delta k\, x - \delta\omega\,t)/2$ acting as envelope to a faster oscillation whose phase is $(\bar{k} x - \bar{\omega}t)$. The phase velocity of that faster oscillation is $$ \frac{\bar{\omega}}{\bar{k}}. $$ The envelope meanwhile moves along at the velocity $\delta\omega/\delta k$ and that velocity is called the group velocity.

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