# Relation between phase velocity and $\delta\omega$ and $\delta k$

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 $$\delta\omega=\delta k=0.06$$?

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

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.