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Suppose we have two waves, with the same amplitude but different wavelengths and velocities and we add them $${y}_{1}=A{\mathrm{sin}}\left[\frac{2\pi}{{\lambda}_{1}}\left(x-{{\text{v}}}_{1}t\right)\right]$$ $${y}_{2}=A{\mathrm{sin}}\left[\frac{2\pi}{{\lambda}_{2}}\left(x-{{\text{v}}}_{2}t\right)\right]\text{.}$$ Then $${y}_{1}+{y}_{2}=A\left({\mathrm{sin}}\left[\frac{2\pi}{{\lambda}_{1}}\left(x-{{\text{v}}}_{1}t\right)\right]+{\mathrm{sin}}\left[\frac{2\pi}{{\lambda}_{2}}\left(x-{{\text{v}}}_{2}t\right)\right]\right)\text{.}$$ Lets rewrite using wave number and angular frequency $${y}_{1}+{y}_{2}=y=A\left({\mathrm{sin}}\left[\left({k}_{1}x-{\omega}_{1}t\right)\right]+{\mathrm{sin}}\left[\left({k}_{2}x-{\omega}_{2}t\right)\right]\right)\text{.}$$ Now we will use ${\mathrm{sin}}(\theta +\phi )+{\mathrm{sin}}(\theta -\phi )=2{\mathrm{sin}}\theta {\mathrm{cos}}\phi $ and set $$\theta +\phi ={k}_{1}x-{\omega}_{1}t$$ $$\theta -\phi ={k}_{2}x-{\omega}_{2}t\text{.}$$ We can rearrange to get $$2\theta =({k}_{1}+{k}_{2})x-({\omega}_{1}+{\omega}_{2})t$$ $$2\phi =({k}_{1}-{k}_{2})x-({\omega}_{1}-{\omega}_{2})t\text{.}$$ By substituting we can then see that $$y=2A\left({\mathrm{cos}}\left[\frac{{k}_{1}-{k}_{2}}{2}x-\frac{{\omega}_{1}-{\omega}_{2}}{2}t\right]\times {\mathrm{sin}}\left[\frac{{k}_{1}+{k}_{2}}{2}x-\frac{{\omega}_{1}+{\omega}_{2}}{2}t\right]\right)\text{.}$$ Now set $$\Delta k={k}_{1}-{k}_{2}$$ $$\Delta \omega ={\omega}_{1}-{\omega}_{2}$$ $$k=\frac{{k}_{1}+{k}_{2}}{2}$$ $$\omega =\frac{{\omega}_{1}+{\omega}_{2}}{2}$$ and we can rewrite the wave as $$y=2A{\mathrm{cos}}(x\frac{\Delta k}{2}-t\frac{\Delta \omega}{2}){\mathrm{sin}}(kx-\omega t)\text{.}$$
The above equation shows beats. For example you can set $t=0$ and see that you get $$y=2A{\mathrm{cos}}(x\frac{\Delta k}{2}){\mathrm{sin}}(kx)\text{.}$$ Likewise you could pick $x=0$ and get the same figure, but now the horizontal axis is time $$y=2A{\mathrm{cos}}(-t\frac{\Delta \omega}{2}){\mathrm{sin}}(-\omega t)$$ or $$y=2A{\mathrm{cos}}(t\frac{\Delta \omega}{2}){\mathrm{sin}}(-\omega t)\text{.}$$ You get a traveling wave that has an oscillating amplitude.
When we look at $$y=2A{\mathrm{cos}}(x\frac{\Delta k}{2}-t\frac{\Delta \omega}{2}){\mathrm{sin}}(kx-\omega t)$$ we see that there are two velocities. One, referred to as the phase velocity, is the speed of the individual wavecrests: $${{\text{v}}}_{p}=\frac{\omega}{k}=\nu \lambda \text{.}$$ The group velocity is the velocity of the envelope $${{\text{v}}}_{g}=\frac{\Delta \omega}{\Delta k}\to \frac{d\omega}{dk}$$ Energy and momentum normally move with the group velocity.
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