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A wave packet in one dimension with a dispersion relation $\omega$ is a function of the form: $$f(x,t):=\sum_{i=1}^n\lambda_i\exp[i(k_i x+\omega[k_i] t)]\hspace{1cm}\forall i\leq n\colon\lambda_i,k_i\in\mathbb{R}.$$

For simplicity I consider a superposition of only 2 waves: $$\psi(x,t):=\exp[i k_1(x+\omega[k_1] t)]+\exp[i k_2(x+\omega[k_2] t)]$$

with the dispersion relation: $$\omega:k\mapsto\frac{k^2}{2}\hspace{1cm}\Rightarrow v_g:=\frac{\partial\omega}{\partial k}=k,\hspace{0.3cm}v_p:=\frac{\omega}{k}=\frac{k}{2}.$$

The wave has 2 wavenumbers, so I get different group and phase velocities for $k_1$ and $k_2$. So my questions are:

How can I find the 2 velocities at which a phase of the wave (phase velocity) and an amplitude (group velocity) of the wave actually move? Do they even have something to do with the group and phase velocity from above? Basically I want to reproduce this animation: https://upload.wikimedia.org/wikipedia/commons/b/bd/Wave_group.gif

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  • $\begingroup$ I think you can use this equation $\begin{aligned}f_{1}=A\sin \left( k_{1}x-\omega t\right) \\ f_{2}=A\sin \left( k_{2}x-\omega t\right) \\ f_{1}+f_{2}=2A\cos \left( \dfrac {k_{1}-k_{2}}{2}\cdot \left( x-vt\right) \right) \cdot \sin \left( \dfrac {k_{1}+k_{2}}{2}\cdot \left( x-vt\right) \right) \end{aligned}$ $\endgroup$ – Eli Apr 20 at 6:32

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