Group velocity for wave pack

Question

I was reading about the derivation for group velocity on the wiki page here. As far as I understood, what's done here is based on these steps

1. Get a basic solution for the wave equation, that is $exp(i(kx-\omega t))$
2. Write some general solution using the assumption that its Fourier transform at $t=0$ is localized
3. Consider that $\omega=\omega(k)$ and using some assumptions one can get the envelope and therefor its velocity.

What I don't understand here is related to step 3. where on the wiki site the author writes $\omega(k)\approx\omega_0+(k-k_0)\frac{d \omega}{dk}(k_0)$ as if we don't know how $\omega(k)$ looks like. However, this comes after solving the wave equation for step 1. where we already established that $\omega=ck$ as is mentioned here, where $c$ is the velocity from the differential equation.

So, my question is why was $\omega(k)\approx\omega_0+(k-k_0)\frac{d \omega}{dk}(k_0)$ needed in this scenario and if I'm missing something, what is it. In other words, did the author actually assume that he does not know something that he knows in order to get a more 'general' result?

If I'm not missing something, then where could I find a better and also general way (if it exists) of defining the group velocity?

Edit:

I felt that a clarification is needed regarding the question.

I am not implying that there are no other relation other than $\omega=ck$. My problem arose from the derivation of the solution of wave equation using separation of variables. By using that procedure, one can derive the basic solution $exp(i(kx-\omega t))$, where $k$ is introduced as the constant needed for separation of variables. Solving the ODE for the space coordinate gives the $exp(ikx)$ part, while solving for the temporal coordinate gives the other part $exp(i(ck)x)$. Only now, one can make a notation $\omega=ck$ such that the temporal part becomes something like $exp(i\omega t)$. This reasoning is the one used in the second link I used.

I will try to reformulate the question. Why am I allowed to consider the same solution $exp(i(kx-\omega t))$ that depends strictly on the above mentioned notations, even if I change the parameters $\omega$, $k$?

Answer 1

For general waves (such as gravity or capillary waves on the surface of water) it is not true that $\omega=ck$. In other words they do not satisfy the same wave equation as for sound or for waves on a string. For example waves on the surfae of deep water have $\omega(k)=\sqrt{gk}$. Group velocity is different from the phase velocity $v_{\rm phase} = \omega/k$ when this is so. So you first need to find $\omega(k)$ for your particlular species of wave and then find $$ v_{\rm group} = \frac{\partial \omega}{\partial k}. $$ Again, for deep water waves, $$ v_{\rm group} = \frac 1 2 \frac \omega k $$ so the group velocity is half that of the wave crests, which move at $v_{\rm phase}= \omega/k= \sqrt {g/k}$.