quantum mechanics - Interpreting group velocity of free particle wave packet

I am trying to understand the concept of group velocity of a free particle wave packet:

$$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \phi(k)e^{ikx}e^{-\frac{i \hbar k^2 t}{2m}}dk.$$

Where

$$\omega(k) = \frac{\hbar k^2}{2m}.$$ Assuming that $\phi(k)$ is narrowly peaked about some particular value $k_0$ we can Taylor expand $\omega(k)$ about $k_0$ to get $$\omega(k) = \frac{\hbar k_0}{2m} + \frac{\hbar k_0}{m}(k-k_0)+ \frac{\hbar}{2m}(k-k_0)^2.$$ Then after defining variable $\kappa := k-k_0$ we get $$\Psi(x,t) \approx \frac{1}{\sqrt{2 \pi}}e^{\frac{-i \hbar k_0^2 t}{2m}}e^{\frac{i \hbar k_0^2 t}{m}} \int^{\infty}_{-\infty}\phi(\kappa + k_0)e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}dk.$$

This integral is the superposition of waves of the form

$$e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}.$$ However notice that each of these waves have the same speed $\frac{\hbar k_0}{m}$. Then in terms of the energy $$v_{\text{group}} = \frac{d \omega(k_0)}{dk} = \frac{\hbar k_0}{m} = \sqrt{\frac{2 E_0}{m}}.$$ Hence for stationary state $\Psi_{k_0}$ we have that $$v_{\text{group}} = 2 v_{\text{phase}}.$$

Apparently we can then evaluate the group velocity for each $k$, hence $v_{\text{group}} = \frac{d \omega(k)}{dk}$.

Question: I understand the above calculation. But qualitatively I don't understand how the group velocity can be a function of $k$ as opposed to just a single value (independent of $k$) which describes how fast the envelope propagates? I know that each stationary state has it's own phase velocity but I thought that superimposing these waves of different speeds produces an envelope which travels at a single velocity $v_{\text{group}}$?