For any system supporting dispersive waves, the question of the speed of propagation of disturbances turns out to be rather subtle. There are two different speeds which are relevant to different aspects of the question. The simplest way to see that something interesting is going on is to consider a simple mixture of two different sinusoidal waves with slight different frequencies and wavenumbers. Suppose we have the mixture
$$y(x,t) = \sin (k_1 x -\omega_1 t) + \sin (k_2 x -\omega_2 t) \tag{1}$$
where $\omega_1$ and $\omega_2=\omega_1 + 2\Delta \omega$ are two nearby frequencies, and $k_1$ and $k_2 = k_1 + 2\Delta k$ are the corresponding wavenumbers calculated from the dispersion relation of the system.
We can apply a familiar trigonometrical identity to eq. (1) to deduce that
$$y(x,t) = 2 \sin(\bar{k} x – \bar{\omega} t) \cos(\Delta k~x – \Delta \omega~t) \tag{2}$$
where $\bar{k}=(k_1 + k_2)/2$ and $\bar{\omega} = (\omega_1 + \omega_2)/2$ are the average wavenumber and frequency, respectively. The term $\sin(\bar{k} x – \bar{\omega} t)$ describes a “carrier wave” at the average of the two frequencies, shown in red in Fig. 1. It travels at a speed
$$c_p = \dfrac{\bar{\omega}}{\bar{k}} \tag{3}$$
known as the phase velocity, appropriate to a single sinusoidal wave at frequency $\bar{\omega}$. The remaining term $2\cos(\Delta k~x – \Delta \omega~t)$ describes an amplitude modulation envelope of this carrier wave, shown in blue dashes in Fig. 1. This modulation pattern propagates at a different speed
$$c_g = \dfrac{\Delta \omega}{\Delta k} \approx \dfrac{d \omega}{dk} \tag{4}$$
known as the group velocity. In section 4.4 this behaviour was illustrated by several animations, featuring different relationships between phase velocity and group velocity.
To understand these animations, we can calculate the phase and group velocities for the dispersive systems introduced in section 4.4.1. The first of these was a bending beam, which had dispersion relation
$$\omega = \dfrac{k^2}{\alpha^2} , \tag{5}$$
where $\alpha=(m/EI)^{1/4}$ in terms of the bending rigidity $EI$ and mass per unit length $m$ of the beam. So the phase velocity is
$$c_p=\dfrac{\omega}{k} = \dfrac{k}{\alpha^2} = \dfrac{\sqrt{\omega}}{\alpha} \tag{6}$$
while the group velocity is
$$c_g = \dfrac{d \omega}{dk} = \dfrac{2k}{\alpha^2} = \dfrac{2\sqrt{\omega}}{\alpha} = 2 c_p . \tag{7}$$
Next, we looked at a stretched string with bending stiffness, which had the dispersion relation
$$\omega = c_0 k \sqrt{1 + \beta k^2}\tag{8}$$
where
$$c_0=\sqrt{P/m}\tag{9}$$
is the (non-dispersive) wave speed on the string in the absence of bending stiffness, and
$$\beta = \dfrac{EI}{P} ,\tag{10}$$
$P$ being the string tension.
The phase velocity is now
$$c_p = c_0 \sqrt{1+ \beta k^2} ,\tag{11}$$
while the group velocity is
$$c_g = c_0\left[ \sqrt{1+ \beta k^2} + \dfrac{\beta k^2}{\sqrt{1+ \beta k^2}} \right] = c_0~\dfrac{1+2 \beta k^2}{\sqrt{1+ \beta k^2}} .\tag{12}$$
To illustrate the resulting behaviour, Fig. 2 shows a plot of the ratio of group velocity to phase velocity as a function of frequency $\omega$, for the particular case with $c_0 = 1$ and $\beta = 1$. At low frequency the ratio is 1, because the waves are approximately non-dispersive, but as frequency rises the ratio tends towards 2, the value we have just seen for the bending beam with no tension.
As a last example we looked at water waves, with a dispersion relation
$$\omega^2 = gk .\tag{13}$$
Now the phase velocity is
$$c_p = \dfrac{\omega}{k} = \dfrac{g}{\omega} = \sqrt{\dfrac{g}{k}} ,\tag{14}$$
while the group velocity is
$$c_g = \dfrac{d \omega}{dk} =\dfrac{1}{2}~\sqrt{\dfrac{g}{k}} = c_p/2 .\tag{15}$$
To finish this subsection, I will give a slightly more sophisticated analysis of group velocity. The repetitive beating pattern shown in Fig. 1 did not, in all honesty, look very much like the pattern of ripples after a stone was thrown into water. Does the analysis really hold for a more realistic case? Figure 3 shows a snapshot of what part of the ripple pattern might look like. The train of waves shows a slowly-varying wavelength, getting shorter as we move from left to right. We can conveniently capture something like this through a “phase function” $\psi(x,t)$ by writing the displaced water surface displacement in the form
$$\zeta(x,t) = \zeta_1(x,t)~e^{i \psi(x,t)} \tag{16}$$
where $\zeta_1(x,t)$ is a slowly-varying amplitude. The wave crests occur where $\psi = 0,~2\pi,~4\pi,\cdots$
Around a particular position $x_0$ and time $t_0$ the phase function can be approximated, using a Taylor expansion, by linear variation:
$$\psi(x,t) \approx \psi(x_0,t_0) + (x~-~x_0)~\left[\dfrac{\partial \psi}{\partial x}\right]_{(x_0,t_0)}$$
$$+ (t~-~t_0)~\left[\dfrac{\partial \psi}{\partial t}\right]_{(x_0,t_0)} . \tag{17}$$
Substituting this expression in eq. (16), we can see that it describes a travelling wave of the form
$$\zeta = \zeta_1 ~e^{ik(x-x_0) – i\omega (t-t_0)} \tag{18}$$
with local wavenumber and frequency given by
$$k = \left[\dfrac{\partial \psi}{\partial x}\right]_{(x_0,t_0)},~~~\omega = -\left[\dfrac{\partial \psi}{\partial t}\right]_{(x_0,t_0)}. \tag{19}$$
By a standard result about partial derivatives, it follows that these must obey
$$\dfrac{\partial k}{\partial t} + \dfrac{\partial \omega}{\partial x} = 0 , \tag{20}$$
or since $\omega=\omega(k)$
$$\dfrac{\partial k}{\partial t} + c_g~\dfrac{\partial k}{\partial x} = 0 \tag{21}$$
where
$$c_g=\dfrac{d \omega}{dk} \tag{22}$$
as before. This equation has the form of a convective derivative: it says that the value of wavenumber remains constant along lines in the $(x,t)$ plane satisfying
$$x – c_g t = \mathrm{constant} , \tag{23}$$
which is the result we want.