= Solution
For the propagating mode of an air-filled or evacuated <rectangular electromagnetic waveguide>, the longitudinal <dispersion relation> is
$$
\omega^2=c^2k_z^2+\omega_c^2.
$$
The <phase velocity> and <group velocity> along the guide are consequently
$$
v_p=\frac{\omega}{k_z}=\frac{c}{\sqrt{1-(f_c/f)^2}},\qquad
v_g=\frac{d\omega}{dk_z}=c\sqrt{1-(f_c/f)^2}.
$$
With <waveguide cutoff frequency> $f_c=10\ \mathrm{GHz}$ and carrier <frequency> $f=30\ \mathrm{GHz}$, $\sqrt{1-(f_c/f)^2}=\sqrt8/3$. Thus
$$
\boxed{v_p=\frac{3c}{\sqrt8}=3.18\times10^8\ \mathrm{m\,s^{-1}},\qquad
v_g=\frac{\sqrt8\,c}{3}=2.83\times10^8\ \mathrm{m\,s^{-1}}.}
$$
The pulse envelope travels at the <group velocity> for a sufficiently narrow range of <frequencies>. The <phase velocity> exceeding the <speed of light> does not describe superluminal signal transport. The check $v_pv_g=c^2$ follows directly from the <dispersion relation>. A different uniform filling medium would require replacing $c$ by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.
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