= Solution
For a fixed transverse <normal mode>, define its cutoff <angular frequency>
$$
\omega_{c,n}=\frac{n\pi v}{b},\qquad
\omega^2=\omega_{c,n}^2+v^2k_x^2.
$$
The longitudinal <phase velocity> and <group velocity>, for propagation in $+x$, are
$$
\boxed{v_p=\frac\omega{k_x}=\frac{v}{\sqrt{1-(\omega_{c,n}/\omega)^2}},\qquad
v_g=\left.\frac{d\omega}{dk_x}\right|_n
=\frac{v^2k_x}{\omega}=v\sqrt{1-(\omega_{c,n}/\omega)^2}.}
$$
Thus $v_pv_g=v^2$, $v_p>v$ and $v_g<v$. The large longitudinal <phase velocity> reflects the geometry of the transverse <standing wave> and does not imply faster transport of <energy> than the membrane <wave speed>.
The <cutoff frequency of a membrane waveguide> for mode $n$ is
$$
\boxed{f_{c,n}=\frac{\omega_{c,n}}{2\pi}=\frac{nv}{2b}.}
$$
At cutoff $k_x=0$ and $v_g=0$; the longitudinal <phase velocity> has no finite value. Below cutoff,
$$
k_x=i\gamma_n,\qquad \gamma_n=\sqrt{(n\pi/b)^2-\omega^2/v^2},
$$
and a bounded field excited at $x=0$ decays as $e^{-\gamma_nx}$. It is an <evanescent wave>, rather than a freely propagating mode carrying net time-averaged <energy flux> down an indefinitely long uniform strip. A finite section can still transmit through coupling of evanescent fields.
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