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The cutoff frequency of a membrane waveguide is the lowest frequency at which a specified transverse normal mode can propagate longitudinally. For fixed edges separated by $b$, ${\omega }_{c,n}=n\pi v/b$ and $f_{c,n}={\omega }_{c,n}/(2\pi )$. Below cutoff the longitudinal wavenumber is imaginary and a bounded excitation is an evanescent wave. Above it, $v_g=v\sqrt{1-({\omega }_{c,n}/\omega {)}^2}$ and $v_p{v}_g=v^2$.

For the $n$th transverse mode, substitute $q_n=n\pi /a$ into the membrane waveguide dispersion relation:Longitudinal propagation requires real nonzero $k$, henceBelow the cutoff frequency of a membrane waveguide, $k=i\kappa$, with $\kappa =\sqrt{(n\pi /a{)}^2-\sigma {\omega }^2/T}$, and a bounded excitation decays as $e^{-\kappa x}$ rather than carrying a travelling wave along an infinite guide. At equality $k=0$, the displacement has no longitudinal phase advance and the group velocity is zero; it is a cutoff oscillation, not a propagating mode carrying finite longitudinal power.

For a fixed transverse normal mode, define its cutoff angular frequencyThe longitudinal phase velocity and group velocity, for propagation in $+x$, areThus $v_p{v}_g=v^2$, $v_p>v$ 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 $n$ isAt cutoff $k_x=0$ and $v_g=0$; the longitudinal phase velocity has no finite value. Below cutoff,and a bounded field excited at $x=0$ decays as $e^{-{\gamma }_{n}x}$. 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.