 Solution

A mountain width is not an exactly specified sinusoidal wavelength. Taking its characteristic horizontal inverse length as $|k|\sim 1/a$ with $a=10\mathrm{km}$ gives $|k|\sim 10^{-4}{\mathrm{m}}^{-1}$. Since $N/U=10^{-3}{\mathrm{m}}^{-1}$, the stationary vertical wavenumber isThis is close to the long-horizontal-scale/hydrostatic estimate $2\pi U/N=6.28\mathrm{km}$. If instead the quoted width is identified with a sinusoidal wavelength ${\lambda }_x=10\mathrm{km}$, then $|k|=2\pi /{\lambda }_x$ and ${\lambda }_z\simeq 8.1\mathrm{km}$. Thus several kilometres, roughly $6$–$8\mathrm{km}$ under these two common width conventions, is the justified estimate; the terrain profile would fix its actual forcing spectrum.

For the inverse-width choice, both $k$ and $m$ are negative for upward energy radiation. The stationary phase lines $kx+mz=\text{constant}$ tilt upstream with increasing height, while energy propagates upward and slightly downstream. Numerically $\widehat{{\mathbf{c}}_g}\simeq (0.10,0.995)\mathrm{m}{\mathrm{s}}^{-1}$, with ray inclination about $84.3^{\circ }$. Intrinsic phase propagation points upstream and downwards; the ground pattern itself remains fixed. The original sketch below uses this explicit convention.

A ground observer may see stationary lenticular or wave clouds where ascent cools sufficiently moist air, with continuing air flow through a cloud whose position remains almost fixed. Descending regions can clear or warm. An aircraft crossing the pattern encounters alternating updraughts and downdraughts; gliders can exploit the lift. Strong amplitudes, shear or rotors may cause turbulence, but a linear gravity wave can also be smooth, and clear-air waves need not produce any visible cloud. Mountain height, humidity and nonlinear stability are not supplied, so neither a vertical-wind amplitude nor a definite atmospheric precipitation forecast follows.