= Solution
Use a uniform westerly <Barotropic Rossby wave> model on a local <beta plane>. Its dispersion is $\omega=Uk-\beta k/(k^2+l^2)$. A zonal wavenumber-two pattern at latitude $\phi$ has $k_2=2/(a\cos\phi)$, so stationarity requires
$$
\boxed{U=\frac{\beta}{k_2^2+l^2},\qquad\beta=\frac{2\Omega\cos\phi}{a},\quad\phi=45^\circ.}
$$
For the zonally varying, meridionally uniform idealization $l=0$, this is $U=\Omega a\cos^3\phi/2$. Meridional structure, deformation radius, mean-flow shear and full spherical geometry change the condition; the latitude and zonal wavenumber alone do not fix a unique observed jet speed.
For a shorter zonal perturbation $n>2$ with the same $l=0$ assumptions,
$$
\boxed{c_n=U\left(1-\frac4{n^2}\right)>0,\qquad c_{g,n}=U\left(1+\frac4{n^2}\right).}
$$
Its phase travels eastward relative to the stationary long wave, but westward relative to the air at speed $4U/n^2$; its packet propagates downstream faster than the mean flow in this idealization. For example $n=6$ gives $c_n=8U/9$ and $c_g=10U/9$.
Within the meandering flow, poleward parcel motion lowers relative vorticity and equatorward motion raises it through $D\zeta/Dt\simeq-\beta v$. Synoptic anomalies can reinforce or oppose the background ridges/troughs according to phase, exchange momentum with it, and transfer energy downstream. Linear superposition alone does not cause mutual growth; stronger interactions require shear, baroclinicity or nonlinear PV advection. No unique amplification rate follows from the supplied information.
\Image[/past-exam-of-the-atmospheric-physics-course-of-the-university-of-cambridge/2014/paper-1-ap2014-rossby-speeds.png]
{title=In the stated idealized model shorter Rossby-wave phases move east while packets propagate faster downstream}
{source=paper-1-ap2014-rossby-speeds.py}
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