= Solution
For positive <angular frequency>, the first three branches are
$$
\boxed{\omega_n(k)=c_m\sqrt{k^2+(n\pi/a)^2},\qquad n=1,2,3.}
$$
Each branch is even in $k$, has minimum $\omega_n(0)=n\pi c_m/a$, and starts with zero slope. At large $|k|$, all approach the straight asymptote $\omega=c_m|k|$. Their <group velocities> are $v_{g,n}=c_m^2k/\omega_n$, whereas their longitudinal <phase velocities> are $v_{p,n}=\omega_n/k$ for $k\ne0$, giving $v_{p,n}v_{g,n}=c_m^2$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2-b9-dispersion.png]
{title=Three lowest membrane-waveguide dispersion branches, with their cutoffs and common large-wavenumber asymptote}
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