= Solution
Substitute the traveling <normal mode> $u_j=u\exp[i(kja-\omega t)]$. Combining the opposite-neighbour phases gives
$$
m\omega^2=2K(1-\cos ka)+2G(1-\cos2ka),
$$
so the positive-frequency branch is
$$
\boxed{\omega_0(k)=2\sqrt{\frac{K\sin^2(ka/2)+G\sin^2(ka)}{m}},
\qquad |k|\leq\pi/a.}
$$
For positive <spring constants> this is a stable acoustic branch. At long wavelengths,
$$
\omega_0^2\simeq\frac{a^2(K+4G)}m k^2,\qquad
\boxed{c_s=a\sqrt{\frac{K+4G}{m}}.}
$$
The next-nearest-neighbour bond spans twice the distance, hence contributes the factor four. Relative to $G=0$, positive $G$ increases the sound speed by $\sqrt{1+4G/K}$. The chain still has one acoustic branch and no optical branch.
The group speed near $k=0$ and the maximum frequency need not change in the same way. The zone-edge value is always $\omega_0(\pi/a)=2\sqrt{K/m}$, because the next-nearest-neighbour sites then move together. For $G>K/4$ the frequency maximum moves inside the zone, with $\cos(ka)=-K/(4G)$; the small-$k$ sound speed nevertheless keeps increasing.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2013-further-neighbour-phonons-and-viscous-decay.png]
{title=One acoustic branch with further-neighbour springs, and three viscous-decay regimes}
{description=Original calculated dispersions for stated values of $G/K$, all normalized using the same $K,m,a$. The right panel shows a single normal coordinate with initial displacement one and initial <velocity> zero, for damping ratios $\gamma/\omega_0=0.2,1,2$. It illustrates that a nonoscillatory overdamped response does not have the same envelope as an underdamped <phonon>.}
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