= Solution
Write $q=ka$ and use $2\sqrt{C_1/M}$ as the <frequency> unit. The nearest-neighbour chain and its next-nearest-neighbour extension have
$$
\boxed{\omega_1=2\sqrt{\frac{C_1}{M}}\,|\sin(q/2)|},\qquad
\boxed{\omega_2=2\sqrt{\frac{C_1}{M}}\left[\sin^2(q/2)+\frac12\sin^2q\right]^{1/2}}.
$$
These are each a single acoustic branch, repeated over reciprocal-lattice periods. There is no added optical branch merely because a longer-range spring is introduced. The cutoff $C_n=0$ for $|n|>1$ specifies the first chain as nearest-neighbour despite the source's repeated word in its description.
For small $|q|$, $\omega_1\simeq a\sqrt{C_1/M}|k|$ and $\omega_2\simeq a\sqrt{(C_1+4C_2)/M}|k|$. Thus
$$
\boxed{\frac{v_{s,2}}{v_{s,1}}=\sqrt3}.
$$
The maximum is not necessarily at the zone edge. Put $x=\sin^2(q/2)$, so $0\le x\le1$. The second squared dimensionless <frequency> is
$$
x+2x(1-x)=3x-2x^2=\frac98-2\left(x-\frac34\right)^2.
$$
It reaches its maximum at $x=3/4$, or $q=\pm2\pi/3$ in the first zone. The first chain peaks at $q=\pm\pi$. Therefore
$$
\omega_{1,\max}=2\sqrt{C_1/M},\qquad
\omega_{2,\max}=\frac3{\sqrt2}\sqrt{C_1/M},\qquad
\boxed{\frac{\omega_{2,\max}}{\omega_{1,\max}}=\frac3{2\sqrt2}=1.061}.
$$
At $q=\pm\pi$ both chains have the same <frequency>, but the second chain has already passed its interior maximum. The <next-nearest-neighbour monatomic-chain frequency maximum> is therefore crucial to a correct sketch. Periodic repetition gives the requested curves for all <wavevectors>.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2015-long-range-chain-dispersion.png]
{title=Nearest- and next-nearest-neighbour monatomic-chain dispersions in repeated zones; the second chain peaks inside the first zone}
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