 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 haveThese 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 ${\omega }_2\simeq a\sqrt{(C_1+4C_2)/M}|k|$. ThusThe maximum is not necessarily at the zone edge. Put $x={\sin }^2(q/2)$, so $0\le x\le 1$. The second squared dimensionless frequency isIt reaches its maximum at $x=3/4$, or $q=\pm 2\pi /3$ in the first zone. The first chain peaks at $q=\pm \pi$. ThereforeAt $q=\pm \pi$ 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.