= Solution
With atomic displacement $u_j$, the harmonic potential energy is $\tfrac12K\sum_j(u_{j+1}-u_j)^2$. <Newton's second law> gives
$$
m\ddot u_j=K(u_{j+1}+u_{j-1}-2u_j).
$$
For a periodic chain of $N$ atoms, its <normal modes> are $u_j=Ue^{i(kja-\omega t)}$, with $k=2\pi s/(Na)$ and one representative per <Brillouin zone>. Substitution yields the <monatomic-chain dispersion relation>
$$
-m\omega^2=K(e^{ika}+e^{-ika}-2),\qquad
\boxed{\omega(k)=2\sqrt{\frac Km}\left|\sin\frac{ka}{2}\right|.}
$$
A real general displacement is a superposition of these <normal modes> and their complex conjugates. The $k=0$ mode is a rigid translation. At small $|k|$, the <acoustic phonon branch> is linear, with <speed of sound>
$$
\boxed{v=\lim_{k\to0^+}\frac{d\omega}{dk}=a\sqrt{\frac Km}.}
$$
For $0<k<\pi/a$, the <group velocity> is $a\sqrt{K/m}\cos(ka/2)$ and vanishes at the zone boundary. The dashed initial tangent in the sketch has slope $v$.
\Image[/past-exam-of-the-general-relativity-and-cosmology-and-condensed-matter-physics-paper-of-part-b-of-the-physics-course-of-the-university-of-oxford/2011-monatomic-chain-dispersion.png]
{title=Monatomic-chain dispersion and its sound-speed tangent}
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