= Solution
For a long periodic chain of length $\mathcal L=Na$, neighboring allowed signed <Wavevector>[wavevectors] differ by $2\pi/\mathcal L$. Counting both signs once gives a <Density of states>[state density] per length $dk/\pi$ on the positive half-zone. Therefore the <Phonon density of states>[density per angular-frequency interval] and per length is
$$
g(\omega)=\frac1\pi\left|\frac{dk}{d\omega}\right|
=\boxed{\frac1{\pi v_s\sqrt{1-(\omega/\omega_m)^2}}},\qquad0<\omega<\omega_m.
$$
It is zero outside this band, starts at $g(0)=1/(\pi v_s)=7.958\times10^{-5}$ s/m and has an integrable inverse-square-root <Van Hove singularity> as $\omega\to\omega_m^-$. The state-count normalization is
$$
\int_0^{\omega_m}g(\omega)\,d\omega=\frac{\omega_m}{2v_s}=\frac1a.
$$
Thus the scalar chain has one vibrational mode per <Chemical atom>[atom], with no extra factor of two for the already counted propagation directions. If the horizontal variable were ordinary frequency $\nu$, the density would be $g_\nu(\nu)=2\pi g_\omega(2\pi\nu)$. A finite chain has discrete lines instead of an actual infinite peak; the smooth divergent curve is the large-chain continuum description.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2016-monatomic-chain-density-of-states.png]
{title=Vibrational state density per length and angular-frequency interval}
{description=Original continuum state count including both wavevector signs. The upper-edge singularity is integrable and its normalization gives one scalar mode per atom.}
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