 Solution

For a long periodic chain of length $\mathcal{L}=Na$, neighboring allowed signed wavevectors differ by $2\pi /\mathcal{L}$. Counting both signs once gives a state density per length $dk/\pi$ on the positive half-zone. Therefore the density per angular-frequency interval and per length isIt is zero outside this band, starts at $g(0)=1/(\pi v_s)=7.958\times 10^{-5}$ s/m and has an integrable inverse-square-root Van Hove singularity as $\omega \to {\omega }_m^{-}$. The state-count normalization isThus the scalar chain has one vibrational mode per 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.