= Solution
Write the real <displacement> as $U_n(t)=\operatorname{Re}[u e^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store <energy> in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical <energy> is
$$
\boxed{E=\frac12M\sum_n\dot U_n^2+\frac12\alpha\sum_n(U_n-U_{n+1})^2.}
$$
The printed complex <displacement> is a calculation device: squaring the complex <phasor> itself would not produce physical <energy>. The chain sketch in the accompanying figure shows <masses> at <equilibrium> spacing $a$, with longitudinal <displacements> along the chain.
The two spring <forces> on site $n$ give
$$
M\ddot U_n=\alpha(U_{n+1}+U_{n-1}-2U_n).
$$
Substituting the normal-mode <phasor> and cancelling its nonzero amplitude gives
$$
-M\omega^2=\alpha(e^{iqa}+e^{-iqa}-2),\qquad
\boxed{\omega^2=\frac{4\alpha}{M}\sin^2\left(\frac{qa}{2}\right).}
$$
The long-wavelength limit is acoustic, $\omega\simeq a\sqrt{\alpha/M}|q|$, and the first <Brillouin zone> ends at $|q|=\pi/a$. The maximum <angular frequency> is \b[$\omega_m=2\sqrt{\alpha/M}$].
Time averaging uses $\langle[\operatorname{Re}(ze^{i\omega t})]^2\rangle=|z|^2/2$. Consequently
$$
\frac{\langle T\rangle}{N}=\frac14M\omega^2u^2,\qquad
\frac{\langle V\rangle}{N}=\frac14\alpha u^2|1-e^{-iqa}|^2
=\frac12\alpha(1-\cos qa)u^2.
$$
The <dispersion relation> makes these two contributions equal, giving
$$
\boxed{\langle E\rangle/N=\frac14M\omega^2u^2+
\frac12\alpha(1-\cos qa)u^2=\frac12M\omega^2u^2.}
$$
This is equality of time-averaged kinetic and <potential energies> in a harmonic <normal mode>; the instantaneous <energy> partition need not be equal.
Use <periodic boundary conditions> on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi)$. Using <folded wave-number counting for a monatomic chain> combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression is
$$
\boxed{g(q)=\frac{2L}{2\pi}=\frac{Na}{\pi},\quad 0<q<\pi/a.}
$$
This is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.
For positive $q$, $\omega=\omega_m\sin(qa/2)$ and
$$
\frac{d\omega}{dq}=\frac a2\sqrt{\omega_m^2-\omega^2},\qquad
\boxed{g(\omega)=g(q)\left|\frac{dq}{d\omega}\right|
=\frac{2N}{\pi\sqrt{\omega_m^2-\omega^2}},\quad0<\omega<\omega_m.}
$$
The <density of states of a monatomic harmonic chain> vanishes outside the band, is constant to leading order near zero, and has an integrable <Van Hove singularity> at the upper band edge where <group velocity> vanishes. Normalization checks the directional counting:
$$
\int_0^{\omega_m}g(\omega)\,d\omega
=\frac{2N}{\pi}[\arcsin(\omega/\omega_m)]_0^{\omega_m}=N.
$$
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