= Solution
Use periodic <boundary conditions> temporarily to count the allowed <wavevectors>. Each state occupies $(2\pi)^3/V$ in reciprocal space. For one acoustic polarization, the spherical shell between $k$ and $k+dk$ contains
$$
dN=\frac{V}{(2\pi)^3}4\pi k^2\,dk.
$$
Both positive and negative <wavevector> directions are already included in that sphere; no further factor of two is needed. With the linear <dispersion relation> $k=\omega/v_s$,
$$
g_1(\omega)=\frac{dN}{d\omega}
=\frac{V\omega^2}{2\pi^2v_s^3}.
$$
A three-dimensional monatomic <crystal> has three acoustic branches. Under the stated common-speed approximation,
$$
\boxed{g(\omega)=\frac{3V\omega^2}{2\pi^2v_s^3}.}
$$
This is a <phonon density of states> per unit <angular frequency>, not per unit ordinary <frequency>. If longitudinal and transverse speeds differ, the isotropic expression becomes $g=V\omega^2(v_L^{-3}+2v_T^{-3})/(2\pi^2)$. In the <Debye model>, imposing $\int_0^{\omega_D}g(\omega)\,d\omega=3N$ sets the cutoff; the linear approximation is not continued to arbitrarily high <frequency> in a finite <lattice>.
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