= Solution
Label the two displacements in cell $n$ by $u_n,v_n$, with cell length $2a$. Choose the $K$ spring within a cell and the $G$ spring between cells. The equations are
$$
m\ddot u_n=-(K+G)u_n+Kv_n+Gv_{n-1},\qquad
m\ddot v_n=-(K+G)v_n+Ku_n+Gu_{n+1}.
$$
A <Bloch wave> $(u_n,v_n)=(u,v)e^{i(2aqn-\omega t)}$ gives a two-by-two <eigenvalue> problem, whose determinant is
$$
(K+G-m\omega^2)^2-(K+Ge^{-2iaq})(K+Ge^{2iaq})=0.
$$
Thus the <alternating-spring chain dispersion relation> is
$$
\boxed{\omega_\pm^2(q)=\frac{K+G\pm\sqrt{K^2+G^2+2KG\cos(2aq)}}m.}
$$
Periodic boundaries require $e^{2iaqN}=1$, so $q=\pi j/(Na)$ modulo $\pi/a$. There are $N$ distinct wavenumbers in the <Brillouin zone> $-\pi/(2a)\leq q<\pi/(2a)$ and two branches, accounting for $2N$ modes. The minus branch is acoustic, with zero frequency and in-phase motion at the centre; the plus branch is optical, with opposite cell displacements.
Expanding the square root near zero gives
$$
\boxed{\omega_-(q)\sim a\sqrt{\frac{2KG}{m(K+G)}}|q|,\qquad
\omega_+(q)\sim\sqrt{\frac{2(K+G)}m}
\left[1-\frac{KG a^2q^2}{2(K+G)^2}\right].}
$$
At the zone boundary the square root is $K-G$, so the acoustic maximum is $\sqrt{2G/m}$ and the optical minimum is $\sqrt{2K/m}$. The frequency gap is therefore
$$
\boxed{\Delta\omega=\sqrt{2K/m}-\sqrt{2G/m}.}
$$
Quantizing each harmonic mode gives <phonons>, bosonic excitations of energy $\hbar\omega_\pm(q)$ and crystal momentum $\hbar q$ defined modulo a reciprocal lattice vector. Their <group velocity> is $d\omega/dq$; acoustic <phonons> carry long-wavelength sound, while optical <phonons> have a nonzero centre frequency. There is one longitudinal polarization per branch here, and <phonon> number is not conserved in thermal equilibrium.
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