 Solution

Let the nearest-neighbour spacing be $d$, so the primitive unit cell has length $\ell =2d$. Denote the two longitudinal displacements in cell $n$ by $u_n,v_n$, with spring $K$ within each cell and spring $K^{\prime }$ between cells. Hooke's law and Newton's second law giveFor normal modes $u_n=u{e}^{i(qn\ell -\omega t)}$ and $v_n=v{e}^{i(qn\ell -\omega t)}$, the two amplitudes satisfyIts vanishing determinant gives the alternating-spring chain dispersion relationThe minus sign is the acoustic phonon branch, with ${\omega }_{-}(0)=0$ and $u=v$ at the origin. The plus sign is the optical phonon branch, with $u=-v$ at the origin. For positive springs the acoustic phonon branch rises from zero to $\sqrt{2\min (K,K^{\prime })/m}$ at the Brillouin zone edge; the optical phonon branch falls from $\sqrt{2(K+K^{\prime })/m}$ to $\sqrt{2\max (K,K^{\prime })/m}$. Their long-wavelength limits areThus the leading frequency in the strong-bond limit of the alternating-spring chain is $\sqrt{2K/m}$, the internal stretching frequency of a pair with reduced mass $m/2$.