Consider a linear chain of \(2N\) atoms with the same spacing \(a\) and force constant \(\beta \) but of two different masses \(m,M\), \(M>m\), placed alternately. Label the light atoms by even integers \(2n\), \(n=0,\ldots ,N-1\) and the heavy atoms by odd integers \(2n-1\), \(n=1,\ldots ,N\). Denoting their displacements from equilibrium by the variables \(U_{2n}\) and \(V_{2n-1}\), Newton’s law of motion gives \[ \begin {aligned} m\ddot {U}_{2n} &= \beta [V_{2n-1}+V_{2n+1}-2U_{2n}],\\ M\ddot {V}_{2n-1} &= \beta [U_{2n}+U_{2n-2}-2V_{2n-1}]. \end {aligned} \] Determine the modes of vibration of the system and sketch the resulting pair of dispersion curves. [Assume periodic boundary conditions \(U_0=U_{2N}\) and \(V_1=V_{2N+1}\).]
Question
Math 6420 • PDE
Fall 2004 • Question 2