THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

92 5. Vibrating BarsSettingc 2 = E/ρ (5.6)we now obtain the one-dimensional longitudinal wave equation:∂ 2 ξ∂t 2= c2 ∂2 ξ∂x 2 (5.7)Equation (5.7) corresponds to Equation (4.3) for the transverse motion of thestring, with the longitudinal displacement ξ assuming the role of the transversedisplacement y. We note that Equation (5.7) is identical to Equation (2.3), andwe have derived in this section the wave equation (5.7) that applies to acousticpropagation in a linearly elastic solid.5.3 Solutions of the Longitudinal Wave EquationThe format of the general solution to Equation (5.7) is identical with that of thesolution to Equation (4.3), i.e.,ξ = f (ct − x) + g(ct + x) (5.8)The square root of Equation. (5.6) gives us the wave propagation velocity c:√Ec =(5.9)ρwhich indicates that c is a property of the bar material.Let us write the solution (5.8) in the form of a complex harmonic solutionξ = Ae i(ωt−kx) + Be i(ωt+kx) (5.10)where A and B represent complex amplitude constants and k = ω/c is the wavenumber. We now assume that the bar is rigidly fixed at both ends; the boundaryconditions ξ(x, t) becomes ξ = 0atx = 0 and at x = L at all times t. Applyingthe condition ξ(0, t) = 0 yields A =−B, and Equation (5.10) revises toThe stipulation ξ(L, t) = 0 results inor equivalentlywhich means thatξ = Ae iωt(e−ikx −e ikx )e −ikt − e ikt =2 sin kLisin kL = 0= 0k n L = nπ, n = 1, 2, 3,...(5.11)