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

146 7. Pipes, Waveguides, and ResonatorsThen,L ′ = L + 2(0.85a) = L + 1.7a (flanged out end)L ′ = L + (0.85 + 0.6)a = L + 1.5a (unflanged outer end) (7.57)Now consider the neck of the resonator to be fitted with an air-tight piston. Whenthe piston travels a distance δ, the volume of the cavity changes by V =−δ A.Then,ρρ=−V V= AδVand from the thermodynamic relationp = ρ 0 c 2 ppCombining the last two equations gives the following:p = ρ 0 c 2 Aδ(7.58)VThe force f = pAnecessary to execute the displacement is ρ 0 c 2 A2δ. The effectiveVstiffness S (from the spring formula f = Sδ)isS = ρ 0 c 2 A2(7.59)VThe fluid moving in the neck radiates sound into the surrounding medium in thesame manner as an open-ended pipe. For wavelengths much larger than the radiusof the neck, the radiation resistance [cf. Equations (7.12) and (7.16)] isR r = ρ 0 c k2 A 2(flanged)(7.60a)2πorR r = ρ 0 c k2 A 2(unflanged)(7.60b)4πThe sound wave impinging on the neck opening is represented as an instantaneousdriving force with a pressure amplitude P:f = APe iωt (7.61)We then can write the differential equation for the inward displacement δ of thefluid “plug” in the neck asm d2 δdt + R dδ2 rdr + Sδ = APeiωt (7.62)This last equation is analogous to that of a sinusoidally driven oscillator which hasanalogous solutions. Solution of Equation (7.62) gives a complex displacement δ.The real part of the driving force represents the actual driving force AP cos ωt,and the real part of the complex displacement represents the actual displacement.