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

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

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

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134 7. Pipes, Waveguides, and ResonatorsThe impedance becomes infinite whenorψtan 2 kL + (ψ 2 − 1) tan kL − ψ = 0ψ = cot kLLet us briefly examine the situation for a constant driving force at x = 0. Thevanishing of Z n0 = f/u 0 connotes that the speed amplitude at the point of forceapplication (x = 0) is infinite, the condition for mechanical resonance. Converselythe input impedance reaching infinity means that the speed amplitude approacheszero, which describes the condition of antiresonance.To obtain the condition of resonance in a pipe driven at x = 0 and sealed with arigid cap at x = L,welet|Z nL /ρ 0 cS| approach infinity in Equation (7.4), givingZ n0=−icot kLρ 0 cSThe reactance becomes zero when cot kL = 0,k n L − (2n − 1)π/2 n = 1, 2, 3,...and so we obtain the set of resonant frequencies as for the forced-fixed string:f n = 2n − 14With the odd harmonics of the fundamental constituting the resonance frequencies,the driven closed pipe contains a pressure antinode at x = L and a pressure node atx = 0. This means that the driver must present a vanishing mechanical impedanceto the tube.cL7.4 The Open-Ended PipeConsider a pipe driven at x = 0 but open-ended at the other end x = L. The assumptionthat Z nL = 0atx = L (which would lead to resonances at f n = 1 / 2 nc/L)is not valid, because the open end of the pipe radiates into the surrounding air. Theappropriate condition is Z nL = Z r , where Z r is the radiation impedance at theopen end of the tube. Also, the presence of a flange at the open end affects theexit impedance. Consider the case of a flange at the open end of a circular pipe ofradius a. The flange is large with respect to the wavelength of the sound, which, inturn, is considerably larger than the tube radius (λ ≫ a). This situation resemblesa baffled piston in the low-frequency limit. From theory (Kinsler and Frey, 1962)Z nLρ 0 cS = (ka)2 + i 8 ka2 3 π(7.12)where the real component r = (ka) 2 /2 and the imaginary component ψ = 8ka/3πare both much less than unity and r ≪ ψ. Under these conditions the solution to

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