Abstract
The well known simple explanation of the Herschel-Quincke interference tube is inadequate. It is herein shown that the ratio of the transmitted to incident acoustic energy is [4(sin(α3+α2)2)(cos(α3α2)2)]{[12cos(α3+α2)+cos(α3α2)]2+4sin2(α3+α2)}12 wherein α2 and α3 are the phase changes over the two branches of the tube. This ratio is zero not only when α3α2=(2n+1)π, as formerly explained, but also when α3+α2=2nπ, provided α3α22n1π, where n and n1 are independent integers.

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