A tube of certain diameter and of length $$48$$ cm is open at both ends. Its fundamental frequency of resonance is found to be $$320$$ Hz. The velocity of sound in air is $$320$$ $$m/s$$. One end of the tube is now closed, considering the effect of end correction, calculate the lowest frequency of resonance for the tube (in$$ $$Hz).

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Infinity Learn · Accepted Answer

The displacement curves of longitudinal waves in a tube open at both ends is shown in Figs. (a) and (b). Let D be the diameter of the tube. We know that antinodes occur slightly outside the tube at a distance $$0.3$$ D from the tube end.In case of open organ pipe, the distance between two antinodes isgiven by λ$$2=48+2$$×$$0.3D$$          …(i)We$$ have, λ$$=vn=32000320=100cmSubstituting$$ the value of λ in Eq. $$(i), we $$get1002=48+0.6D$$⇒$$50=48+0.6D$$⇒$$D=103cm=3.33cmIn$$ case of closed organ pipe, λ$$4=48+0.3$$×$$103=49$$ or λ$$=4$$×$$49=196cmHence$$ the lowest frequency is, $$f=$$νλ$$=32000196=163.3Hz$$

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