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).

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 λ2=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, λ4=48+0.3×103=49 or λ=4×49=196cmHence the lowest frequency is, f=νλ=32000196=163.3Hz