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Question

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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Solution

The displacement curves of longitudinal waves in a tube open at both ends and when closed at one end 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 is given by $\frac{\mathrm{\lambda}}{2}=48+2\times 0.3\mathrm{D}$ …(i)

We have, $\mathrm{\lambda}=\frac{\mathrm{v}}{\mathrm{n}}=\frac{32000}{320}=100\mathrm{cm}$

Substituting the value of $\mathrm{\lambda}$ in Eq. (i), we get

$\frac{100}{2}=48+0.6\mathrm{D}\Rightarrow 50=48+0.6\mathrm{D}\Rightarrow \mathrm{D}=\frac{10}{3}\mathrm{cm}=3.33\mathrm{cm}$

In case of closed organ pipe,

$\frac{\mathrm{\lambda}}{4}=48+0.3\times \frac{10}{3}=49\text{or}\mathrm{\lambda}=4\times 49=196\mathrm{cm}$

Hence the lowest frequency is, $\mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda}}=\frac{32000}{196}=163.26\mathrm{Hz}$

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