Question

Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced.
A open organ pipe $85cm$ long is sounded. If the velocity of sound is $340m/s$, what is the fundamental frequency of vibration of the air column?

Answer 1

1)Formation of stationary waves can be seen in the provided diagram,antinodes will be created at the open ends and nodes are in between them.

Modes of Vibrations

a) For fundamental mode of vibrations,$L=\frac{{\lambda }_1}{2};$

$\therefore {\lambda }_1=2L$

$V={\lambda }_1{f}_1;$

$\therefore V=2L{f}_1----\left(1\right)$

b) For the second harmonic or first overtone,

$L={\lambda }_2$

$V={\lambda }_2{f}_2\therefore V=L{f}_2----\left(2\right)$

c) For the third harmonic or second overtone,

$L=3\times \frac{{\lambda }_3}{2}\therefore {\lambda }_3=\frac{2}{3}L$

$V={\lambda }_3{f}_3\therefore V=\frac{2}{3}L{f}_3----\left(3\right)$

From (1), (2) and (3) we get,

$f_1:f_2:f_3......=1:2:3:......$

i.e. for a cylindrical tube, open at both ends, the harmonics excitable in the tube are all integral multiples of its fundamental.

$\therefore Inthegeneralcase,\lambda =\frac{2L}{n},wheren=1,2,......$

$Frequency=\frac{V}{\lambda }=\frac{nV}{2l},wheren=1,2,......$

2) Using $V=2L{f}_1$ (Fundamental mode of vibration)

$f_1=\frac{340}{2\times 0.85}=200$ Hz