Question

How are Stationary waves formed in closed pipes and open pipes? Explain the various modes of vibrations and obtain relations for their frequencies.

Answer 1

Part 1 )

The pipe open at one end and close at other end is known as closed pipe. If the waves with some frequency are sent through the closed pipe, the waves gets reflects from closed end. When the incident and reflected waves with same frequency and in opposite direction superimposed the stationary waves formed in the closed pipe.

Let, $l$ be the length of pipe $v$ be the velocity of sound.

The first harmonic will form only when there is a node at closed end and anti-node at open end of pipe. Here, the length of pipe is equal to one fourth of the wavelength $\lambda$.

$l=\frac{{\lambda }_1}{4}$

$\lambda =4l$ ...........(1)

The fundamental frequency ${\nu }_1$ will be given by:

${\nu }_1=\frac{v}{{\lambda }_1}$

${\nu }_1=\frac{v}{4l}$ .............from(1)

The third harmonic or first overtone will form only when there is a node at closed end and anti-node at open end of pipe. Here, the length of pipe is equal to three fourth of the wavelength. Hence,

$l=\frac{3{\lambda }_1}{4}$

$\lambda =\frac{4}{3}l$ ...........(2)

The third harmionic frequency ${\nu }_3$ will be given by:

${\nu }_3=\frac{v}{{\lambda }_3}$

${\nu }_3=\frac{v}{\frac{4}{3}l}$ .............from(2)

${\nu }_3=\frac{3v}{4l}$

Similarly, the next overtone will be fifth harmonic with frequency,

${\nu }_5=\frac{5v}{4l}$ .........(3)

The fundamental frequency is $v/4L$ and the higher frequencies are odd harmonics of the fundamental frequency, i.e. $3f_1:5f_1:7f_1...$,where $f_1$ is fundamental frequency.

Part 2)

For pipe copen at both the ends.

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.

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

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