Name: Relation Between Pressure and Speed of Sound Waves
Uploaded: 2019-07-19
Duration: 7 min 29 s
Description: Sound waves can be thought of as pressure waves because the regions of compressions and rarefactions have different pressures. Let's explore the speed and pr...

Question

Explain the formation of stationary waves in an air column enclosed in an open pipe. Derive the equations for the frequencies of the harmonics produced. A closed organ pipe

70 cm

long is sounded. If the velocity of the sound is

331 m / s

, what is the fundamental frequency of vibration of the air column

?

Accepted Answer

When a source of sound is kept at the open end of a pipe of an open pipe, compression and rarefactions are se up in the pipe. At the end a rarefaction is formed and returns as a compression. Thus, incident and reflected waves superimpose, and stationary wave is formed. An open pipe has antinodes at both ends. The wavelength of the wave is given by $λ = \frac{2 L}{n}$

Where $L$ = length of pipe $n$ = integer for mode number

First normal mode of vibration:

In this mode there will be two anti nodes and one node corresponding to $n = 1$ .

$λ_{1} = \frac{2 L}{1} or L = \frac{λ_{1}}{2}$ and frequency $ν = \frac{v}{λ_{1}} = \frac{v}{2 L}$

This is the lowest (fundamental frequency) or first harmonic.

Second normal mode of vibration:

In this case there will be three antinode and two nodes and the wavelength is corresponding to $n$ $= 2$ , the wavelength is given by:

$λ_{2} = \frac{2 L}{2} = L, frequency = ν_{2} = \frac{2 v}{2 L} = 2 ν_{1}$

This is double of the fundamental and called second harmonic or first overtone.

Third normal mode of vibration:

In this case there will be $4$ antinodes and $3$ nodes. The wavelength is given by

$λ_{3} = \frac{2 L}{3} or L = \frac{3 λ_{3}}{2} and frequency = ν_{3} = \frac{v}{λ_{3}} = \frac{3 v}{2 L} = 3 ν_{1}$

This is third harmonic or second overtone.

Thus, it is clear that both even and odd harmonics are produced in open organ pipe.

Given $:$

Length of the closed organ pipe, $l = 70 cm$

Velocity of sound, $v = 331 m s^{- 1}$

We know that,

The fundamental frequency of closed organ pipe is

$ν = \frac{v}{4 l}$

where $v$ is the velocity of sound, $l$ is length.

Substitute all the values in above equation, we get

$ν = \frac{331}{4 \times 0.7} = 118.21 Hz$

So, the fundamental frequency of vibration of the air column is $118.21 Hz$ .

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