4.3 WAVE EQUATION SOLUTION

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Let us consider the solution of the wave equation for the case of sound

waves being generated by motion of a plane wall, as shown in Fig. 4-2.

The motion of the wall (at x ј 0) may be represented by:

XрtЮ ј Xm cos !t (4-29)

where Xm is the peak amplitude of motion, and ! ј 2_f is the circular

frequency for the motion. Using the complex notation, remembering that

84 Chapter 4

FIGURE 4-2 Sound wave generated by a vibrating wall. XрtЮ is the instantaneous

velocity of the wall.

Copyright © 2003 Marcel Dekker, Inc.

we are actually using only the real part, Eq. (4-29) may be written in an

alternative form:

XрtЮ ј Xm ej!t (4-30)

The resulting sound wave generated by the simple harmonic motion of

the plane wall should also have a simple harmonic form. Let us consider the

following solution form for the displacement:

_рx; tЮ ј рxЮ ej!t (4-31)

The quantity рxЮ is the amplitude function, which is dependent on the

x-coordinate only, for one-dimensional waves.

The spatial and time derivatives from Eq. (4-31) are found, as follows:

@2_

@t2 ј j2!2 рxЮ ej!t ј _!2 рxЮ ej!t (4-32)

@2_

@x2 ј

d2

dx2 ej!t (4-33)

If we make these substitutions into the wave equation, Eq. (4-14), we obtain

the following ordinary differential equation:

d2

dx2 ј _

!2

c2 рxЮ (4-34)

If we introduce the wave number, k ј !=c ј 2_ f =c ј 2_=_, we obtain the

final form of the equation to be solved:

d2

dx2 ю k2 ј 0 (4-35)

The general solution of Eq. (4-35) is:

рxЮ ј Ae_j!t ю B ej!t (4-36)

where A and B are constants of integration to be determined from the

boundary conditions. The general solution for plane waves is found by

making the substitution from Eq. (4-36) into Eq. (4-31):

_рx; tЮ ј Aejр!t_kxЮ ю B ejр!tюkxЮ (4-37)

The general solution involves two terms:

(a) One term containing р!t _ kxЮ, which represents a sound wave

moving in the юx-direction (positive x-direction), and

Transmission of Sound 85

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(b) Another term containing р!t ю kxЮ, which represents a sound

wave moving in the _x-direction (negative x-direction).

This behavior may be easily recognized if we consider the motion of

the peaks in the sound waves. Consider the term containing р!t _ kxЮ, for

which the real part is cosр!t _ kxЮ. One peak of the wave at an arbitrary

time to occurs at position xo, where:

!to _ kxo ј 0

A short time later (at t ј to ю _tЮ, the position x1 of the peak is described

by:

!рto ю _tЮ _ kx1 ј 0

If we subtract these two relations, we find the location to which the peak of

the sound wave has moved during the small time _t:

x1 ј xo ю р!=kЮ_t ј xo ю c_t

We note that x1 > xo, so the peak of the sound wave has moved in the

юx-direction. By going through a similar procedure, we may show that the

other part of the particle displacement expression, involving р!t ю kxЮ,

corresponds to a sound wave moving in the _x-direction.

For this example, suppose that the sound wave is transmitted into a

very large (infinite) space so that there is no sound reflected back to the

vibrating plate. In this case, we will have no sound wave moving in the

_x-direction back toward the plate. To achieve this condition, we must have

B ј 0.

At the surface of the vibrating plate, the fluid must follow the motion

of the plate:

_р0; tЮ ј XрtЮ ј Xm ej!t ј Aejр!t_kxЮjxј0 ј Aej!t (4-38)

Thus, the other constant of integration is A ј Xm. The final expression for

the instantaneous particle displacement caused by the vibrating plate is:

_рx; tЮ ј Xm ejр!t_kxЮ (4-39)

According to the complex notation convention, only the real part actually

represents the particle displacement:

_рx; tЮ ј Xm cosр!t _ kxЮ ј Xm cosЅ2_ft _ 2_рx=_Ю_ (4-40)

We note that the peak amplitude of the particle displacement is j_j ј Xm,

and the rms particle displacement is _ ј Xm=

ffiffiffi2p .

86 Chapter 4

Copyright © 2003 Marcel Dekker, Inc.

Let us evaluate the other quantities for a plane sound wave. The

instantaneous particle velocity may be found from the velocity–displacement

relation, Eq. (4-18):

uрx; tЮ ј

@_

@t ј j!Xm ejр!t_kxЮ ј !Xmejр!t_kxю_=2Ю (4-41)

Note that we have written j ј e j_=2. If we take the real part of Eq. (4-41), we

obtain the actual variation of the instantaneous particle velocity:

uрx; tЮ ј !Xm cosр!t _ kx ю _=2Ю (4-42)

The magnitude of the particle velocity is related to the magnitude of the

particle displacement:

juj ј !Xm ј !j_j ј 2_f j_j (4-43)

By comparing Eq. (4-42) and Eq. (4-40), we observe that the particle velocity

leads the particle displacement by         ј 1

2 _ radians ј 908.

Next, let us find the expression for the instantaneous acoustic pressure

by applying Eq. (4-20):

pрx; tЮ ј __oc2 @_

@x ј юjk_oc2Xm ejр!t_kxЮ (4-44)

We note that ! ј kc and j ј ej_=2:

pрx; tЮ ј _oc!Xm ejр!t_kxю_=2Ю (4-45)

Taking the real part of Eq. (4-45), we obtain:

pрx; tЮ ј _oc!Xm cosр!t _ kx ю _=2Ю (4-46)

If we compare Eq. (4-46) and Eq. (4-42), we observe that the acoustic

pressure and particle velocity are related:

pрx; tЮ ј _ocuрx; tЮ (4-47)

The characteristic acoustic impedance is defined by:

Zo ј _oc (4-48)

The magnitudes of the acoustic pressure and particle velocity for a plane

wave are related by the following:

j pj ј Zojuj (4-49)

We also note that the phase angle between the acoustic pressure and particle

velocity for a plane wave is       ј 08, or these quantities are in-phase.

Transmission of Sound 87

Copyright © 2003 Marcel Dekker, Inc.

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