ABSTRACT
To explain the propagation of sound through a medium, we start with a 1dimensional situation.
Consider an innitesimal element of length dx and cross-section dS. Let's assume the element is rigid in all directions except for the x-direction. Also, let's dene a longitudinal compressive sound wave travelling in the x-direction. Suppose that the centre of the element is displaced by a distance u as a result of a sound pressure p. Then, the displacements of the boundaries are
and
, respectively. The dierence in volume, therefore, is u @u
@x
dx
u+
@u
@x
dx
dS = @u
@x dxdS: (4.1)
The volumetric strain
= "x + "y + "z = @u @x
dxdS
dxdS = @u
@x : (4.2)
By denition, the bulk modulus of elasticity is given by
= stress
strain =
p
= p
@u=@x (4.3)
or
p = @u
: (4.4)
p(u 12 dx) at one side of the element and p(u+ 12 dx) at the other side. Hence, the force acting on the element is expressed by
p(u+ 12 dx) p(u 12 dx) dS = @p@x dxdS: (4.5) This force accelerates the mass dxdS, with being the density. The acceleration equals the second time derivative of the displacement in the x-direction u. Therefore, the following balance must hold:
@p @x
dxdS = @2u
@t2 dxdS; (4.6)
which gives the equation of motion
@p
@x = @
2u
@t2 : (4.7)
Substituting (4.4) for p into (4.7) results in
@ 2u
@x2 = @
2u
@t2 (4.8)
or
@2u
@t2 =
@2u
@x2 ; (4.9)
which is the displacement wave equation of a periodic uctuation u in the xdirection at a speed
c =
r
: (4.10)
Taking the partial derivative with respect to x of (4.7) gives
@2p
@x2 = @
3u
@x @t2 ; (4.11)
whereas taking the second derivative with respect to t of (4.4) gives
@2p
@t2 = @
3u
@x @t2 : (4.12)
Combining (4.11) and (4.12) gives the linear wave equation
@2p
@t2 =
@2p
@x2 = c2
@2p
@x2 : (4.13)
For an adiabatic change in an ideal gas,
PV = constant; (4.14)
where P is the absolute pressure, V the volume of an element, and the ratio of specic heats. Dierentiating with respect to V gives
V dP
dV + PV 1 = 0: (4.15)
Dividing this through by V 1 gives
V dV
dP = P: (4.16)
The volumetric strain is
= dV V
(4.17)
and the stress associated with it is
p = dP: (4.18)
Hence, using (4.3), (4.16) can be written as
= P: (4.19)
Therefore, the speed of sound in air
c =
r
=
s P
: (4.20)
In an ideal gas,
PV = nRT ; (4.21) where n is the amount of substance, R is the gas constant, and T is the absolute temperature. Replacing n by mM and R by M
R, in which M is the molar mass and R the specic gas constant, we obtain
PV = m RT = V RT ; (4.22)
or P
= RT ; (4.23)
Thus, for an ideal gas,
c =
q RT : (4.24)
Consider the wave equation (4.13)
@2p
@t2 = c2
@2p
@x2 (4.25)
and a solution of the form
p = f1(ct x) + f2(ct+ x): (4.26)
Then, @2p
@t2 = c2 (f 001 + f
00 2 ) (4.27)
and @2p
@x2 = f 001 + f
00 2 : (4.28)
For the displacement equation (4.9), a similar solution is considered:
u = g1(ct x) + g2(ct+ x): (4.29)
If we concentrate on the wave travelling in the positive direction (progressive wave), described by g1, for the moment ignoring g2, at t = 0, u0 = g1(x). Also, at t = 1, u1 = g1(cx). The displacement u1 caused by p1 must have the same form as the displacement u0 caused by p0, only at a distance 1c further on. Hence,
g1(x0) = g1(c x1) (4.30) and thus
x0 = c x1 (4.31) or
x1 = x0 + c: (4.32)
For the wave travelling in the negative direction (regressive wave), described by g2, the corresponding equations are
g2(x0) = g2(c+ x1) (4.33)
and
x1 = x0 c: (4.34) Consider a plane single-frequency (monotonous) progressive wave, for which the pressure deviation from the ambient constant value is described by
p = p0 cos h ! t x
i = p0 cos (!t kx) : (4.35)
k = !
