EL PAPEL DE LA ESCUELA EN LA PREVENCION

An absorber consists of two rigid resistive sheets separated by 2 inches and backed by a 4 inch air layer in front of a rigid wall. The normalized flow resistances of the sheets are θ₁= 1 and θ₂= 2, the latter being closest to the wall (i.e., 4 inches from the wall).

(a) What is the combined transmission matrix of the two sheets and the air layer?

(b) Calculate the normal incidence absorption coefficient.

(c) If the placement of the two sheets are interchanged, will that influence the absorption coefficient?

5. Transmission matrix for an air column Following the outline in the text, prove Eq. 4.117.

Chapter 5

The Wave Equation

5.1 Fluid Equations

In the introductory discussion in Chapter 3 we simply used the impulse-momentum relation to illustrate the basic idea involved in the dynamics of wave motion. To go further, the differential equations of fluid motion are more appropriate and we proceed accordingly.

The thermodynamic state of a fluid is described by three variables, such as pressure, density, and temperature and the motion by the three components of velocity. Thus, there is a total of six variables which have to be determined as functions of space and time to solve a problem of fluid motion. Therefore, six equations are needed.

They are conservation of mass (one equation), conservation of momentum (three equations, one for each component), conservation of energy (one equation), and one equation of state for the fluid.

In describing the motion of a fluid, we shall use what is known as the Eulerian description. The velocity and the thermodynamic state (such as pressure) at a fixed position of observation are then recorded as functions of time. Different fluid par-ticles pass the observer as time goes on. (In the Lagrangian description, the time dependence is expressed in a coordinate frame that moves and stays with the fluid element under consideration.)

5.1.1 Conservation Laws

The conservation of mass in the Eulerian description simply states that the net mass influx into a control volume, fixed with respect to the laboratory coordinate frame, must be balanced by the time rate of change of the mass within the volume. We consider first one-dimensional motion in the x-direction and let the velocity and density at x and time t be u(x, t) and ρ(x, t).

The mass flux j (x, t) = ρ(x, t)u(x, t) is the mass passing through unit area per unit time at x. Similarly, the efflux at x+
x is obtained by replacing x by x +
x in j.

Thus, the net mass influx to the control volume is j (x)− j (x +
x) = −9∂j/∂x)
x in the limit as
x→ 0.

149

Conservation of mass requires that the influx of mass must be balanced by the time rate of change of the mass ρ
x in the control volume. and we obtain

∂ρ/∂t= −∂j/∂x or ∂ρ/∂t + ρ∂u/∂x = 0. (5.1) In the last step, the term u∂ρ/∂x has been neglected in ∂j/∂x since it is much smaller (by a factor u/c) than ρ∂u/∂x. This follows, for example, if we express ∂ρ/∂x as ρκ∂p/∂x, where κ = 1/ρc²is the compressibility, discussed in Chapter 3. In a plane wave, u = p/ρc so that ∂u/∂x = (1/ρc)∂p/∂x. The ratio of the neglected term u∂ρ/∂x and ∂ρ/∂x is then seen to be of the order of u/c. The neglected term is of second order in the field variables (product of two first order perturbations) and the omission results in the linearized version of the equation.

To obtain the corresponding equation for conservation of momentum, we proceed in an analogous manner, replacing the mass flux by the momentum flux. The momentum density in the fluid is ρu and the influx into the control volume at x is G= (ρu)u. The corresponding efflux at the other side of the box is G(x+
x) = G(x)+(∂G/∂x)
x, making the net influx equal to−(∂G/∂x)
x.

There is a contribution also from the thermal motion which is expressed by the pressure in the fluid (recall that the pressure is of the order of ρc²which should be compared with the convective momentum flux ρu²). This results in a rate of momentum influx p(x) at x and an efflux p(x+
x) = p(x)+(∂p/∂x)
x at x +
x, making the net influx contribution from pressure (−∂p/∂x)
x.

The total influx is now (−∂p/∂x − ∂G/∂x)
x and this must equal the time rate of change of the momentum (∂ρu/∂t)
x contained in the box, i.e.,

ρ∂u/∂t= −∂ρu²/∂x− ∂p/∂x. (5.2) From an argument analogous to that used in the linearization of the conservation of mass equation, we find that the term ∂ρu²/∂x ≡ ∂G/∂x can be neglected in comparison with ∂p/∂x (in Chapter 10 on nonlinear aspects of acoustics it is retained) and that ∂ρu/∂t can be replaced by ρ∂u/∂t.

Thus, the linearized form of the momentum equation is

ρ∂u/∂t+ ∂p/∂x = 0. (5.3)

It is left as a problem to show that the omitted (nonlinear) terms are smaller than the linear by a factor of the order of u/c which normally is much less than one.

The momentum equation (5.3) contains the variables u and p and the mass equation (5.1) the variables ρ and u. The latter can also be expressed in terms of p and u since

∂ρ/∂t = (1/c²)∂p/∂t(recall that c²= dP /dρ = γ P /ρ).

Then, in terms of the compressibility κ = 1/ρc², the linearized form of Eq. 5.1 can be written

κ∂p/∂t+ ∂u/∂x = 0. (5.4)

The fluid equations 5.1 and 5.3 can readily be generalized to three dimensions. In the mass equation, the term ∂u/∂x has to be replaced by ∂ux/∂x+ ∂uy/∂y+ ∂uz/∂z which can also be expressed as div u.

The momentum equations for the three components of the velocity can be con-densed into a vector equation in which grad p= ˆx∂p/∂x + ˆy∂p/∂y + ˆz∂p/∂z, where ˆx, ˆy, ˆz are the unit vectors in the x, y, z directions. Thus, the linearized fluid equa-tions take the form

Acoustic equations κ ∂p/∂t= −div u

ρ ∂u/∂t= −grad p (5.5)

where we have introduced the compressibility κ= 1/ρc².

For harmonic time dependence, the corresponding equations for the complex amplitudes p(ω) and u(ω) are, with ∂/∂t → −iω,

iωκ p= div u (5.6)

iωρ u= grad p. (5.7)

5.1.2 The Wave Equation

From Eqs. 5.1 and 5.3, we can eliminate u by differentiating the first with respect to tand the second with respect to x to obtain a single equation for p,

∂²p/∂x²− (1/c²)∂²p/∂t²= 0. (5.8) In three dimensions, we differentiate the mass equation (5.5) with respect to t and take the divergence of the momentum equation (5.5). Then, with div grad p= ∇²p, it follows that

Acoustic wave equation

∇²p− (1/c²)∂²p/∂t²= 0 (5.9) which replaces Eq. 5.8. For harmonic time dependence this equation reduces to

∇²p(t )+ (ω/c)²p(t )= 0. (5.10) Plane Waves

The general solution to the one-dimensional wave equation is a linear combination of waves traveling in the positive and negative x-direction, respectively, and can be expressed as

p(x, t )= p+(t− x/c) + p−(t+ x/c), (5.11) where p₊ and p₋ are two independent functions. The validity of the solution is checked by direct insertion of this expression into Eq. 5.8.

For harmonic time dependence,

p(x, t )= A cos(ωt − kx − φ1)+ B cos(ωt + kx − φ2), (5.12) where A, B, φ1, and φ2are constants.

It follows from Eqs. 5.6 and 5.7 by eliminating u that the wave equation is valid also for the complex amplitude p(ω),

∇²p(ω)+ (ω/c)²p(ω)= 0. (5.13)

The general one-dimensional solution is then

Complex pressure amplitude. Plane wave solution

p(x, ω)= A e^ikx+ B e^−ikx (5.14) representing the sum of waves traveling in the positive and negative x-directions. The constants A and B are now complex, defining the magnitude and phase angles of the two waves. They are determined by the known complex amplitudes at two positions (boundary conditions).

The corresponding velocity field follows from the momentum equation (5.7) and is given by

Complex velocity amplitude. Plane wave solution

ρc u(x, ω)= A e^ikx− B e^−ikx (5.15) [ρc: Wave impedance. A, B: Complex constants. ω: Angular frequency. k= ω/c.]

The sum of several traveling waves of the same frequency, direction, and wave speed can always be represented as a single traveling wave. The complex amplitude of this wave is then the sum of the complex amplitudes of the individual waves.

Spherical Waves

So far we have been dealing with plane waves, such as encountered in a duct with rigid walls with a plane piston as a sound source. The next simplest wave is the spherically symmetrical. Such a wave, like the plane wave, depends only on one spatial coordinate, in this case the radius.

The prototype source of the spherically symmetrical wave is a pulsating sphere which takes the place of the plane piston for plane waves. At large distances from the source, a spherical wave front can be approximated locally as plane, and the relation between pressure and velocity is then expected to be the same as for the plane wave, i.e., p= ρcu, so that the intensity becomes I = (1/2)p²/ρc. The total emitted power from the source is then = 4πr²I, and it follows that the intensity decreases with distance as 1/r²and the pressure as 1/r. As will be shown below, this r-dependence of the pressure turns out to be valid for all values of r.

To proceed, we need to express the wave equation in terms of the radial coordi-nate r. To do that, we recall that if A is a vector, the physical meaning of div A is the

‘yield’ of A per unit volume, where the yield is the integral of the (outward) normal component of A over the surface surrounding the volume. Our volume element in this case is S(r) dr where S(r) is the spherical surface S(r)= 4πr². The outflow of A from

By direct insertion into the equation, we find that the solution for the pressure in an outgoing wave can be written in the form

p(r, t )= a

r p(a, t− t^), (5.17)

where p(a, t) is the pressure at r = a and t^ = (r − a)/c is the delay time of wave travel from a to r.

For harmonic time dependence, with p(a, t)= |p(a)| cos(ωt), we get p(r, t )= |p(a)|a

r cos[ωt − k(r − a)]. (5.18) The velocity field follows from ρ∂ur/∂t = −∂p/∂r. In the case of harmonic time dependence such that p(a, t)= |p(a)| cos(ωt) the velocity field becomes

u_r(r, t )=a The first term represents the far field and dominates at distances many wavelengths from the source, i.e., kr >> 1, where k= 2π/λ. It is in phase with the pressure field and is simply p(r, t)/ρc. The second is the near field which dominates for kr << 1.

With sin[ωt − k(r − a)] written as cos[ωt − k(r − a) − π/2], we see that this velocity lags behind the pressure by the phase angle π/2.

The complex amplitudes of pressure and velocity that correspond to Eqs. 5.18 and 5.19 are

Complex pressure and velocity amplitudes; spherical wave p(r, ω)= (A/r) e^ikr

ρc u_r(r, ω)= (A/r) e^ikr(1+ i/kr) (5.20) [k= ω/c. A = (p(a, ω)a exp(−ika)].

The complex velocity amplitude in this case is obtained from momentum equation

−iωρ ur(r, ω)= −∂p/∂r.

The constant A is now complex and incorporates a phase factor exp(−ika). In Section 5.1.2 we return to this problem in an analysis of the sound generated by a pulsating sphere in which case the velocity at the surface of the sphere rather than the pressure is given.

To obtain the complex amplitudes for an incoming rather than outgoing wave, we merely replace ikr by−ikr in Eq. 5.20.

5.1.3 Problems

1. Linearization

Show that the omitted terms in the linearization of the momentum equation (5.3) are of the order of u/c, where u is the particle velocity.