$$\begin{aligned} Y\_i(\widehat{\mathbf{p}}\_i - \widehat{\mathbf{p}}\_r) &= \left(\sum\_{n=1}^N Y\_n\right) \widehat{\mathbf{p}}\_t \quad \text{where} \quad Y\_n = \frac{A\_n}{\rho\_n c\_n} \\ \text{and} \quad \widehat{\mathbf{p}}\_t &= \widehat{\mathbf{p}}\_i + \widehat{\mathbf{p}}\_r = \widehat{\mathbf{p}}\_1 = \widehat{\mathbf{p}}\_2 = \dots = \widehat{\mathbf{p}}\_n \end{aligned} \tag{10.107}$$

The amplitude reflection and transmission coefficients are the corresponding generalizations of Eq. (10.105).

$$R = \frac{\hat{\mathbf{p}\_r}}{\hat{\mathbf{p}\_i}} = \frac{Y\_i - \sum\_{n=1}^N Y\_n}{Y\_i + \sum\_{n=1}^N Y\_n} \quad \text{and} \quad T = \frac{\hat{\mathbf{p}\_t}}{\hat{\mathbf{p}\_i}} = \frac{2Y\_i}{Y\_i + \sum\_{n=1}^N Y\_n} \tag{10.108}$$

The power transmission coefficients must be different for different outlet tubes. For the nth outlet tube, h i Πt,n t ¼ ð Þ ½ ℜe bpnUb n h i ¼ ð Þ ½ Yn pt j j2 . Meanwhile, h i Πi t ¼ ð Þ ½ Yi bpi j j2 . We can use T ¼ bpt=bpi in Eq. (10.108) to write the power transmission coefficient into the nth tube:

$$T\_{\Pi,n} = \frac{\langle \Pi\_n \rangle\_t}{\langle \Pi\_i \rangle\_t} = \frac{Y\_n}{Y\_i} \frac{\left| \widehat{\mathbf{p}}\_\mathbf{n} \right|^2}{\left| \widehat{\mathbf{p}}\_\mathbf{i} \right|^2} = \frac{4Y\_i Y\_n}{\left( Y\_i + \sum\_{n=1}^N Y\_n \right)^2} \tag{10.109}$$

By using Eqs. (10.108) and (10.109), one can verify that energy is conserved to exhibit the reassuring result:

$$R\_{\rm II} + \sum\_{n=1}^{N} T\_{\rm II,n} = R^2 + \sum\_{n=1}^{N} T\_{\rm II,n} = 1 \tag{10.110}$$

An interesting application of the pressures in branching systems to the human cardiovascular system, in particular the iliac bifurcation of the aorta, is discussed by Lighthill [1].

#### 10.8.2 Tuned Band-Stop Filter

Based on the diagrams in Fig. 10.15, it appears that our interest is focused on the junction between semi-infinite pipes where we expect unidirectional propagation in the +x direction and an acoustic

admittance for right-going traveling waves, as expressed in Eq. (10.86) or Eq. (10.107), so that Yn ¼ A/ ρmc for all n 1. There is no such restriction on Yn. In the following application, we will use the formalism of Eq. (10.109) to calculate the transmission coefficient for sound that propagates along a duct that uses a Helmholtz resonator as a side branch to produce a band-stop filter that is shown schematically in Fig. 10.16.

The lumped element approach of Chap. 8 can be exploited to analyze this band-stop filter, also known as a "trap," that can be useful for suppression of a single frequency in ducts, like those produced by a fan's blade-passage frequency.

As calculated in Sect. 8.5.1, the series combination of an inertance and a compliance can have a vanishingly small input acoustical impedance when XC and XL cancel each other at the Helmholtz frequency, ωo. If a Helmholtz resonator is attached to a duct as a side branch, shown schematically in Fig. 10.16, all of the incident volume velocity is diverted to the branch leaving nothing to produce a transmitted pressure at the Helmholtz frequency through the duct beyond the branch to the right of the junction.

As shown in Fig. 10.16, a Helmholtz resonator that combines a compliance with internal volume, V, and a neck of cross-sectional area, AH, with an effective length (see Sect. 8.5.2), Leff, is connected to a duct of otherwise uniform cross-sectional area, A. This situation can be incorporated into Eq. (10.109) by letting N ¼ 2, with Yi ¼ Y1 ¼ Yt ¼ R1 ac ¼ A=ρmc and Y2 ¼ 1/jXH whereXH ¼ XL þ XC ¼ ωL (1/ωC) and ωo ¼ 1= ffiffiffiffiffiffi LC p .

$$X\_H = aL - \frac{1}{a\epsilon \mathcal{C}} = aL\left(1 - \frac{\alpha\_o^2}{\alpha^2}\right) = \alpha\_o L \left(\frac{\alpha}{\alpha\_o} - \frac{\alpha\_o}{\alpha}\right) \tag{10.111}$$

Substitution into Eq. (10.109) provides the power transmission coefficient.

$$\begin{split} T\_{\Pi,1} &= \frac{4Y\_i Y\_1}{|Y\_i + Y\_1 + \mathbf{Y}\_2|^2} = \frac{4Y\_i^2}{\left| 2Y\_i + \frac{1}{jX\_H} \right|^2} \\ &= \frac{1}{1 + \frac{\alpha^\*}{4X\_H^2}} = \frac{1}{1 + \frac{1}{4\alpha\_o^2 L^2} \left( \frac{\alpha}{\alpha\_o} - \frac{\alpha\_o}{\alpha} \right)^2} \end{split} \tag{10.112}$$

Introduction of an exponential time constant, τL/R ¼ L/Rac, can simplify and symmetrize the expression for the transmission coefficient.

Fig. 10.17 Power transmission coefficient, TΠ,1, for a duct with a Helmholtz resonator as a side branch. The solid line corresponds to(2ωoτL/R) ¼ 1.0, with the dashed line for (2ωoτL/R) ¼ 0.5 and the dotted line for (2ωoτL/R) ¼ 4.0

$$T\_{\Pi,1} = \frac{1}{1 + \frac{1}{\left(2a\_o \tau\_{L/R}\right)^2} \left(\frac{\alpha}{a\_o} - \frac{\alpha\_o}{a}\right)^2} = \frac{\left(2a\_o \tau\_{L/R}\right)^2 \left(\frac{\alpha}{a\_o} - \frac{\alpha\_o}{a}\right)^2}{\left(2a\_o \tau\_{L/R}\right)^2 \left(\frac{\alpha}{a\_o} - \frac{\alpha\_o}{a}\right)^2 + 1} \tag{10.113}$$

This transmission coefficient is plotted for three values of the dimensionless product of twice the Helmholtz resonance frequency times the exponential time constant, 0.5 (2ωoτL/R) 4.0, in Fig. 10.17, as a function of the frequency ratio, f/fo, on a logarithmic axis.

If the frequency is either much lower or much higher than the Helmholtz frequency, then the squared frequency ratio difference term in Eq. (10.113), [(ω/ωo) (ωo/ω)]2 , dominates, and nearly all of the power is transmitted past the branch: TΠ ffi 1. If ω ¼ ωo, then the frequency term in the numerator is zero, so no power is transmitted in the limit that the dissipation in the Helmholtz resonator can be neglected.

At the Helmholtz frequency, ωo, the energy conservation condition in Eq. (10.110) requires that all of the power be reflected. With the admittance of the Helmholtz resonator at resonance being infinite (in the absence of dissipation), the phase-inverted reflection from the junction is what would be expected since, from the left, the situation would be indistinguishable from the case where At Ai in Eq. (10.105).

#### 10.8.3 Stub Tuning

One final high-pass filter application can be very useful if steady flow needs to be removed from acoustic propagation through the duct. Sirens can have very high efficiency [62], but they require steady gas flow. That steady flow can be diverted from a duct while allowing the high-amplitude acoustic pressure wave to be transmitted through the duct if a thin membrane is placed across the duct and a vent tube that is one-quarter wavelength long is placed between the siren and the membrane.

If the membrane is sufficiently thin and flexible, the sound will pass through nearly unattenuated (see Sect. 11.1.1). If the "stub" is one-quarter wavelength long, at the siren's frequency, then it will act as a 1:1 transformer (see Sect. 3.8.1 and Fig. 3.10), so the nearly zero low-frequency acoustical impedance of the open end will present a nearly infinite acoustical impedance at the duct end for sound at the siren's frequency; all the sound goes down the duct, and all the siren's gas flow goes out the stub.

#### 10.9 Quasi-One-Dimensional Propagation (Horns)

Thus far, this chapter has examined propagating plane waves in ducts and standing plane waves in resonators, both having constant cross-sectional areas, A, that are assumed to have linear dimensions that are much smaller that the wavelength of sound, ffiffiffi A p λ. The spatial dependence of such waves has been specified by a single coordinate. In all cases, the evolution of the waves depended only upon the x coordinate, and all of the wave's acoustic variables (i.e., pressure, density, temperature, and particle velocity) have been uniform over planes that are perpendicular to the x axis. This geometry is responsible for the term "plane wave" that we use to designate such behavior.

In fact, the dependence on a single spatial coordinate is not unique to one-dimensional propagation. In Chap. 12, when the radiation of sound in a three-dimensional unbounded fluid medium is analyzed, spherically diverging sound waves are also characterized by a single spatial coordinate, the radial distance, r, from the omnidirectional source's acoustic center. In the nondissipative limit, energy is conserved so that the pressure amplitude at a distance, r, is inversely proportional to that distance. The integral of the time-averaged intensity, hIit ¼ h[p2 (r, t)]it/2ρmc, over any spherical surface of area, A (r) ¼ 4πr 2 , centered on the source, remains constant.

For such spherically symmetric wave propagation, all fluid particle motion is radial. There would be no fluid motion that would cross the boundary of any cone with its apex centered at the source. For that reason, any conical horn of infinite length, with rigid surfaces, would support identical wave motion within its constant apex angle.17 This argument should be reminiscent of the analysis of nodes of standing waves on pendula in Sect. 3.6.2, as well as arguments regarding membrane mode shapes that exploited nodal lines and nodal circles to calculate the behavior of those modes in Sect. 6.2.4 and Prob. 2 in Chap. 6 for membrane shapes that are not bounded by a rectangle or circle.

#### 10.9.1 Semi-infinite Exponential Horns

Bolstered by such arguments, it is attractive to extend the analysis of one-dimensional propagation in a duct of uniform cross-sectional area to horns that have a monotonic change in cross-sectional area with distance, A(x), which is a function of only the x coordinate, if we assume that the area changes slowly over distances comparable to the wavelength of the sound wave propagating through the horn.

Long before the development of electroacoustics and before an acoustical theory for horns existed, horns were recognized as an apparatus for concentrating sound energy and for improving the coupling between vibrating surfaces to the surrounding fluid medium. Several (amusing?) historical implementations are shown in Fig. 10.18. In the early days of electroacoustics, when the power

17 Within the cone, the pressure amplitudes would be larger than those of a spherically spreading wave if the source's volume velocity was the same in both cases. If the cone subtends a solid angle, Ω, then the pressures would be enhanced by a factor of 4π/Ω, assuming the additional load would not reduce the source's volume velocity.

Fig. 10.18 (Top left) Two horns from the author's personal collection. The "Triumph" straight horn, located behind the Edison Standard Phonograph, has a 2400 (61 cm) diameter opening that reduces to a ½" (1.3 cm) diameter at the apex for an area ratio of 2300:1. That horn is 3600 (91 cm) long. Placed on that phonograph is a very small solid stepped exponential horn that is 1¾00 (5.5 cm) long that was used by Prof. W. L. Nyborg to concentrate ultrasonic energy for streaming [63] and cellular (biological) cavitation experiments in fluids (courtesy of Prof. Richard Packard, a former Nyborg master's student). (Bottom left) Two curved Edison phonograph horns. (Top right) A horn-based aircraft detection and localization system used by the Imperial Japanese Army in 1936, known as the "Wartuba." (Center right) Another aircraft system used for stereo localization, photographed at Bolling Air Force Base, near Washington, DC, in 1921. (Bottom right) Personal listening device worn by German soldiers in 1917 that is combined with binoculars so that the aircraft can be seen as well as heard

available using vacuum tube audio amplifiers was limited and motion pictures added sound tracks, horns were required to increase the efficiency of loudspeakers to ensonify the large volumes of theaters and auditoria [64].

To formulate a quasi-one-dimensional model of horns, it is necessary to modify the linearized one-dimensional continuity Eq. (10.1) that was derived in Sect. 8.2.1, based on the geometry specified in Fig. 8.2. In analogy with Eqs. (8.10)–(8.12), the change in the mass of fluid within a differential slab of volume, dV ¼ A(x) dx, will be due to the difference in mass of fluid that enters A(x) with velocity along the x direction of u(x) and the mass that exits through an area, A(x þ dx), with velocity, u(x þ dx), as shown in Fig. 10.19.

$$\begin{aligned} \dot{m} &= (\rho\_m \mu A)\_x - (\rho\_m \mu A)\_{x+dx} = -\left(\frac{\hat{\mathcal{O}}(\rho\_m \mu A)}{\hat{\mathcal{O}}x}\right)\_x d\mathbf{x} \\ \frac{\partial \rho\_1}{\partial t} + \frac{\rho\_m}{A(x)} \left(\frac{\hat{\mathcal{O}}(u\_1 A)}{\hat{\mathcal{O}}x}\right) &= \frac{\hat{\mathcal{O}}\rho\_1}{\hat{\mathcal{O}}t} + \frac{\rho\_m u\_1}{A(x)} \left(\frac{\hat{\mathcal{O}}A}{\hat{\mathcal{O}}x}\right) + \rho\_m \left(\frac{\hat{\mathcal{O}}u\_1}{\hat{\mathcal{O}}x}\right) = 0 \end{aligned} \tag{10.114}$$

The linearized Euler Eq. (10.2) remains unchanged. The linearized equation of state (10.4) still relates adiabatic pressure and density changes, but the square of the sound speed here will now be subscripted,

co, to remind us that it is the equilibrium thermodynamic sound speed, c2 o ¼ ð Þ ∂p=∂ρ s , as distinguished from the phase speed, cph ¼ ω /k, since those two speeds will not be the same, as they were for the truly one-dimensional problem in Eq. (10.19).

$$
\rho\_m \frac{\partial \mu}{\partial t} + \frac{\partial \rho}{\partial \mathbf{x}} = \rho\_m \frac{\partial \mu}{\partial t} + c\_o^2 \frac{\partial \rho}{\partial \mathbf{x}} = 0 \tag{10.115}
$$

The second term in the continuity Eq. (10.114) involves the derivative, (1/A)(∂A/∂x) ¼ ∂ ln (A)/∂x (see Sect. 1.1.3), suggesting that a particularly simple solution to these coupled first-order differential equations might exist if the cross-sectional area of the horn varied as an exponential function of distance, x, along the horn's axis.

$$A(\mathbf{x}) = A\_o e^{2\mathbf{x}/h} \tag{10.116}$$

A flare constant, h, has been introduced to scale the rate of change in area from its initial area, Ao A (0). Using the same nondissipative argument for energy conservation made in the previous discussion of the conical horn, the amplitude of the acoustic pressure variation along the horn's axis must also decrease exponentially since the product, p2 (x)A(x), must remain constant. Since the horn is semiinfinite, we need only to consider wave motion that propagates in the + x direction.

$$p\_1(\mathbf{x}, t) = \Re e \left[ \hat{\mathbf{p}}\_\mathbf{o} e^{-\mathbf{x}/\hbar} e^{j(\alpha t - \mathbf{x})} \right] \text{ and} \\ \rho\_1(\mathbf{x}, t) = \Re e \left[ \frac{\hat{\mathbf{p}}\_\mathbf{o}}{c\_o^2} e^{-\mathbf{x}/\hbar} e^{j(\alpha t - \mathbf{x})} \right] \tag{10.117}$$

Our symbol for the wavenumber, κ, is chosen to remind ourselves that it has yet to be determined. The initial acoustic pressure phasor at the throat of the horn is bpo.

Substitution of Eqs. (10.116) and (10.117) into the coupled differential Eqs. (10.114) and (10.115) allows them to be converted to coupled algebraic equations.

$$\begin{aligned} j a \rho \rho\_m \hat{\mathbf{u}} - c\_o^2 \left(\frac{1}{h} + j \kappa\right) \hat{\mathbf{p}} &= 0\\ \rho\_m \left[\frac{2}{h} - \frac{1}{h} - j \kappa\right] \hat{\mathbf{u}} + j a \hat{\mathbf{p}} &= 0 \end{aligned} \tag{10.118}$$

As before, when seeking the dispersion relation, like that in Eq. (10.19), the determinant of the coefficients of the acoustic field variables, bρ andbu, must vanish if nontrivial solutions to Eq. (10.118) exist.

$$
\begin{vmatrix}
jao\rho\_m & -\frac{c\_o^2}{h}(1+j\kappa h) \\
\frac{\rho\_m}{h}(1-j\kappa h) & jo
\end{vmatrix} = 0\tag{10.119}
$$

We now can construct the necessary dispersion relation between wavenumber, κ, and frequency, ω.

$$\rho a^2 = \frac{c\_o^2}{h^2} + c\_o^2 \kappa^2 \quad \Rightarrow \quad \kappa = \pm \frac{a\nu}{c\_o} \sqrt{1 - \frac{c\_o^2}{\left(\alpha h\right)^2}} \Rightarrow \quad c\_{ph} = \frac{a\nu}{\kappa} = \frac{c\_o}{\sqrt{1 - \frac{c\_o^2}{\left(\alpha h\right)^2}}}\tag{10.120}$$

It is clear from Eq. (10.120) that the solution no longer corresponds to wave motion for frequencies below a cut-off frequency, ωco ¼ 2πfco ¼ co/h. The phase speed, cph, can be re-written in terms of the ratio of that cut-off frequency, ωco, to the drive frequency, ω.

$$c\_{ph} \equiv \frac{a\nu}{\kappa} = \frac{c\_o}{\sqrt{1 - \frac{a\nu^2}{a^2}}} = \frac{c\_o}{\sqrt{1 - \frac{f\_o^2}{f^2}}} \quad \text{with} \quad a\_{co} = 2\pi f\_{co} = \frac{c\_o}{h} \tag{10.121}$$

When κ becomes imaginary, the acoustic pressure in Eq. (10.117) decays exponentially, and the disturbance becomes more localized at the apex (throat) of the horn as the frequency decreases below ωco. For an infinite exponential horn in air that changes its diameter by a factor of e ¼ 2.72 over 1 m (i.e., h ¼ 1.0 m), the cut-off frequency, fco ffi 55 Hz.

We also neglected dissipation, so the transmitted power is independent of distance. That constant transmitted power can be calculated at the horn's apex by using Euler's Eq. (10.115) and Eq. (10.117) to calculate the gas particle velocity, buð Þ0 , and the associated volume velocity, Ubð Þ¼ 0 Að Þ0 buð Þ0 .

$$\hat{\mathbf{U}}(\mathbf{0}) = \hat{\mathbf{u}}A\_o = \frac{A\_o}{\rho\_m c} \left[ \sqrt{1 - \left(\frac{f\_{co}}{f}\right)^2} - j \left(\frac{f\_{co}}{f}\right) \right] \hat{\mathbf{p}}(\mathbf{0}) \text{ if } f > f\_{co} \tag{10.122}$$

The transmitted power, hΠit, depends on the time-averaged product of pressure and volume velocity (see Sect. 1.5.4).

$$
\langle \Pi \rangle\_t = \frac{1}{2} \Re e \left[ \hat{\mathbf{p}}(0) \hat{\mathbf{U}}^\*(0) \right] = \frac{\hat{\mathbf{p}}(0)}{2} \frac{A(0)\hat{\mathbf{p}}(0)}{\rho\_m c\_o} \sqrt{1 - \left(\frac{f\_{co}}{f}\right)^2} \text{ for} f > f\_{co} \tag{10.123}
$$

Using Eq. (10.122), the magnitude of the volume velocity at the apex, Ubð Þ0 , can be used to simplify Eq. (10.123) and compare the time-averaged power radiated down the horn to the power radiated by a simple source generating a spherically spreading wave due to a volume velocity at the surface of a spherical-pulsating source, Ubð Þa , given in Eq. 12.24.

$$\left| \hat{\mathbf{U}}(\mathbf{0}) \right| = \frac{\hat{\mathbf{p}}(\mathbf{0})A(\mathbf{0})}{\rho\_m c\_o} \left[ \left( \sqrt{1 - \frac{f\_{co}^2}{f^2}} \right)^2 + \left( \frac{f\_{co}^2}{f^2} \right)^2 \right]^{\prime \natural} = \frac{\hat{\mathbf{p}}(\mathbf{0})A(\mathbf{0})}{\rho\_m c\_o} \tag{10.124}$$

$$
\langle \Pi \rangle\_t = \frac{1}{2} \frac{\rho\_m c\_o}{A(0)} \left| \hat{\mathbf{U}}(\mathbf{0}) \right|^2 \sqrt{1 - \frac{f\_{co}^2}{f^2}} \tag{10.125}
$$

For a spherical source radiating in an infinite three-dimensional fluid, the time-averaged power can be expressed in a similar form that excludes the frequency dependence in the square root and substitutesλ2 /π for the horn's cross-sectional area at its apex, A(0).

$$
\langle \Pi \rangle\_t = \frac{1}{2} \frac{\rho\_m c\_o}{\left(\lambda^2/\pi\right)} |\mathbf{U}(a)|^2 \tag{10.126}
$$

This result shows that the horn can act as a transformer to increase the power radiated by a physically compact source of volume velocity above ωco since A(0)  λ2 /π. This is the reason that

horns are used routinely to improve the radiation efficiency of pistons with diameters that are much smaller than the wavelength of the sound they must produce. Since the propagation is reversible, it is also the reason that the horns, like those shown in Fig. 10.22 (right), were used to concentrate sound energy and deliver it to the ear of a hearing-impaired listener with increased amplitude (i.e., the "ear trumpet").

The power transmitted by a piston down a duct, h i Πduct t ¼ ð Þ ½ ρmco Ubð Þ0 2 , with the same crosssectional area as the piston's effective area, Apist, can be compared to the power transmitted down an infinitely long exponential horn with the same initial area, Apist ¼ A(0). As shown in Fig. 10.23, that ratio approaches unity monotonically for frequencies above cut-off (Fig. 10.20).

These results assume that a one-dimensional description of the horn used to derive the continuity Eq. (10.114) provides a sufficiently good approximation to the acoustic behavior of the medium exhibiting the wave-like behavior within the horn. Thus far, the conditions under which such an approximation might be valid have not been addressed. The surfaces of constant phase for a geometry like that depicted in Fig. 10.22 must intersect the horn's surface at a right angle. If the horn's crosssectional area changes too rapidly, then the wave fronts will not "cling" to the horn's surface [65]. For an infinite horn, the only physical parameters that characterize its geometry are the initial area, A(0), and its flare constant, h. For this quasi-one-dimensional analysis to be accurate, ro ffi ffiffiffiffiffiffiffiffiffi Að Þ0 p h, where ro is the effective radius of the horn's throat at x ¼ 0.

#### 10.9.2 Salmon Horns\*

As shown in Sect. 10.2, it is also possible to combine the first-order continuity and Euler equations with the equation of state to produce a wave equation for a horn in the quasi-one-dimensional limit that is a homogeneous second-order partial differential equation.

$$\frac{1}{A} \frac{\partial}{\partial \mathbf{x}} \left( A \frac{\partial p(\mathbf{x}, t)}{\partial \mathbf{x}} \right) - \frac{1}{c\_o^2} \frac{\partial^2 p(\mathbf{x}, t)}{\partial t^2} = \mathbf{0} \tag{10.127}$$

Our investigation of the exponential horn was motivated by the fact that (1/A)(∂A/∂x) ¼ ∂ ln (A)/∂x equaled a constant for the exponential change in cross-section that was assumed in Eq. (10.116). A more general family of solutions was suggested by Salmon [66] who required that the entire dependence of Eq. (10.127) on area be a positive constant.18

$$\frac{\partial^2 \left( p\sqrt{A} \right)}{\Im x^2} - \left( \frac{a}{c\_o} \right)^2 \left( 1 - \beta^2 \right) \left( p\sqrt{A} \right) = 0 \quad \text{where} \quad \beta^2 = 1 - \left( \frac{a c\_{co}}{a \nu} \right)^2 \tag{10.128}$$

Solutions to this time-independent Helmholtz equation are parameterized by two constants: the flare constant, h, and the constant, C, which controls the superposition of the two solutions to that secondorder differential equation.

$$\begin{aligned} A(\mathbf{x}) &= A(\mathbf{0}) \Big[ \cosh \left( \frac{\mathbf{x}}{h} \right) + C \sinh \left( \frac{\mathbf{x}}{h} \right) \Big]^2 \\ \text{and } p\_1(\mathbf{x}, t) &\propto \Re e \left[ \frac{\widehat{\mathbf{p}}(\mathbf{0})}{\sqrt{A(\mathbf{x})}} e^{j(\mathbf{w}t - \mathbf{x}\mathbf{x})} \right] \end{aligned} \tag{10.129}$$

As for any linear solution, the amplitude, bpð Þ0 , is arbitrary until the initial conditions at x ¼ 0 are specified. For C ¼ 1, the exponential horn is recovered. With C ¼ 0, the horn's shape is catenoidal. In the limit that h ! 1 and C ¼ (h/xo), the horn's shape is conical with an apex angle, ϕ ¼ tan1 (ro/xo), for the apex of the cone located at x ¼ 0 < xo and an initial area, Að Þ¼ 0 πr2 o, assuming the cone has circular cross-sections. For 0 < C < ffiffiffi 2 p , the relative transmission coefficient, Πhorn/Πduct, plotted in Fig. 10.23, has a maximum at ( f/fco)max. Under that condition, with ( f/fco)>( f/fco)max, the relative power transmission coefficient asymptotically approaches Πhorn/Πduct ¼ 1.

$$\left(\frac{\Pi\_{\text{hom}}}{\Pi\_{\text{dim}}}\right)\_{\text{max}} = \frac{1}{2C\sqrt{1-C^2}} \quad \text{for } 0 < C < \sqrt{2} \text{ at } \left(\frac{f}{f\_{co}}\right)\_{\text{max}} = \sqrt{\frac{1-C^2}{1-2C^2}}\tag{10.130}$$

For C ¼ 0.5, (Πhorn/Πduct)max¼ 1.15 at ( f/fco)max ¼ 1.22, providing fairly uniform transmission above the cut-off frequency.

#### 10.9.3 Horns of Finite Length\*

Our analyses of horns of infinite length (i.e., no reflected wave) have introduced several useful concepts, particularly the existence of the low-frequency cut-off, ωco, but real horns are never infinitely long. Finite-length horns fall into two general categories: (i) The horns that terminate musical instruments depend on reflection from the "bell" to define the standing wave that determines the instrument's pitch. (ii) The horns used to couple sources to the surrounding space try to avoid resonances that reduce the uniformity of radiated sound as a function of frequency above cut-off.19

Horns of finite length will exhibit resonances at frequencies above cut-off, at least until the radiation impedance at the bell is sufficient to couple most of the energy out of the horn, thus nearly eliminating

18 If β 2 is negative, then there is a family of sinusoidal horn shapes (e.g., a globe terminated in a cusp) that describe the shape of the bell of the flute commonly associated with Indian snake charmers or the English horn, first used in the by Rossini, in 1829, in the opera, William Tell. [See B. N. Nagarkar and R. D. Finch, "Sinusoidal horns," J. Acoust. Soc. Am. 50(1) 23–31 (1971).]

19 A nonuniform power transmission coefficient does not necessarily reduce the value of guided-wave enhancement of the coupling to an electrodynamic transducer, since human perception of low-frequency musical content is not particularly sensitive to such a nonuniform response. The best example of such a psycho-acoustic tolerance may be the success of the Bose Wave™ radio that employs a long serpentine duct that is driven by the rear of the forward-radiating speakers as shown in US Pat. No. 6,278,789 (Aug. 21, 2001).

the back-reflected wave from the bell. Although an exact calculation will be postponed until Sect. 12.8.3, a reasonable rule-of-thumb is that the resonances will be suppressed once the circumference of the bell exceeds the wavelength of the sound.

At reasonably high frequencies, horn-coupled loudspeakers can both be efficient and provide fairly uniform radiation over a significant range of frequencies. Above 1.0 kHz, λ < 35 cm, so the bell of a horn that is only 10 cm in diameter will provide acceptable performance (see Problem 17). The engineering design of such horn-coupled electrodynamic loudspeakers (commonly called compression drivers) is beyond the scope of this treatment and can involve considerations of directionality, the compliance of the space between the driver's diaphragm and the start of the horn, as well as the nonlinear distortion [67, 68] that can be produced due to the very high acoustic pressures near the throat. Several textbooks with a greater focus on audio engineering provide detailed guidance [69, 70].

Resonances in horns are particularly pronounced at low frequencies. The lowest note on a double bass or an electric bass guitar is usually E1 ¼ 41.2 Hz, based on A4 ¼ 440 Hz. The corner horn shown in Fig. 10.21 has a catenoidal shape with a flare constant, h ¼ 1.37 m, selected to make fco ¼ co/ 2πh ¼ 40 Hz. It is driven by an Axon Model 6S3 direct radiator loudspeaker. The front of that speaker radiates midrange frequencies, and the rear of the speaker drives the horn.

Also shown in Fig. 10.21 (right) is a tweeter for high frequencies with a passive cross-over network attached to the triangular top of the horn. The resonance frequencies and corresponding mode shapes were calculated using a DELTAEC model, shown in Fig. 10.22, that represent the catenoidal horn as eight CONE segments driven at constant voltage by the electrodynamic speaker with parameters listed in the VSPEAKER segment (#1). The bell of the horn is terminated with an OPNBRANCH segment (#11) that simulates radiation loading.

Fig. 10.21 (Right) Corner horn (this horn design, suggested by I. Rudnick, is particularly simple because it involves cutting a single sheet of plywood with a hand-held circular saw that has its blade set at 45, creating a smooth cut that will produce a leak-tight seal against the walls of a room. Even though the plywood is fairly thin (3/800 ffi 9.5 mm), it produces a rigid boundary since it is curved initially along one axis making it very stiff against bending along an orthogonal direction. The plywood is held against the walls with a single turnbuckle anchored in the corner. The place where the turnbuckle should be attached to the plywood is found by applying force to one location that pushes the plywood smoothly against the wall. A horizontal "stringer" is then screwed and glued to the plywood to accept the hook. The stringer's position is visible in Fig. 10.21 due to the six flat-head screws that can be seen forming a line about one-sixth of the way above the bell.) created by a single sheet of plywood that is cut to produce a catenoidal change in cross-sectional area from 173 cm2 at the top to 0.152 m2 at the opening over a length of 80 ffi 2.4 m. (Left) Plots of the acoustic pressure (solid black line) that is in-phase with the volume velocity (dashed blue line) within the horn created when the speaker, at the throat, is driven with an electrical input of 10 Vpk ¼ 7.07 Vac. At x ¼ 0, the volume velocity is equal to that produced by the rear of the loudspeaker. From top to bottom: f1 ¼ 77 Hz, f2 ¼ 139 Hz, f3 ¼ 204 Hz, and f8 ¼ 618 Hz

#### Exercises

	- (a) Acoustic density, ρ1(x, t)

Fig. 10.22 Screenshot of a DELTAEC model for the corner horn shown in Fig. 10.21 (right) that calculated the resonance frequencies and produced the mode shapes in Fig. 10.21 (left). The resonance condition was imposed by the RPN target (Seg. #2) that seeks a condition where the volume velocity and pressure are in-phase at the horn's throat

	- (a) Acoustic density, ρ1(x, t)
	- (b) Acoustic particle velocity, v1(x, t)
	- (c) Kinetic energy density, (KE)
	- (d) Potential energy density, (PE)
	- (e) Time-averaged intensity,hIit
	- (a) Time-averaged intensity, hIit
	- (b) Intensity level, re: 1.0 x 1012 W/m2
	- (c) Peak particle speed, v1
	- (d) Peak-to-peak particle displacement, 2x1
	- (e) Peak acoustic temperature change, T1

$$\text{in the }\star\text{x direction:}\\h(\mathbf{x},t) = h\_o + \left|\widehat{\mathbf{h}}\right|\sin\left(\alpha t - k\mathbf{x}\right).$$

$$\frac{\partial h\_1(\mathbf{x}, t)}{\partial t} + h\_o \frac{\partial \nu\_1(\mathbf{x}, t)}{\partial \mathbf{x}} = 0 \tag{10.131}$$

$$\frac{\partial \nu\_1(\mathbf{x}, t)}{\partial t} = -\frac{1}{\rho\_m} \frac{\partial}{\partial \mathbf{x}} [\rho\_m g h\_1(\mathbf{x}, t)] = -\mathbf{g} \frac{\partial h\_1(\mathbf{x}, t)}{\partial \mathbf{x}}\tag{10.132}$$

	- (a) Hydrostatic pressure. If the transducer is located 50 m below the surface, what is the static water pressure, pm, remembering that the pressure on the surface is one atmosphere, po ¼ 101.3 kPa.
	- (b) Cavitation level. What is the intensity level, re: 1 μParms, of a pressure wave that would create a peak negative pressure,j j bp ¼ pm, and cause an instantaneous rupture in the water?
	- (a) Approximate sound speed change. Get an approximate result by ignoring the change in γ.
	- (b) Polytropic coefficient. Use γair ¼ 1.4 and γCO2 ¼ 1.3. Will the change in sound speed due to the change in γmix have the same sign as that due to the change in Mmix?
	- (c) Sound speed change. Use Eq. (10.25) to calculate the change in γ, and use that result to obtain a more accurate result for the change in sound speed.

20 Fritz Haber was also one of the world's most infamous chemists, having developed the gas that was used to murder prisoners in the Nazi death camps.

of 1.5%. For the purposes of this problem, you may take the mean molecular mass of air to be Mair ¼ 29.0 gm/mole and the mean molecular mass of hydrogen to be Mhydrogen ¼ 2.0 gm/mole. Both gases can be treated as consisting primarily of diatomic molecules, and the temperature in the garage can be taken as 10 C.

$$\frac{1}{\rho} \left( \frac{\mathfrak{d}p}{\mathfrak{d}\left(\mathbb{1}\backslash\_{\rho}\right)} \right)\_{s} = -\rho \left( \frac{\mathfrak{d}p}{\mathfrak{d}\rho} \right)\_{s} \tag{10.133}$$

	- (a) Bulk modulus of water. If the density of water is ρ Water ¼ 1000 kg/m3 and the sound speed in the water is cWater ¼ 1500 m/s, what is the value of the bulk modulus of water?
	- (b) Effective bulk modulus of a PVC pipe. Consider a PVC pipe with a mean radius, a ¼ 8.0 cm, and a wall thickness, t ¼ 7.0 mm. Simple elasticity theory relates the change in the pipe radius, Δr, to the increase of pressure, Δp, within the pipe.

$$
\Delta r = \frac{a^2}{tE} \Delta p \tag{10.134}
$$

For PVC, the Young's modulus, EPVC ¼ 3.40 x 109 Pa. Consider a short length of this pipe, and calculate the effective bulk modulus by calculating the change in volume, ΔV, of the pipe due to a change in the internal pressure, Δp, within the pipe.

To test this device, it will be placed in a gas-tight cylindrical cavity with a volume of 2.5 liters ¼ 2.5 103 m3 . Assume the mean pressure in the cavity pm ¼ 101 kPa, the mean temperature is Tm ¼ 300 K, so that cair ¼ 347 m/s and γair ¼ 1.403. Under those conditions, ρair ¼ 1.173 kg/m3 , the specific heat of the air is cp ¼ 1005 J/kg-K, the viscosity of air is μ ¼ 1.85 105 Pa-sec, the air's thermal expansion coefficient at constant pressure is βp ¼ (Tm)– 1 ¼ 3.33 103 K1 , and the air's thermal conductivity is κ ¼ 2.62 102 W/m-K.

$$\beta = \frac{1}{T\_m} = \frac{1}{V} \left(\frac{\Im V}{\Im T}\right)\_p \tag{10.135}$$

What is the change in the volume of the gas, δV, caused by its heating during one-quarter cycle?

Fig. 10.25 A standing wave resonator used for reciprocity calibration is capped at each end with the small, reversible, electrodynamic transducers shown in Fig. 10.26. Two ½ inch (1.27 cm) compression fittings are located very close to the ends of the resonator. An Extech™ Model 407736, Type-2 sound level meter is placed in one fitting, and the other is plugged with a solid rod. The small tube at the center allows gases other than air to be used as the calibration medium. At the top of the photo is a custom switch box that allows the swept-sine signal produced by a dynamic signal analyzer to be routed to either reversible transducer #1 or #2 after it has been passed through a precision 1.000 ohm current-sensing resistor

Assuming that the two microphones can be treated as rigid caps and that all dimensions of the "Closed Air Volume" are much smaller than the acoustic wavelength, write an expression for the acoustic transfer impedance Ztr ¼ bpR=UbT , where bpR is the complex pressure amplitude at the "Receiver Microphone," UbT is the volume velocity produced by the "Transmitter Microphone," and the volume of the "Closed Air Volume" is V.

16. Reciprocity calibration in a plane wave resonator. The apparatus shown in Fig. 10.25 is used to produce a reciprocity calibration of the two reversible electroacoustic transducers shown in

Fig. 10.24 A small cavity is created in the space between two reversible condenser microphones

Table 10.5 Summary of measured results for the reciprocity calibration of the two transducers shown in Fig. 10.26 produced using the resonator shown in Fig. 10.25

Fig. 10.29. The resonator tube has an inside diameter of 34.37 mm, and its length is 70.12 cm. Assume the calibration procedure took place in dry air at 20 C and at a mean pressure, pm ¼ 98 kPa.

The data presented in Table 10.4 were produced at the resonance frequency of the second plane wave resonance, f2, by first driving Transducer #1 with current, i1, and measuring Transducer #2's open voltage, V2, along with the C-weighted output of the sound level meter, Vm,1, while also measuring Q2,1. The second data set in that table was obtained when Transducer #2 was driven. Using the dimensions of the resonator, the thermophysical properties of the air and the data in

Table 10.5 make the following calculations:

#### References

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## Reflection, Transmission, and Refraction 11

#### Contents

In Chap. 10, we transitioned from a lumped-element perspective to a formalism which treated fluids as continuous media that include the effects of both compressibility and inertia throughout. We described disturbances from equilibrium using a wave equation and focused attention on planewaves that could be described as propagating along one spatial dimension. In this chapter, we will examine the behavior of such one-dimensional waves propagating through media that are not homogeneous. We start with an examination of the behavior of planewaves impinging on a planar interface between two fluid media with different properties and then extend that analysis to multiple interfaces and to waves that impinge on the interface from an angle that is not perpendicular to that surface. This exploration concludes with consideration of wave propagation through a medium whose properties change slowly and continuously through space.

What does "slowly" mean in the previous sentence? To answer that question, we will break down the general problem of propagation through an inhomogeneous medium into two limiting cases. As before, such limiting cases suggest appropriate (usually simple) analytical approaches that will develop useful intuition for understanding cases that may be intermediate between the simpler limits. In this chapter, the interesting limits depend on the scale of the medium's inhomogeneity relative to the wavelength of the sound that impinges on it (from afar, since we consider only planewaves). If a wave encounters a deviate region with a size that is on the order of its own wavelength or smaller, and that region has a different density, a different compressibility, or both, then we treat the inhomogeneous region as a "scatterer." Those cases will be examined in Chap. 12 when the problem of radiation in three dimensions is analyzed and the scattering body will be driven by the impinging sound wave causing the ensonified region to behave as a radiating "source." That "scattered" sound field will be superimposed on the incident sound field.

In this chapter, we will consider the opposite limit, where the size of the boundary that separates regions with different acoustical properties is much larger than the wavelength of the sound. In fact, many cases to be examined here will assume that the extent of the boundary is infinite. In those cases, the problem is treated as one where the wave incident on such an interface will be both reflected back into the medium from which it originated and be transmitted into the second medium on the other side of the interface.

For the case where the interface is not discontinuous, what we will mean by requiring that the medium's properties (e.g., sound speed and density) "are varying slowly" will again be related to the rate at which the property changes in space relative to the wavelength of the sound. For example, if we specify the change in sound speed with position,dc/dz g, then g will have the units m/s per m, which is equivalent to a frequency. The wave also has a characteristic frequency, f ¼ c/λ, that is the ratio of the sound speed to the wavelength. If g f, then any significant change in sound speed will occur over a very large number of wavelengths.

#### 11.1 Normal Incidence

All of us are familiar with echoes, whether produced by a handclap reflected from a large building; a loud "hello" shouted from a precipice and reflected from a bluff, as diagrammed in Fig. 11.1; or a "flutter echo" produced by some impulsive sound source reverberating between long parallel walls. In those cases, a sound wave in air impinges on a rigid solid surface. The air's particle velocity that accompanies the pressure wave (via the Euler equation) cannot penetrate the solid. To satisfy this rigid boundary condition, we can imagine a sound wave of equal amplitude, but propagating in the opposite direction, coming from within the solid toward the interface, just as was done when examining the reflection of a pulse propagating along a string in Sect. 3.2. As justified in Sect. 12.4.1, the superposition of the two waves cancels the velocity at the interface, satisfying the condition that the interface is impenetrable to the gas, while doubling the acoustic pressure amplitude on that surface.

To compare how a real wave reflects in the "rigid boundary" case (and to determine how "rigidly" the boundary behaves), let's say the magnitude of the incident planewave, bpi j j , approaching the boundary is 94 dBSPL ) bpi j j ¼ ffiffiffi 2 p Pa ¼ 1:0 Parms . Again, we focus on a single-frequency wave and indicate the incident wave's amplitude and phase by the complex phasor, bpi. The time-averaged intensity of a 94 dBSPL planewave in air is h i Iair t ¼ bpi j j2 =ð Þ 2ρc air ¼ 10ð Þ 94=10 1012 W=m2 ffi 2:5 mW=m2. (If this is not instantaneously obvious, you need to review Sect. 10.5.1.)

Let's also say the wave was reflected from a concrete wall. Since the pressure amplitude at the wall is double that of the sound wave far from the wall, and that the pressure is continuous across the interface, we can calculate the intensity of the sound that entered the wall. The density of concrete is about 2600 kg/m3 and the speed of compressional waves in concrete is about 3100 m/s, so (ρc)concrete ¼ 8.06 MPa-s/m. Since the pressure at the wall is twice that of the wave in air, bpt j j ¼

Fig. 11.1 The sound produced by this person is nearly a planewave (wave fronts shown in blue), characterized by the wavenumber, k ! i, when it is reflected by a large flat rigid surface. The duration of the sound pulse is sufficiently short that the reflected wave, characterized by k ! r, and the incident wave only interact near the surface. To satisfy the condition that no fluid can enter or leave the solid interface, we imagine another planewave of equal amplitude traveling back (with wave fronts shown in *red*) toward the source. Where the incident and reflected wave superimpose at the rigid surface, the pressure is doubled, but the slope of the pressure, (dp/dx)x ¼ 0, evaluated at the interface, is zero. By Euler's equation, the total acoustic particle velocity normal to the surface vanishes, as required

2 bpi j j ¼ 2 ffiffiffi 2 p Pa, the time-averaged intensity of the sound penetrating the concrete is h i Iconcrete t ¼ bpt j j2 =ð Þ 2ρc concrete ¼ 0:5μW=m2, less than 0.02% of the intensity of the sound in air.

The equations developed in this chapter can be used to show that the amplitude of a wave in air that is reflected from a concrete wall is nearly the same as the incident wave, but not quite. Our assumption that the echo had the same amplitude as the incident sound wave was just as solid as the concrete.

We can also compare the particle velocity of the sound in air with that of the compressional wave within the concrete. In dry air, a 94 dBSPL sound wave would have a particle velocity amplitude of bvi j j ¼ bpi j j=ð Þ ρc air ¼ 3:3 mm=s. In the concrete, the pressure of the transmitted wave is twice that in air (far from the wall), but bvt j j ¼ 2 bpi j j=ð Þ ρc concrete ¼ 0:35 μm=s. Again, we see that our assumption of a rigid and immobile boundary was good to about one part in ten thousand (0.01%). We can check our intensity results using Eq. (10.36), since we also know h iI t ¼ bpt j j bvt j j=2, since bpt and bvt are in-phase for our assumed traveling wave. In dry air at one atmosphere, pm ¼ 101,325 Pa, and letting Tm ¼ 7 C, (ρc)air ¼ 423 Pa-s/m. This gives hIairit ¼ 2.4 mW/m2 and in the concrete hIconcreteit ¼ 0.5 μW/m2 , in good agreement with the earlier calculation, as must be the case.

Using Eq. (9.38), the power dissipated per unit area due to the thermal relaxation losses at the interface can be calculated and compared to the incident intensity, since the concrete wall will behave as an isothermal boundary.1 Since this loss mechanism is frequency dependent, let's do the calculation for f ¼ 1 kHz. From the DELTAEC Thermophysical Properties of air at 300 K and 1 bar ¼ 105 Pa,

1 To determine if the concrete wall forces the air that it contacts to behave isothermally, we can calculate the heat capacity per unit area that is contained within a layer of the material that is one thermal penetration depth thick, ρ cPδκ. Using the thermophysical properties available in DELTAEC, (ρ cPδκ)air ¼ 0.10 J/m2 - C. Approximate values for concrete are ρ ¼ 2600 kg/m3 , cP ¼ 880 J/kg-C, and κ ¼ 0.29 W/m-C, so at 1 kHz, δκ ¼ 6.35 μm and (ρ cPδκ)Concrete ¼ 14.5 J/ m2 o C (ρ cPδκ)air ¼ 0.10 J/m2 - C. For a more detailed discussion, see Eq. (59) for εs in G. W. Swift, "Thermoacoustic engines," J. Acoust. Soc. Am. 84(4), 1145–1180 (1988).

Fig. 11.2 Coordinate system for two fluids that can be thought of as oil floating on water. The dashed line is normal to the plane interface between the two fluids at y ¼ 0. In this diagram, it is assumed that the sound originates from the oil which has a specific acoustic impedance of zi ¼ ρici. A wave will be transmitted across the interface into a second medium (water) with a specific acoustic impedance zt ¼ ρtct

δκ ¼ 84.6 μm, corresponding to a thermal absorption of 0.76 μW/m2 . This is a miniscule loss (0.03%), although thermal relaxation at the interface is responsible for a slightly larger decrease in the reflected amplitude, at this frequency, than the intensity of the sound that enters the concrete. Why do we not need to calculate the viscous boundary layer dissipation?

Now that we have the behavior of a common reflection situation "under our belts," we can start our formal investigation of reflection and transmission of planewaves impinging normally on an interface between any two dissimilar media. For these calculations, keep in mind a very simple case of two immiscible liquids in a uniform gravitational field. Let's assume that we have a sea of oil floating on a sea of water, as shown in Fig. 11.2, with the "up" direction being þy and y ¼ 0 at the interface between the two liquids. The mass density of the oil, ρoil, is less than the density of water. To simplify the specification of variables on either side of the interface, we will designate parameters of the wave and of the medium above the interface (y > 0) with the subscript, i (incident), and those below the interface (y < 0) with the subscript, t (transmitted).

Let's imagine that a single-frequency plane sound wave originates far above the interface and is propagating in the –y direction. That wave excites the interface with a pressure amplitude, bpi j j. We can express the pressure amplitude as a function of y and t above the interface

$$p\_i(\mathbf{y}, t) = \Re e \left[ \hat{\mathbf{p}}\_{\mathbf{l}} e^{j(\mathbf{o} \cdot \mathbf{t} + k\_i \mathbf{y})} \right] \quad \text{for} \quad \mathbf{y} > \mathbf{0}. \tag{11.1}$$

That incident plane pressure wave also has an associated fluid particle velocity, bvi, that is related to the pressure by the Euler equation.

$$\nu\_i(\mathbf{y}, t) = \Re e \left[ \frac{\widehat{\mathbf{p}}\_{\mathbf{i}}}{\rho\_i c\_i} e^{j(\alpha t + k\_i \mathbf{y})} \right] = \Re e \left[ \frac{\widehat{\mathbf{p}}\_{\mathbf{i}}}{\mathbb{Z}\_i} e^{j(\alpha t + k\_i \mathbf{y})} \right] \quad \text{for} \quad \mathbf{y} > \mathbf{0} \tag{11.2}$$

In the right-hand term of Eq. (11.2), zi is the specific acoustic impedance of the oil. Using our generic expression, Eq. (10.21), for the sound speed, c, in a fluid with an adiabatic bulk modulus, Bs, c ffiffiffiffiffiffiffiffiffi Bs=ρ p , we see that the specific acoustic impedance is a combination of the fluid's density and compressibility,z ¼ ρc ¼ ffiffiffiffiffiffiffi ρBs p .

As for the echo case, there will be a wave that is reflected at the interface that will travel in a direction opposite to that of the incident wave.

$$p\_r(\mathbf{y}, t) = \Re e \left[ \hat{\mathbf{p}}\_\mathbf{r} e^{j(\mathbf{z} \cdot \mathbf{t} - k \mathbf{y})} \right] \quad \text{for} \quad \mathbf{y} > \mathbf{0} \tag{11.3}$$

The presence of the incident wave will excite motion in that interface which will also generate a wave propagating into the water as well as the wave that is reflected back into the oil. To demonstrate the necessity for the simultaneous existence of the incident, reflected, and transmitted waves, we need to consider the properties of the interface between the two media.

Since we will treat these media and the waves they contain as a linear system, we can guarantee that the reflected and transmitted waves both have the same frequency as the incident wave:ω¼ωi¼ωr¼ωt. As we have seen for the simple harmonic oscillator, at steady state, the forced linear system can only respond at the forcing frequency. It is the incident wave that is forcing the motion of the interface. Since the frequencies of all of the waves must be the same, and the sound speeds in the two media might differ, the wavelength of the sound in the water will be different than that in the oil: λt ¼ ct/f ¼ 2π/kt.

$$p\_t(\mathbf{y}, t) = \Re e \left[ \hat{\mathbf{p}}\_t e^{j(\alpha\_t t + k\_t \mathbf{y})} \right] \quad \text{for} \quad \mathbf{y} < \mathbf{0} \tag{11.4}$$

Our second requirement at the interface is that the two fluids always remain in contact. There are cases where the amplitude of the incident wave is sufficiently large to produce a vapor cavity at the interface, but we will limit our attention here to wave amplitudes that are insufficient to create such cavitation effects [1]. In most practical cases, the boundaries do not separate, so the normal velocities of the two fluids must match at the interface for all times.

$$\nu\_i(0, t) + \nu\_r(0, t) = \nu\_t(0, t) \tag{11.5}$$

Our final requirement is dictated by the fact that the interface, y ¼ 0, is only a mathematical construct—it has no mass. Since that interface is massless, Newton's Second Law of Motion would guarantee that any pressure difference across the interface would produce a non-zero force that would create infinite fluid accelerations. To guarantee that does not happen, we require that the pressure be continuous across that interface for all time.

$$p\_i(0, t) + p\_r(0, t) = p\_t(0, t) \tag{11.6}$$

Those three conditions are sufficient to completely determine the behavior of the waves at the interface when we recognize that vt(y, t) is related to pt(y, t) by the specific acoustic impedance of the water, zt. Taking the ratio of Eq. (11.6) to Eq. (11.5), the amplitude ratio of the reflected and transmitted waves to the amplitude of the incident wave can be calculated.

$$\frac{p\_i(0,t) + p\_r(0,t)}{\nu\_i(0,t) + \nu\_r(0,t)} = \frac{p\_r(0,t)}{\nu\_t(0,t)} = z\_t \tag{11.7}$$

Recognizing that the incident and reflected waves are both in the oil, but are traveling in opposite directions, the magnitude of the specific acoustic impedance, |zi|, is the same for both waves, but their signs are opposite. This leads to the further simplification of Eq. (11.7).

$$z\_i \frac{p\_i(0, t) + p\_r(0, t)}{p\_i(0, t) - p\_r(0, t)} = -z\_t \tag{11.8}$$

Again, because we are limiting ourselves to linear acoustics, it is only the ratios of the amplitudes that have any significance. If we double the amplitude of the incident wave, the amplitudes of the reflected and transmitted waves must also double. It is therefore reasonable to define a pressure reflection coefficient, R bpr=bpi, as we did in Eq. (10.105). With that definition and a little algebraic manipulation, Eq. (11.8) can be re-written in a compact form to represent the magnitude of the pressure reflection coefficient, R.

$$R \equiv \frac{\widehat{\mathbf{p}}\_{\mathbf{r}}}{\widehat{\mathbf{p}}\_{\mathbf{i}}} = \frac{z\_{l} - z\_{i}}{z\_{l} + z\_{i}} = \frac{\left(\swarrow\_{i}\right) - 1}{\left(z\swarrow\_{i}\right) + 1} \tag{11.9}$$

This result makes sense and is reminiscent of Eq. (10.105). If the two media have the same specific acoustic impedances, there is no reflection. Plugging in the specific acoustic impedances for air and concrete produces the same result as we obtained for the "echo" example that began this treatment and produced nearly perfect reflection: R ¼ 0.9999.

In the case of oil over water, as shown in Fig. 11.2, then zi < zt so R is positive and the phase of the reflected wave will be the same as the incident wave, although their amplitudes will differ, similar to the in-phase pulse reflection in Fig. 3.4. If the situation were reversed so that the incident wave originated in the water, then zi > zt so R is negative and the phase of the reflected wave will be inverted, as shown for the pulses in Fig. 3.3.

Defining the pressure transmission coefficient in a similar way, T bpt=bpi , the pressure continuity boundary condition of Eq. (11.6) can now be written as 1 þ R ¼ T.

$$T \equiv \frac{\widehat{\mathbf{p}}\_{\mathbf{i}}}{\widehat{\mathbf{p}}\_{\mathbf{i}}} = \frac{2z\_{\iota}}{z\_{\iota} + z\_{i}} = \frac{2\left(\bigvee\_{i}\right)}{\left(z\bigvee\_{i}\right) + 1} \tag{11.10}$$

The factor of two is reassuring, since we expected pressure doubling at the interface in the echo example where zt zi, and we saw the same factor of two in Eq. (10.105).

For oil above water, we can use ρoil ¼ 950 kg/m3 , coil ¼ 1540 m/s, zoil ¼ (ρc)oil ¼ 1.463 106 Pa ‐ s/m, and ρwater ¼ 998 kg/m3 , cwater ¼ 1481 m/s, and zwater ¼ (ρc)water ¼ 1.478 106 Pa-s/m. The specific acoustic impedance of these two liquids is very close, so for a planewave originating in the oil, R ¼ 0.005 and T ¼ 1.005. If the planewave originated in the water, R ¼ 0.005 and T ¼ 0.995. The fact that R is negative for sound traveling from water into oil indicates that the wave reflected back into the water from the interface is reflected with a 180 phase shift with respect to the incident wave.

If Fig. 11.2 represented air over water, as on a lake, then the impedance contrast would be similar to the echo example, if we assume dry air at one atmosphere, pm ¼ 101,325 Pa, and Tm ¼ 20 C, then (ρc) air ¼ 413 Pa-s/m. If the wave originated in the air, the results would be similar to the echo example. If the wave originated in the water, R ¼ 0.9994, so again the reflection is almost perfect, but the reflected wave is 180 out-of-phase with incident wave. The pressure transmission coefficient is T ¼ 0.00056, so we call that interface, in such a situation (i.e., the wave originating in the medium of higher specific acoustic impedance), a pressure release boundary.

#### 11.2 Three Media

If we had two gases in contact, then we would need a membrane to keep them from diffusing into each other. Similarly, we might be interested in the transmission of sound between two living spaces separated by a wall. There is no reason why we could not extend the same style of argument just employed with one interface between two media. We would need five waves instead of three since the

Fig. 11.3 For a planewave that propagates through three media, bounded by two parallel interfaces, two additional waves need to be included for propagation through the central region. In this diagram, two waves are "trapped" inside the central medium, between y ¼ 0 and y ¼ L, that has a specific acoustic impedance of zc ¼ ρccc. The wavevector of the incident planewave, k ! i, and the reflected wavevector, k ! r, are both in the upper medium with zi ¼ ρici. The transmitted wave represented by the wavevector, k ! t , is in the lower medium, with specific acoustic impedance zt ¼ ρtct. All wave vectors are colinear and normal to both interfaces

central region, with specific acoustic impedance, zc, can support both waves moving up and moving down, as shown in Fig. 11.3.

$$p\_{\rm up}(\mathbf{y},t) = \Re e \left[\hat{\mathbf{p}}\_{\rm up} e^{j(a\,t - k\_{\rm c}\mathbf{y})}\right] \quad \text{and} \quad p\_{\rm down}(\mathbf{y},t) = \Re e \left[\hat{\mathbf{p}}\_{\rm down} e^{j(a\,t + k\_{\rm c}\mathbf{y})}\right] \tag{11.11}$$

Again, linearity guarantees that these two new waves also have a frequency ω ¼ ωup ¼ ωdown, so their wavelengths and wavenumbers depend upon the speed of sound in the central medium: λc ¼ cc=f ¼ 2π= k ! c . We would need to impose the continuity of normal particle velocity (11.5) and the continuity of pressure (11.6) on two planes, y ¼ 0 and y ¼ L, if the central medium has thickness, L. We are again only interested in the amplitude ratios (i.e., specification of the incident amplitude, bpi, is still arbitrary) and can calculate the complex pressure reflection coefficient, R.

$$\mathbf{R} = \frac{\left(1 - \boldsymbol{\angle\angle\_{\boldsymbol{\omega}}}\right)\cos\left(k\_c L\right) + j\left(\boldsymbol{\angle\angle\_{\boldsymbol{\omega}}} - \boldsymbol{\angle\varphi}\_{\boldsymbol{\omega}}\right)\sin\left(k\_c L\right)}{\left(1 + \boldsymbol{\angle\varphi}\_{\boldsymbol{\omega}}\right)\cos\left(k\_c L\right) + j\left(\boldsymbol{\angle\varphi}\_{\boldsymbol{\omega}\_t} + \boldsymbol{\angle\varphi}\_{\boldsymbol{\omega}\_c}\right)\sin\left(k\_c L\right)}\tag{11.12}$$

The power transmission coefficient will be a scalar, as it was in Eq. (10.109), and will be related to the reflection coefficient through energy conservation, as it was in Eq. (10.110).

$$T\_{\rm II} = \frac{4}{2 + \left(\varphi\_{\rm \overline{\gamma}} + \varphi\_{\rm \overline{\omega}}\right) \cos^2(k\_c L) + \left(\varepsilon\_{\rm \overline{\gamma}\omega}^2 + \varepsilon\varphi\_{\rm \overline{\omega}}\right) \sin^2(k\_c L)}\tag{11.13}$$

In many cases, the incident and transmitted media are the same, zi ¼ zt, simplifying the power transmission coefficient.

$$T\_{\rm II} = \left[1 + (\%) \left(z\_{\%} - z\_{\%} \right)^2 \sin^2(k\_c L) \right]^{-1} \quad \text{if} \quad z\_i = z\_t \tag{11.14}$$

#### 11.2.1 A Limp Diaphragm Separating Two Gases

If we apply Eq. (11.12) to the case of two different gases separated by a membrane with zc zi and zc zt, which is so thin (i.e., L λc) that (zc/zt) sin (kcL) 1and cos(kcL) ffi 1, then the pressure amplitude reflection coefficient is the same as (11.9), so R ffi (zt zi)/(zt þ zi), making the diaphragm effectively transparent and the pressure reflection coefficient dependent only upon the relative specific acoustic impedances of the gases.

Let us consider a limp latex diaphragm, having a thickness of 0.00600 ¼ 150 μm, separating air from molecular hydrogen, H2. At 1.0 kHz, with clatex ≌ 1000 m/s, (kcL) ¼ ωL/clatex ffi 0.001. Using (ρc)latex ffi 50 kPa-s/m, (zc/zt) sin (kcL) ffi 0.1 1, the simplified reflection coefficient of Eq. (11.9) will provide a decent approximation.

For ideal gases, ð Þ ρc gas ¼ ffiffiffiffiffiffiffiffiffiffiffiffi γpmρm p . Since the diaphragm is assumed to be limp, the pressures of the gases on either side must be identical (otherwise the limp membrane would bulge and then burst), soð Þ zt=zi gas ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Mt=Mi p , where M is the molecular mass of each gas. For air and hydrogen, the ratio of specific acoustic impedances is about ffiffiffiffiffi 15 p , or its reciprocal, if the planewave originates on the air side, making R ¼ 0.59, with the sign dependent upon whether the planewave originates on the air side (þ) or the H2 side ().

#### 11.2.2 An Impedance Matching Antireflective Layer

An important special case covered by Eq. (11.12) occurs when the thickness of the intermediate layer is one-quarter of a wavelength to create a perfectly transmitting condition that repeats for integer multiples of half-wavelengths being added to the original quarter wavelength, resulting in an odd-integer multiple of one-quarter of the wavelength of sound, L ¼ (2n 1)λc/4; n ¼ 1, 2, 3, ...; hence, (kc L) ¼ (n½)π. In that case, cos(kcL) ffi 0 and sin(kcL) ¼ 1. In addition, if zc ¼ ffiffiffiffiffiffi zizt p , then R ¼ 0 for frequencies near f ¼ (n ½)cc/2L, and the intermediate medium is known as a quarterwavelength impedance matching layer.

In high-quality optics, like camera lenses, this effect is used to make antireflective coatings that optimize the transmission of light through the lens. Since this effect is wavelength-dependent, those optical coatings look magenta when they are optimized for transmission of green light (520 nm ≲ λgreen ≲ 570 nm), since red (620 nm ≲ λred ≲ 740 nm) and blue (450 nm ≲ λblue ≲ 495 nm) will have non-zero optical reflection coefficients.

Impedance matching layers are very useful for ultrasonic transducers used in medical imaging to optimize the transmission from the transducer material (typically a piezoelectric ceramic) to flesh and to maximize the return signal's excitation of the transducer during detection.

#### 11.2.3 The "Mass Law" for Sound Transmission Through Walls

If we consider a solid wall separating two rooms, both containing air, then zc zi ¼ zt. In that case, the power transmission coefficient of Eq. (11.14) is further simplified.

$$T\_{\rm II} \cong \left[\frac{2z\_i}{z\_c \sin\left(k\_c L\right)}\right]^2 \quad \text{if} \quad z\_c \gg z\_i = z\_t \tag{11.15}$$

For reasonably small wall thicknesses, L λc ¼ cc/f, sin(kcL) ffi kcL, so the reduction of sound intensity transmission through the wall depends upon the surface mass density of the wall, ρS ¼ ρcL.

$$T\_{\Pi} \cong \left(\frac{2}{k\_c L} \frac{z\_i}{z\_c}\right)^2 \cong \left[\frac{z\_i}{\pi f(\rho\_c L)}\right]^2 \quad \text{if} \quad k\_c L \ll 1 \tag{11.16}$$

For this reason, in architectural acoustics, this result is known as the "mass law" and is particularly important for multi-occupant dwellings (e.g., apartment buildings) because the transmission of sound depends inversely on the square of both the surface mass density of the wall and of the frequency of the sound.2 [2] This inverse-square dependence upon frequency explains why the bass from a neighbor's stereo in an adjoining apartment is more annoying than conversation.

#### 11.2.4 Duct Constriction/Expansion Low-Pass Filters

We can exploit Eq. (11.14) for the power transmission coefficient to analyze the one-dimensional propagation through a duct that has an expansion chamber or constriction as shown in Fig. 11.4. Following the analysis in Sect. 10.8, and recognizing that such a system has the same sound speed and fluid density throughout, we can copy the five-wave procedure used in Sect. 11.2, but require the continuity of volume velocity, instead of particle velocity, to produce the analog of Eq. (11.14) where A replaces zi ¼ zt and Ac replaces zc [3].

$$T\_{\rm II} = \frac{1}{1 + \frac{\left[\left(\mathbf{A}\_{\rm \%}\right) - \left(\mathbf{A}\_{\rm A\_c}\right)\right]^2}{4} \sin^2(kL)}\tag{11.17}$$

It is worthwhile to notice that this coefficient for the power transmission is symmetric with respect to the area ratio, Ac/A, and its reciprocal, A/Ac. That means that the result does not depend upon whether the area change is produced by an expansion chamber, Ac/A > 1, shown in Fig. 11.4 (left), or by a constriction, A/Ac > 1, shown in Fig. 11.4 (right).

Fig. 11.4 (Left) A compliance with volume, V ¼ AcL, is placed in a tube of otherwise uniform cross-sectional area, A, to create a low-pass filter. (Right) A constriction with cross-sectional area, Ac < A, is placed in a tube of otherwise uniform cross-sectional area, A, to produce a low-pass filter

2 It is important to remember that this "mass law" is a low-frequency result, as indicated in Eq. (11.16). At higher frequencies, other effects can become significant.

Fig. 11.5 The expansion/constriction duct transmission loss, TΠ, for small kL, in dB ¼ 10log10TΠ for area ratios of Ac/A ¼ 2 or 0.5 (solid line), Ac/A ¼ 5 or 0.2 (dotted line), and Ac/A ¼ 10 or 0.1 (dashed line) plotted against a normalized frequency, f , that is scaled by the length of the length of the expansion or constriction, L, and the sound speed, c: f ¼ πf L/c, which is also f ¼ πL/λ ¼ kL/2

If we start by examining the limit where L λ or kL 1, we see that either the expansion chamber or the constriction acts as a low-pass filter as shown in Fig. 11.5 for Ac/A ¼ 2, Ac/A ¼ 5, and Ac/A ¼ 10.

$$T\_{\rm II} = \left\{ 1 + \frac{\pi^2 L^2}{c^2} \left[ \left( \mathbb{A}\_{\%} \right) - \left( \mathbb{A}\_{\%} \right) \right]^2 f^2 \right\}^{-1} \quad \text{for} \quad kL \ll 1 \tag{11.18}$$

As expected from Eq. (11.18), the transmission for kL < 1 is reduced by 20 dB/decade, in accordance with the f 2 frequency dependence. It is clear from Fig. 11.5 that the frequency at which the transmission is reduced by 3 dB is dependent upon the area ratio. For Ac/A ¼ 2 or 0.5, f -3dB ¼ 0.665; for Ac/A ¼ 5 or 0.2, f -3dB ¼ 0.208; and for Ac/A ¼ 10 or 0.1, f -3dB ¼ 0.101.

As frequency increases, the requirement that kL < 1 will be violated. In Fig. 11.5, that means that f  ½. This restriction indicates that the frequency bandwidth of the expansion/constriction strategy is rather limited as a low-pass filter. From Eq. (11.17), we see that the transmission coefficient becomes one (i.e., 0 dB, no attenuation), when kL ¼ π or f ¼ π/2, and returns to one for all successive kL ¼ nπ as shown in Fig. 11.6. The maximum transmission loss, (TΠ)max, occurs when L is equal to an odd multiple of quarter wavelength, (2n 1)λ/4, making f ¼ (2n 1)c/4L.

$$(T\_{\rm II})\_{\rm max} \cong -10 \log\_{10} \left[ (\%) \left( \frac{A\_c}{A} + \frac{A}{A\_c} \right)^2 \right] \tag{11.19}$$

For Ac/A ¼ 5 or 0.2, (TΠ)max¼ 8.3 dB; and for Ac/A ¼ 10 or 0.1, (TΠ)max¼ 14.1 dB.

Fig. 11.6 The expansion/constriction duct transmission loss, TΠ, in dB ¼ 10log10TΠ. Shown are area ratios of Ac/A ¼ 2 or 0.5 (solid line), Ac/A ¼ 5 or 0.2 (dotted line), and Ac/A ¼ 10 or 0.1 (dashed line) plotted against a normalized frequency, 0  kL ¼ (2πL/λ)  π. This behavior repeats for integer multiples of π with no loss for kL ¼ 0 or π and maximum loss for kL ¼ π/2

As will be shown in Sect. 13.5.4, for frequencies above f1,1 ¼ 0.578 c/a, where a ¼ ffiffiffiffiffiffiffiffi A=π p is the radius of the circular duct, the planewave nature of propagation in the duct can no longer be guaranteed. That further limits the utility of such a strategy. Despite these limitations, elaborate combinations of expansion chambers and constrictions are used to attenuate sound in automotive exhaust mufflers using techniques that go well beyond those discussed here [4].

#### 11.3 Snell's Law and Fermat's Principle

The previous results for planewaves normally incident on an interface between two media can be extended to planewaves that impinge on the interface from some arbitrary angle, θi, with respect to the normal direction that characterized the surface. The incident wavevector will now have two components, k ! i ¼ kybey þ kxbex, where bey and bex are unit vectors in the +y and þ x directions.

$$\begin{aligned} p\_i(\mathbf{x}, \mathbf{y}, t) &= \mathfrak{Re} \left[ \hat{\mathbf{p}}\_{\mathbf{t}} \exp \mathbf{j} \left( \boldsymbol{\alpha}t + \left| \vec{k}\_i \right| \mathbf{y} \cos \theta\_i - \left| \vec{k}\_i \right| \mathbf{x} \sin \theta\_i \right) \right] \\ p\_r(\mathbf{x}, \mathbf{y}, t) &= \mathfrak{Re} \left[ \hat{\mathbf{p}}\_{\mathbf{t}} \exp \mathbf{j} \left( \boldsymbol{\alpha}t - \left| \vec{k}\_i \right| \mathbf{y} \cos \theta\_r - \left| \vec{k}\_i \right| \mathbf{x} \sin \theta\_r \right) \right] \\ p\_t(\mathbf{x}, \mathbf{y}, t) &= \mathfrak{Re} \left[ \hat{\mathbf{p}}\_{\mathbf{t}} \exp \mathbf{j} \left( \boldsymbol{\alpha}t + \left| \vec{k}\_t \right| \mathbf{y} \cos \theta\_i - \left| \vec{k}\_t \right| \mathbf{x} \sin \theta\_t \right) \right] \end{aligned} \tag{11.20}$$

We can make a simple geometrical argument regarding the directions of the reflected and transmitted planewaves by adopting a perspective that was introduced by the Dutch physicist, Christiaan

Fig. 11.7 Diagram showing an incident wave with wavevector, k ! i, intersecting a plane interface at an angle, θi, relative to the normal to the interface. Wave fronts of equal phase are shown perpendicular to the wavevector spaced by one wavelength of the incident sound, λi. That incident wave excites a reflected wave (*green*) and a transmitted wave (*red*). Successive maxima of pressure along the interface are separated by the "trace wavelength," λtrace, that is common to all three waves. Based on the wavelengths in this figure, the sound speed in the lower medium, ct, is greater than the sound speed in the upper medium, ci < ct. When the incident angle equals the critical angle, θi ¼ θcrit, then the direction of k ! t is along the interface (i.e., θt ¼ 90), and we have total internal reflection for all θi θcrit, so no energy is transmitted

Huygens, in 1678,3 that makes a simple argument which matches boundary conditions at the interface. Figure 11.7 shows the incident wavevector, k ! i , with lines normal to k ! i , indicating the planes of constant pressure phase. We will imagine those lines represent the pressure maxima of the waves propagating toward the interface. Those wave fronts impinge on the interface, and we will consider each intersection of those incident wave fronts and the interface to be a source of pressure maxima for both the reflected and transmitted planewaves.

The spacing between those maxima along the interface, which we will call the trace wavelength, λtrace, is easy to calculate from trigonometry and Garrett's First Law of Geometry4 : sin θi ¼ λi/λtrace. For normally incident planewaves, the trace velocity ctrace ¼ fλtrace ¼ 1, since the wave fronts of equal phase intersect the interface so that the phase of the pressure on the interface is the same at all locations at every instant. Since λtrace is the same for both the reflected and transmitted waves, the sound speeds

3C. Huygens, Traitė de la Lumiere (completed in 1678, published in Leyden in 1690).

4 "Angles that look alike are alike."

in the two media will determine the angle that the reflected planewave, θr, and the angle transmitted planewave, θt, make with the direction of the normal to the interface.5

$$
\lambda\_{\text{trace}} = \frac{\lambda\_i}{\sin \theta\_i} = \frac{\lambda\_r}{\sin \theta\_r} = \frac{\lambda\_t}{\sin \theta\_t} \tag{11.21}
$$

The sound speed, ci, is the same for both the incident and reflected planewaves, so θr ¼ θi; the angle of incidence equals the angle of reflection. This case is called specular reflection. If the interface has a surface roughness that is characterized by random assortments of "bumps" that are all much smaller in height and extent than the wavelength of sound, the reflection will be diffuse.

Again, in a linear system, the frequency, ω, with which the incident planewave drives the interface, must also be the frequency of the driven reflected and transmitted planewaves. Since λ ¼ c/f, we can invert Eq. (11.21) to produce a result known as Snell's law. 6

$$\frac{\sin \theta\_i}{c\_i} = \frac{\sin \theta\_r}{c\_i} = \frac{\sin \theta\_t}{c\_t} \tag{11.22}$$

Before moving on to calculation of the pressure reflection and transmission coefficients, it worth noticing that Snell's law could have been derived by insisting that propagation time from one location in the upper medium to some other location in the lower medium is minimized by the same condition on the angles that is expressed in Eq. (11.22). This approach utilizes Fermat's principle (1622)7 : "The actual path between two points taken by a [wave] is the one that is traversed in the least time."

For the actual path to be an extremum of the total transit time, a neighboring path must take the same time; their time difference being at most second order in the deviation of the alternate path from the path of least time. Figure 11.8 provides a diagram of the path of least time (ACB) and a nearby path (AXB).

In the upper medium of Fig. 11.5, the perpendicular XE makes CE the reduction in distance traveled by the nearby path AX. The time savings by traveling via AC is therefore ti ¼ CE/ci. Similarly, in the lower medium, the perpendicular CF makes XF the additional distance traveled by the nearby path XB. The excess time in the lower medium is therefore tt ¼ XF/ct. Since ∡EXC ¼ θi and ∡FCX ¼ θt, we can express the two transit times for the nearby path in terms of the common length XC using simple trigonometry.

$$t\_i = \frac{EC}{c\_i} = \frac{XC\sin\theta\_i}{c\_i} = t\_t = \frac{XF}{c\_t} = \frac{XC\sin\theta\_t}{c\_t} \tag{11.23}$$

Equation (11.23) also yields Snell's law after dividing through by XC.

5 Since the trace wavelength is constant, its reciprocal, the trace wavenumber, 2π/λtrace, is also constant. In the underwater acoustic propagation community, this is commonly called "conservation of the horizontal wavenumber" [12]. Since I associate the term "conservation" with quantities that obey equations like (10.35), I prefer to consider the horizontal wavenumber to be an invariant, like the trace wavelength.

6 Snell's law is known as Descartes' law in France. Willebrord Snellius (1580–1626) was not the first to produce that result. The law was first accurately described by the Persian scientist, Ibn Sahl, in Baghdad, where in 984, he used the law to derive lens shapes that focus light with no geometric aberrations as described in the manuscript On Burning Mirrors and Lenses.

7 The principle of least time was actually first applied to reflections from a mirror much earlier, by Hero of Alexandria, around 60 CE. Since the reflected light propagates at the same speed as the incident light, this is equivalent to the path that is also the shortest geometrical distance.

Fig. 11.8 Three paths from point A in the upper medium to point B in the lower medium. In this diagram, it is assumed that the propagation speed of the wave in the lower medium, ct, is slower than the propagation speed in the upper medium, ci > ct. The path AB (*blue*) is the shortest distance, and the path ACB (*red*) takes the least time. The angles θi and θt can be determined by calculating the additional travel time beyond that required for ACB when going along the path through AXB (*green*) and setting that excess equal to zero [5]

#### 11.3.1 Total Internal Reflection

In passing from a faster medium into a slower medium, ci > ct, the angle that the transmitted wavevector makes with respect to the normal at the interface, θt, is smaller than the incident angle, θi, also measured from the normal. If the incident medium has a slower sound speed than the transmission medium, ci < ct, then it is possible that the angle of transmission, θt, dictated by Snell's law, could become a complex number. In that case, we can define a critical angle, θcrit, where θt ¼ 90 (i.e., the transmitted wave travels along the interface but not into the second medium).

$$
\sin \theta\_{crit} = \frac{c\_i}{c\_t}; \quad \text{if} \quad c\_i < c\_t \tag{11.24}
$$

This is a consequence of the fact that the minimum trace velocity, ctrace ¼ fλtrace, shown in Fig. 11.7, in the lower medium, is limited to ct, which equals ctrace when θt ¼ 90 . Using the fact that cos2 θ þ sin2 θ ¼ 1, the angle of transmission can be expressed in terms of the angle of incidence and the sound speed ratio.

$$\cos\theta\_t = \sqrt{1 - \sin^2\theta\_t} = \sqrt{1 - \left(\mathbf{c}\_{\langle\rangle\_{\rm c}}\right)^2 \sin^2\theta\_i} = -j\sqrt{\left(\mathbf{c}\_{\langle\rangle\_{\rm c}}\right)^2 \sin^2\theta\_i - 1} \tag{11.25}$$

If ct > ci, the first term under the square root becomes imaginary, which is expressed explicitly in the final term at the right of Eq. (11.25).

Substitution of Eq. (11.25) into the expression for the transmitted pressure in Eq. (11.20) produces an exponential decay of the sound with increasing distance away from the interface, for incident angles that exceed the critical angle, producing an inverse characteristic depth, (1/δ).

$$k\_{\varepsilon} = \frac{1}{\delta} = \left| \overrightarrow{k}\_{l} \right| \sqrt{\left(c\_{l}/c\_{i}\right)^{2} \sin^{2}\theta\_{i} - 1} \tag{11.26}$$

$$p\_t(\mathbf{x}, z, t) = \Re e \left[ \hat{\mathbf{p}}\_t e^{+(\mathbf{y}/\delta)} \mathbf{e}^{j\left[\boldsymbol{\alpha}t - \left|\vec{k}\_i\right| \mathbf{x} \sin \theta\_i\right]} \right]; \quad \text{for } \mathbf{y} \le \mathbf{0} \tag{11.27}$$

This solution is a wave that is localized just below the interface. For angles of incidence, θi < θcrit, the transmitted wave fills the half-space of the lower medium, as shown in the lower half of Fig. 11.7. When θi ¼ θcrit, all of that outgoing energy collapses to the surface in a layer with exponential thickness, δ. The thickness of that layer decreases monotonically as θi increases beyond θcrit.

If we apply the Euler equation to Eq. (11.27) to determine the velocity, vt, normal to the surface, we find bvt ¼ jbptð Þ ωρδ , so the pressure and particle velocity are 90 out-of-phase. The intensity leaving the interface is zero since there is no in-phase component of pressure and velocity. That does not mean that the layer contains no energy, just that none of the energy is propagating. If there were some scattering centers (i.e., defects, voids, inclusions of material with some compressibility, and/or density contrast), the energy trapped in that layer could be reradiated (i.e., scattered; see Sect. 12.6) and escape.

The concept of total internal reflection was first recognized in optics. One application of total internal reflection that I find particularly inspiring occurs in the compound eyes of insects and other arthropods. With thousands of lenses, the use of an equal number of irises, like those in avian and mammalian eyes, is not particularly efficient. In the arthropod's compound eye, each lens is connected to the optic nerve by a transparent tube that contains the light by total internal reflection, due to the difference between the index of refraction (i.e., the optical equivalent of our specific acoustic impedance) between the tube and the surrounding fluid. The way that insects control the amount of light that reaches the optic nerve is by a chemical reaction in the fluid that surrounds the tube. Under bright lighting, the fluid will produce a precipitate with small particles that scatters light out of the trapped layer, thereby reducing the amount of light reaching the optic nerve. The physical structure of the compound insect eye is shown in Fig. 11.9.

Proof of this system's efficacy is the fly's ability to avoid getting swatted outdoors in broad daylight and indoors in a darkened room. (We should now pause briefly out of respect for Darwin and the power

Fig. 11.9 (Left) The compound eyes of the blue bottle fly. (Right) Each of the "tubes" is optical fibers, called an ommatidium. They guide light from the lens to the optic nerve endings (4 and 5) using total internal reflection to regulate the transmission of the light. Ant eyes have about six ommatidia and dragonfly eyes have as many as 25,000. This anatomical drawing of a compound eye shows the corneal lens (1) and the crystalline cone (2) which combine to form the dioptric apparatus. At the base is the light-sensitive rhabdom (5 and 6). The space between the tubes (3) contains the pigment cells that will form a precipitate that scatters some of the optical energy out of the trapped layer at the interface if the light is too bright, effectively reducing the field of view (known in optics as the "numerical aperture" of the lens). (Photo and diagram taken from Wikipedia)

of natural selection.) I would be most interested in finding a similar system for controlling the sound amplitude that reaches the auditory nerve in any animal, although the ability to focus sound seems ubiquitous in marine mammals, as shown for the Cuvier's beaked whale in Fig. 11.18.

#### 11.3.2 The Rayleigh Reflection Coefficient

Although we have calculated the angles of reflection and transmission for planewaves of sound impinging on a planar interface from an arbitrary direction, we have yet to calculate the amplitude reflection coefficient in terms of the specific acoustic impedances, as we did for the case of normal incidence at the interface between two media in Eq. (11.9). To produce the equivalent of Eq. (11.9) for oblique incidence (i.e., θi 6¼ 0), we need to generalize the velocity boundary condition of Eq. (11.5), since it is only the normal component of the velocity that needs to be continuous across the boundary to avoid cavitation at the interface.8

$$|\hat{\mathbf{v}}\_{\mathbf{i}}|\cos\theta\_{\mathbf{i}} + |\hat{\mathbf{v}}\_{\mathbf{r}}|\cos\theta\_{\mathbf{r}} = |\hat{\mathbf{v}}\_{\mathbf{i}}|\cos\theta\_{\mathbf{i}}\tag{11.28}$$

Pressure continuity, expressed in Eq. (11.6), is unchanged since pressure is a scalar quantity.

Forming the ratio of those boundary conditions, as we did in Eq. (11.7), leads to the desired result for the ratio of the reflected pressure to the incident pressure R bpr=bpi.

$$R \equiv \frac{\widehat{\mathbf{p}}\_{\mathbf{r}}}{\widehat{\mathbf{p}}\_{\mathbf{i}}} = \frac{\frac{z}{\cos \theta\_{i}} - \frac{z\_{i}}{\cos \theta\_{i}}}{\frac{z\_{i}}{\cos \theta\_{i}} + \frac{z\_{i}}{\cos \theta\_{i}}} \tag{11.29}$$

Once again, we are confronted with an expression for the reflection coefficient that can vanish, providing perfect transmission, if the numerator of Eq. (11.29) is zero.

$$\frac{\cos \quad \theta\_t}{\cos \quad \theta\_i} = \frac{z\_t}{z\_i} \tag{11.30}$$

Using Snell's law to eliminate θt, this angle, called the angle of intromission, θintro, can be expressed in terms of both the impedance and sound speed or mass density ratios.

$$\sin^2(\theta\_{\rm intra}) = \frac{\left(z\_l/z\_i\right)^2 - 1}{\left(z\_l/z\_i\right)^2 - \left(c\_l/c\_i\right)^2} = \frac{1 - \left(z\_l/z\_l\right)^2}{1 - \left(\rho\_1/\rho\_2\right)^2} \tag{11.31}$$

Not every combination of media will have an angle of intromission that depends upon both sound speed and density ratios. For θintro to be real, both the numerator and denominator of Eq. (11.31) must have the same sign.

There are four combinations of impedance and sound speed ratios:

(i) No θcrit and no θintro If both zt/zi < 1 and ct/ci < 1, there is no critical angle, so 0 < R < 1 for all θi. For this case, θi > θt, so (cosθt/ cos θi) > 1 and zt/zi < 1, so there can be no angle of intromission.

8 If we were treating the interface between two viscous liquids or between a viscous liquid and a solid surface, then we would need to impose a "no-slip" boundary condition on the transverse components of particle velocity. This is particularly important for the fluid-solid boundary since the viscous stress on the interface can couple to shear waves into the solid and vice versa. The coupling of shear waves in the solid to viscous waves in liquids has been used to measure fluid density and viscosity.

Total transmission at the angle of intromission has been observed in high-porosity marine sediments, like silty clays, which exhibit a sound speed through its bulk which is lower than that of the interstitial fluid within its pores. When high-porosity sediment is at the water/sediment interface, there can be total transmission of sound into the seafloor. Measurements of the angle of intromission in coastal regions around Italy indicate that the properties of the high-porosity sediments are surprisingly uniform over large areas [6].

#### 11.4 Constant Sound Speed Gradients

To this point in our discussion of the refraction of sound, we have only considered plane boundaries that separate media with a discontinuity in their sound speed, density, and/or specific acoustic impedance. There are many situations of interest where the refraction (or bending) of a planewave of sound can create interesting and significant effects resulting from a gradual change in the acoustical properties of the medium. These effects are easily calculable in the limit that such sound speed changes are a linear function of position when those changes occur over a distance that extends over very many acoustic wavelengths.

Figure 11.10 shows two examples of sound refraction in air that supports a temperature gradient that depends upon the time of day. As expressed by the quadratic dependence of sound speed on the mean temperature of an ideal gas in Eq. (10.23), the change in sound speed, Δc, with height, z, above the ground is related to the changes in the absolute temperature, T.

$$\frac{dc}{dz} = \left(\frac{dc}{dT}\right)\left(\frac{dT}{dz}\right) = \frac{c}{2T}\left(\frac{dT}{dz}\right) \tag{11.32}$$

Similar refractive effects occur in the ocean where the speed of sound in seawater is a complicated function of temperature, salinity, and depth (pressure).

$$\begin{split} c &= 1493.0 + 3(t - 10) - 0.006(t - 10)^2 - 0.04(t - 18)^2 \\ &+ 1.2(\text{S} - \text{35}) - 0.01(t - 18)(\text{S} - \text{35}) + D/61 \end{split} \tag{11.33}$$

In Eq. (11.33), the sound speed, c in m/s, is a function of the temperature, t, in degrees Celsius. The salinity, S, is measured in grams of salt per kilogram of water, and D is the depth below the surface measured in meters [8]. That formula is valid within 0.2 m/s for 2 C  t  +24.5 C, 0.030  S  0.042, and 0  D  1000 m.

Figure 11.11 shows the sound speed variation with depth that was measured over 9 years in the ocean 15 miles (24 km) south-east of Bermuda. Although significant sound speed variation over the 9 years is apparent within roughly 2 km of the surface, it is possible to represent the change in sound speed with depth using a piecewise-linear approximation.

Fig. 11.10 (Left) During the day, the temperature of the atmosphere generally decreases with altitude above the ground causing sound waves to refract upward since the speed of sound is proportional to ffiffiffi T p . (Right) The opposite conditions can occur at night when the temperature of the air in contact with the ground is colder that than the air above it (a "temperature inversion"). This causes sound to be refracted downward. If there is wind, it can also cause the sound speed to vary with height above the ground. (Figures courtesy of T. B. Gabrielson [7])

Fig. 11.11 (Left) These deep-ocean sound speed profiles were taken every 2 weeks over a 9-year period at a location 24 km SE of Bermuda. [9] The solid curve is the average and the dashed curves show the extremes. This typical deep-sea profile may be divided into a number of layers having different properties. At the top, the diurnal layer shows day-night variability and responds to weather changes. Below it, down to the depth of about 300 m, lays the seasonal thermocline that is above the main thermocline, which exhibits the strongest sound speed gradient. Below the thermocline, beneath 1200 m, the deep isothermal layer has a constant temperature of þ4 C. There, the sound speed's increase with depth is dominated by the increasing pressure. (Right) The measured sound speed profile is replaced by a piecewise-linear approximation that provides a constant sound speed gradient for each layer, as indicated by the embedded table. At a depth of approximately 3700 m, the sound speed is the same as it is at a depth of approximately 560 m. Between those two depths, sound can be trapped in the "deep sound channel" in the same way that light is trapped in an optical fiber, except the core of an optical fiber that is about 10 μm and the sound channel that is about 1 km, a height ratio of about 108

#### 11.4.1 Constant Gradient's Equivalence to Solid Body Rotation

The goal in this sub-section is to develop a simple formalism that will permit calculation of the path of planewaves in a medium with a sound speed that is a linear function of height or depth. Furthermore, we would like to be able to track the trajectory of the sound (i.e., follow the direction of the wavevector) through regions of changing sound speed gradient, a process known as ray tracing. To do this, we must first recognize that the propagation of sound through a region of constant sound speed gradient is isomorphic to the solid body rotation of a disk, since the tangential velocity, v !tan, of any point on the rotating disk is proportional to its (linear) distance, r ! , from the axis of rotation.

$$
\overrightarrow{\dot{\boldsymbol{\nu}}}\_{\text{tan}} = \overrightarrow{\boldsymbol{r}} \times \overrightarrow{\boldsymbol{a}} \tag{11.34}
$$

Since this equation refers to a rotating disk, the use of the cross-product might seem to be pedantic excess, but it is intended to remind you of your study of solid body rotational dynamics during your freshman physics course.

If we limit ourselves to linear changes in sound speed with height or depth, then the sound speed gradient is a constant: dc/dz g. In analogy with a rotating disk of radius, R, as diagrammed in Fig. 11.12, we can equate sound speed, c, with the tangential velocity of a point on the rim of the disk, v !rim, and integrate our above definition; it is easy to see that g plays the same role as ω (and has the same units) in Eq. (11.34).

$$\int\_{0}^{V\_{\rm min}} dc = \lg \int\_{0}^{R} dz \quad \Rightarrow \quad R = \frac{\left| \vec{V}\_{\rm rim} \right|}{\mathcal{g}} \tag{11.35}$$

I like to think of the refraction (bending) of the sound wave's trajectory in a constant sound speed gradient by picturing the advance of the planewave fronts as represented schematically in Fig. 11.13.

Fig. 11.12 Shown at the right is a disk that is rotating. The dashed, colored arrows indicate the tangential velocity that increases linearly with distance from the axis of rotation. Such a velocity distribution in space is identical to the assumed constant sound speed gradient that will control the refraction of sound waves

Fig. 11.13 A conceptual sketch of planewave fronts moving through a medium with the velocity gradient shown at the left. The arrows at the top of the wave fronts are always longer than the arrows at the bottom. This turns the wave front downward until the wave front is nearly horizontal. Once horizontal, both ends of the wave front will move with the same speed, and the wave will continue downward without changing direction

To introduce this "disk rotation" approach to analysis of the propagation of planewaves through a medium with constant speed gradients, let's consider the measurement of road noise during the day, when the temperature of the air is decreasing with increasing altitude, as shown in Fig. 11.10 (left).

Assume that you are asked to determine the sound pressure level created by traffic on a highway at a proposed housing location that is 150 m from the highway. By measuring the average temperature at a height of 0.5 m and at 5 m, you determine that the sound speed at 0.5 m is 342.8 m/s and at 5.0 m is 341.3 m/s. How high must your microphone be above the ground, at a distance of 150 m from the source, to ensure that you intercept sound that leaves the highway in an initially horizontal direction?

To begin, let's calculate the sound speed gradient, dc/dz ¼ g ≌ Δc/Δz ¼ (2.8–1.3 m/s)/ 4.5 m ¼ 0.333 s1 . Since the sound speed decreases with height, we know that the wave will be refracted upward, as shown in Fig. 11.10 (left). We'll assume that the noise is generated at the intersection of the tire and the road at z ¼ 0, where co ¼ 343.0 m/s (extrapolating down to the surface from the lowest measurement at 0.5 m above the ground).

We now need to calculate the radius, |R|, of this limiting ray's circular path.

$$|\mathcal{R}| = \frac{c\_o}{g} = 1029 \text{ m} \tag{11.36}$$

A simple trigonometry identity can be used to determine the height, h ¼ 11.0 m, above the ground that a microphone must be placed to receive the sound radiated by the tires. For me, drawing a sketch, like in Fig. 11.14, is always crucial.

$$\begin{aligned} \sin \theta &= \frac{150}{1029} = 0.146 \quad \Rightarrow \quad \theta = 0.146 \,\text{rad} \quad = 8.38^{\circ} \\\ h &= |\mathcal{R}| - |\mathcal{R}| \cos \theta = |\mathcal{R}| (1 - \cos \theta) \cong |\mathcal{R}| \frac{\theta^2}{2} = 11.0 \text{ m} \end{aligned} \tag{11.37}$$

The final expression above made use of the small-angle expansion of cosine in Eq. (1.6).

Although the refractive process was not understood until the twentieth century [7], reports of the existence of an acoustic "shadow zone" appeared much earlier. Below is the account of R. G. H. Kean as he watched the Battle of Gaines's Mill during the American Civil War [10]:

Fig. 11.14 Sketch (not to scale) of the ray path of the sound generated by the tire noise on a road surface. The observation point is 150 m from the sound source. Between the ground and the circular arc is a "shadow zone" that is created by the upward refraction of the sound produced by the sound speed gradient, dc/dz ¼ 0.333 s1 , which was used to calculate the radius, |R| ¼ 1029 m, applying Eq. (11.36). A microphone must be placed at least a distance, h, above the ground to intercept the tire noise. At the surface (h ¼ 0), the sound speed is co ¼ 343.0 m/s. At h ¼ 11.0 m above the surface, the sound speed is 339.3 m/s

I distinctly saw the musket-fire of both lines ... I saw batteries of artillery on both sides come into action and fire rapidly. Yet looking for near two hours, from about 5 to 7 P.M. on a midsummer afternoon, at a battle in which at least 50,000 men were actually engaged, and doubtless at least 100 pieces of field-artillery ... not a single sound of the battle was audible to General Randolph and myself ... [However, the] cannonade of that very battle was distinctly heard at Amhurst Court-house, 100 miles west of Richmond, as I have been most credibly informed.

The process of refraction in a sound speed gradient, as illustrated in a particularly simple form in Fig. 11.14, is entirely reversible. If we were to say that we had a sound source directed 8.4 below the horizontal direction, located 11.0 m above the ground, then we could just as easily say that it would intersect the ground 150 m away, based on Fig. 11.14 and the results of Eq. (11.37).

Since much of the interest in the propagation of sound in a sound speed gradient (linear or otherwise) tends to concern sound waves that are nearly grazing the horizontal axis, the grazing angle is more commonly used in Snell's law than the angle measured with respect to the normal to the surface, which was our previous choice for discussion of reflection and transmission from a plane surface of discontinuity. If we now use the grazing angle, then we can rewrite Eq. (11.22) by letting θ be the angle below the horizontal and co be the sound speed at a location where the sound waves are propagating horizontally (i.e., cos 0 ¼ 1).

$$\frac{c}{\cos \theta} = c\_o \tag{11.38}$$

Let's apply the above version of Snell's law to the example in Fig. 11.14. That ray is horizontal at h ¼ 0, where co ¼ 343.0 m/s. Assuming, as before, a constant sound speed gradient, at the "source height" of 11.0 m, c ¼ 339.3 m/s. Plugging directly into Eq. (11.38), cos θ ¼ c/co, so θ ¼ 0.147 radians ¼ 8.4. Of course, we could have used Eq. (11.37) to calculate the sound speed 11.0 m above the surface, since we already knew that θ ¼ 8.4.

#### 11.4.2 Sound Channels

We will now apply this formalism to a particularly interesting case that occurs in a sound speed profile like the one shown in Fig. 11.11, describing the deep ocean off of Bermuda, which we have approximated by a series of contiguous line segments with constant (but different) sound speed gradients. There is a sound speed minimum of cmin ¼ 1486 m/s, at a depth of 1112 m, between the main thermocline and the deep isothermal layer. As evident from Fig. 11.13, in such a constant gradient (i.e., linear sound speed profile), the sound will "go to the slow," and therefore, we expect that some sound will become trapped around that sound speed minimum in a sound channel.

In this sub-section, we will trace several rays from a submerged source to develop an understanding of the channel's propagation characteristics. Assume that there is a sound source at a depth of 800 m, where the sound speed (by interpolation) is 1507 m/s. We start by calculating the upward angle from the source that will result in a horizontal ray at the top of the upper gradient, z ¼ 560 m, where the sound speed is cmax ¼ 1523 m/s. By Snell's law, as expressed in terms of the grazing angles in Eq. (11.38), θ1 ¼ cos1 (1507/1523) ¼ 0.145; hence, θ1 ¼ 8.31. The radius of the circular path of the ray, |R1| ¼ co/g ¼ (1523 m/s 0.067 s1 ) ¼ 22.73 km. Based on the diagram in Fig. 11.15, the sound ray intersects the top of the channel at a horizontal distance that is r1 ¼ |R1|sin θ1 ¼ 3.3 km from the source.

From the top of the channel (z ¼ 560 m), the ray will return to the channel axis along the same circular path and will intersect the axis (z ¼ 1112 m) at θ2 ¼ cos1 (1486/1523), so θ2 ¼ 0.221 rad ¼ 12.66. The horizontal distance, r2 – r1 ¼ |R1| sin θ2 ¼ 5.0 km.

The ray enters the deep isothermal layer where |R2| ¼ (1486 m/s 0.0143 s1 ) ¼ 103.9 km. (The center of that circle is only a geometrical "construction point," so the fact that it is located above the stratosphere should not be a cause for our concern.) Since we have defined the bottom of the channel as the location where the sound speed in the deep isothermal layer equals the sound speed at the top of the

Fig. 11.15 The diagram of the sound speed vs. depth at the left is based on the piecewise-linear approximation of Fig. 11.11. There is a sound speed minimum at a depth of z ¼ 1112 m that is the axis of a sound channel shown by the horizontal **blue** line. The velocity maxima are shown by the horizontal *green* lines at depths of 560 m and 3700 m, corresponding to cmax ¼ 1523 m/s. The ray paths, ranges, depths, and especially the ray path radii, |R1| and |R2|, are not drawn to scale. The limiting ray paths for a sound source located at a depth of 800 m, and horizontal distance, r ¼ 0, are shown in *red*, launched at an angle of θ1 ¼ 8.31 above the horizontal. The turning point for the ray in the upper layer occurs at a horizontal distance from the source of r1, while the turning point for the ray in the lower layer occurs at a horizontal distance from the source of r3. The ray crosses the channel axis at r2 and enters the lower layer at an angle θ2 below the horizontal

Fig. 11.16 Ray diagram for transmission paths in the deep sound channel for a source located on the channel axis. Depth is in fathoms (1 fathom ¼ 6 feet ¼ 1.829 m), range is in miles (1 mi ¼ 1.609 km), and sound speed is in feet/second (1 fps ¼ 30.48 cm/s), as a function of depth is provided at the right side of the graph. For angles θ > θmax ¼ 12.2, waves escaping the channel are reflected from the air-water interface [11]

main thermocline, cmax ¼ 1523 m/s, Snell's law will guarantee that it will reach that depth with a ray that is again horizontal, after traveling a horizontal distance, r3 – r2 ¼ |R2| sin θ2 ¼ 22.8 km. From that point at horizontal distance from the source, r3, the path should repeat indefinitely with the cycle distance from the top of the channel back to the top of the channel being 2(r3 – r1) ¼ 55.6 km.

To calculate the angular width of the sound radiated by our source located at z ¼ 800 m below the surface, which will be trapped in the channel, we need to follow a ray that leaves the source at an angle below horizontal, θ1, that will also take it to the depth (z ¼ 3700 m) where the sound speed is again cmax. That initially downward-directed ray will also have to cross the sound channel axis at θ2 ¼ 12.66 if it is to become horizontal at z ¼ 3700 m. Again, Snell's law guarantees that (1507 m/s cos θ1) ¼ (1486 m/s cos θ2) ¼ 1523 m/s, so | θ1| ¼ | θ1| ¼ 8.31. That result is easy to visualize since we can imagine a "virtual source" that is located 2r1 away from the original source, where the ray makes an angle of 8.31 with respect to the horizontal. Any ray that leaves the source within 8.31 will be trapped in the sound channel created by the two sound speed gradients.

Again, Snell's law makes it easy to see that an initially horizontal ray leaving our source at a depth of z ¼ 800 m will go up and down in the channel from a depth of z ¼ 800 m to a depth of z ¼ 2581 m, where the sound speed will again be 1507 m/s and the rays will again be horizontal. Figure 11.16 diagrams ray paths from a more complicated sound channel for a source located on the channel axis [11].

The results illustrated in the previous example can be generalized by defining a maximum angle θmax (above or below the horizontal) that will result in the trapping of a wave launched at the channel axis, in terms of the minimum sound speed at the axis, cmin, and the maximum sound speed, cmax, at the bottom and top of the channel.

$$\cos \theta\_{\text{max}} = \frac{c\_{\text{min}}}{c\_{\text{max}}} \tag{11.39}$$

Expanding cos θmax for small angles and defining Δc ¼ cmax cmin, Eq. (11.39) can be approximated as θmax ffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Dc=cmax p . When the source is above or below the height or depth of the channel by a distance, zo, in a layer D thick, Snell's law, and the requirement that a trapped ray at the channel boundary is horizontal, provides the range of trapping angles, θ 1 ¼ θmax ffiffiffiffiffiffiffiffiffiffi zo=D p , when the source is not located on the sound channel's axis. Along the channel axis, zo ¼ D. At the extremes, zo ¼ 0, only the horizontal ray is trapped. The long-range transmission ability of the deep sound channel was used during World War II to rescue aviators that crashed at sea.9 A downed pilot would drop a small explosive charge that was rigged to detonate near the depth of the sound channel axis. If the sound were detected with two or more hydrophones, also located at the axis depth, the reception times could be used to localize the search by triangulation. Successful rescues were made this way using hydrophones connected to shore stations that were thousands of miles from the aircraft impacts. More recently, the same ability to make localizations at sea has been used for missile-impact location [12], and measurement of the time delays has been used to measure the mean temperature of the deep ocean [13].

#### 11.4.3 Propagation Delay\*

It is interesting to calculate the difference in the propagation times for sound traveling along different paths to reach the same horizontal distance from the source. Although the path along the axis is the shortest, it is also going through the medium with the minimum sound speed. The longer (curved) paths travel through water that has a faster sound speed. Does the shorter path beat the faster path?

Let's first consider the ray that is generated on the sound channel axis and travels along a circular path to the top of the channel. The initial angle of such a ray above the horizontal is θmax ffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Dc=cmax p , as already shown. The transit time, Tupper, for that ray can be calculated if we integrate from θ ¼ 0 to θ ¼ θmax, where the sound speed depends upon angle, c (θ ) ¼ cmax cos θ.

$$T\_{\rm upper} = \int\_0^{\theta\_{\rm max}} \frac{|R|d\theta}{c(\theta)} = \int\_0^{\theta\_{\rm max}} \frac{|R|d\theta}{c\_{\rm max} \cos \theta} = \frac{|R|}{c\_{\rm max}} \int\_0^{\theta\_{\rm max}} \frac{d\theta}{\cos \theta} \tag{11.40}$$

There is an analytical solution for the above definite integral, but simply writing the answer provides no useful physical insight.

$$\int\_0^{\theta\_{\text{max}}} \frac{d\theta}{\cos \theta} = \int\_0^{\theta\_{\text{max}}} \sec \theta \, d\theta = \ln \left[ \csc \theta - \cot \theta \right]\_0^{\theta\_{\text{max}}} \tag{11.41}$$

In these problems, the angles are generally small, so the series expansion of the sine and cosine functions and the binomial expansion can both be employed to simplify the integration and its interpretation.

$$\int\_0^{\theta\_{\text{max}}} \frac{d\theta}{\cos \theta} \cong \int\_0^{\theta\_{\text{max}}} \frac{d\theta}{\left(1 - \frac{\theta}{2}\right)} \cong \int\_0^{\theta\_{\text{max}}} \left(1 + \frac{\theta^2}{2}\right) d\theta = \quad \theta\_{\text{max}} + \frac{\theta\_{\text{max}}^3}{6} \tag{11.42}$$

The transit time for the axial ray, Taxial, that goes the same horizontal distance, r ¼ |R| sin θmax, is just Taxial ¼ r/cmin. Since cmin ¼ cmax cos θmax, the same approach can be used.

9 In this context, the channel was known as the SOFAR channel, which stood for sound fixing and ranging.

$$T\_{\rm axial} = \frac{|R|\sin\theta\_{\rm max}}{c\_{\rm max}\cos\theta\_{\rm max}} \cong \frac{|R|}{c\_{\rm max}} \frac{\left(\theta\_{\rm max} - \frac{\theta}{6}\right)}{\left(1 - \frac{\theta\_{\rm max}^2}{2}\right)} \cong \frac{|R|}{c\_{\rm max}} \left(\theta\_{\rm max} - \frac{\theta\_{\rm max}^3}{6}\right) \left(1 + \frac{\theta\_{\rm max}^2}{2}\right) \tag{11.43}$$

We have to be vigilant at this point to be sure we are evaluating Eq. (11.43) to the same level of approximation as we had in Eq. (11.42), which is correct to third order in θmax. The product of the two binomials includes a term that is linear in θmax as well as two terms that are third order.

$$T\_{axial} = \frac{|R|}{c\_{\max}} \left(\theta\_{\max} - \frac{\theta\_{\max}^3}{6} + \frac{\theta\_{\max}^3}{2}\right) = \frac{|R|}{c\_{\max}} \left(\theta\_{\max} + \frac{\theta\_{\max}^3}{3}\right) \tag{11.44}$$

These calculations demonstrate that the (longer) curved path is traversed in less time than the shorter (axial) path. We can use Eq. (11.44) with Eqs. (11.40) and (11.42) to approximate the travel time difference, ΔT.

$$
\Delta T = T\_{\text{axial}} - T\_{\text{upper}} \cong \frac{|R|}{c\_{\text{max}}} \left[ \left( \theta\_{\text{max}} + \frac{\theta\_{\text{max}}^3}{3} \right) - \left( \theta\_{\text{max}} + \frac{\theta\_{\text{max}}^3}{6} \right) \right] = \frac{|R|}{c\_{\text{max}}} \frac{\theta\_{\text{max}}^3}{6} \tag{11.45}
$$

#### 11.4.4 Under Ice Propagation

Sound propagation under Arctic ice provides an interesting variation on the sound channel, since sound can be reflected specularly (i.e., θr ¼ θi) from the ice sheet and the speed of sound under the ice increases monotonically with depth. The lack of solar heating at the surface causes the main thermocline, shown in Fig. 11.11, to be absent under Arctic ice. Typical ray paths under Arctic ice are shown in Fig. 11.17.

Fig. 11.17 Typical sound speed profile (right) and ray paths (left) under Arctic ice. Depth is in fathoms (1 fathom ¼ 6 feet ¼ 1.829 m), range is in kiloyards (1 kyd ¼ 0.9144 km), and sound speed is in feet/second (1 fps ¼ 30.48 cm/s) [14]

#### 11.4.5 Sound Focusing

A final illustration of these refractive processes in constant sound speed gradients, at a much different scale than the global sound propagation in the deep sound channel, comes from the evolution of marine mammals. The Cuvier's beaked whale (Ziphius cavirostris) is a member of the Ziphiidae family of toothed whales. It relies on echolocation, but in an aqueous environment, the excessive hydrodynamic drag and turbulence noise that would be produced by external "ears" (i.e., pinna), like those found on land mammals, is unacceptable. In this whale, the "ear" is located internally behind the jaw (in green), as shown in Fig. 11.18.

Figure 11.19 demonstrates that this region acts like a lens. Parallel rays that enter the "channel" are focused to a single point (see Fig. 11.20). As we have done many times now, we can represent the distribution of sound speeds in the whale's mandible using a piecewise-linear approximation. In this

Fig. 11.18 The sound-sensing organ of the Cuvier's beaked whale is shown by the small *green* area in the drawing of the whale's head at the right. At the left is a tomographic slice through the mandibular fat body which acts as a sound channel (i.e., lens) to focus sounds from the water to the whale's sound-sensing organ. The sound speed of the mandibular fat body (in yellow at the right) shows the sound speed as a function of location (*red* corresponding to 1700 m/s and *light blue* corresponding to 1300 m/s). Diagrams from Soldervilla, et al. [15]

Fig. 11.19 A simplified approximation to the sound channel (i.e., lens) created by the mandibular fat body that assumes that the sound speed gradient is constant, with its minimum value (1340 m/s) along the axis, and its maximum value (1400 m/s) at the upper and lower extremes, 6.0 cm above and below the sound channel axis

Fig. 11.20 (Left) Ray paths for an ordinary lens. (Right) Ray paths for a graded-index (GRIN) lens. In both cases, parallel rays enter the lens and are focused to a single point

case, the sound speed gradient g ¼ 60 m/s 0.06 m ¼ 1000 s1 ; hence, |R| ¼ co/g ¼ 1.4 m. Using Snell's law, cos θmax ¼ 0.957; hence, θmax ¼ 16.8. This makes the "focal length" of the whale's acoustic lens, dfocal ¼ |R| sin θmax ¼ 40.5 cm, just about the extent of the mandibular fat bodies.

The strategy of using a propagation speed profile to act as a lens is very popular in fiber-optic telecommunication systems and fiber-optic sensors. As shown in Fig. 11.20, this application is intended to capture light from an LED or solid-state laser and focus that light into the core of singlemode optical fibers that typically have waveguide (core) diameters on the order of 10 microns or less. In sensor applications [16–18], it usually is used to spread the light from an optical fiber over the surface of some sensing element after which a second GRIN lens injects the modulated light back into the optical fiber.

The graded-index lens was patented by Nippon Sheet Glass in 1968 and given the trademark SELFOC®. According to the accepted evolutionary timeline [19], marine mammals had produced the acoustical equivalent of GRIN lenses for at least 32 million years before the NSG patent was issued.

#### Exercises

	- (a) Transmitted amplitude. What is the amplitude of the pressure that is transmitted into the air?
	- (b) Trace velocity. What is the "trace velocity" of the waves along the interface?

After making the required measurement, the experimenter comes back from lunch and finds that a layer of ethyl alcohol (c ¼ 1150 m/s and ρm ¼ 790 kg/m3 ) has carefully been poured on top of the water in the aquarium. (Good grief!) Now a second beam of sound comes down and hits the bottom at a distance of 22.0 cm from the ultrasonic transducer, as diagrammed in the lower portion of Fig. 11.21. How thick is the layer of alcohol on top of the water, assuming that the two liquids do not mix?

	- sure prms ¼ 100 Pa is incident at 45 on a silt bottom with ρsilt ¼ 2000 kg/m3 and csilt ¼ 2000 m/s.
	- (a) Angle of refraction. What angle does the planewave make with respect to the normal once it enters the silt bottom?
	- (b) Transmitted pressure. What is the effective pressure of the wave in the silt bottom?
	- (c) Angle of intromission. At what angle does the planewave have to approach the silt from the water so that there is 100% transmission into the silt and no wave reflected back into the water?

$$c(z) = 1449 + 0.016z \tag{11.46}$$

The turning depth of the ray is 2.0 km where the sound speed is c(2 km) ¼ 1481 m/s.

#### References

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## Radiation and Scattering 12

#### Contents

Fig. 12.1 Photograph of the sound reinforcement system used by the Grateful Dead [1]. Nearly the entire stage is occupied with discrete vertical line arrays of loudspeakers to radiate the full-frequency spectrum of their music toward the audience with minimal leakage toward the ceiling where there would be excessive reverberation that would degrade the intelligibility

At this point, we have made a rather extensive investigation into the sounds that excite Helmholtz resonators as well as the departures from equilibrium that propagate as plane waves through uniform or inhomogeneous media. We have not, as yet, dealt with how those sounds are actually produced in fluids. Our experience tells us that sound can be generated by vibrating objects (e.g., loudspeaker cones, stringed musical instruments, drums, bells), by modulated or unstable flows (e.g., jet engine exhaust, whistles, fog horns, speech), by electrical discharges in the atmosphere (i.e., thunder), or by optical absorption (e.g., modulated laser beams). In this chapter, we will develop the perspective and tools that will be used for the calculation of the radiation efficiency of various sources and combinations of sources, like the sound reinforcement system shown in Fig. 12.1.

Only sound sources that behave in accordance with linear acoustics will be examined (until Chap. 15). We will find that the entire problem of both radiation and of scattering from small discrete objects can be reduced to understanding the properties of a compact source of sound that is small compared to the wavelength of the sound it is radiating.

"Superposition is the compensation we receive for enduring the limitations of linearity." Blair Kinsman [2]

We will then combine many of these radiation units, called monopoles, in various ways, including the option of having different units in different locations with different relative phases. Through superposition, this will permit construction of other convenient radiators that range from transversely vibrating incompressible objects (represented by two closely spaced monopoles that are radiating 180 out-ofphase) to arrays of discrete radiators (e.g., line arrays of various sizes like those shown in Fig. 12.1). Integration over infinitesimal monopole sound sources will allow modeling of extended vibrating objects (i.e., continuous, rather than discrete) such as loudspeaker cones, lightning bolts, and laser beams.

For convenience, our initial vision of a compact source will be assumed to be a pulsating sphere with radius, a λ. That pulsating sphere produces a sinusoidally varying volume velocity, U1ð Þ¼ a, t ℜe Ubð Þa ejω t h i. To generate such a volume velocity, we will let the radius of that sphere, Fig. 12.2 The loudspeaker enclosure at the right has an irregular shape and provides a total enclosed volume, V. To be treated as a "compact source," we require V1/3 λ ¼ c/ f ¼ 2πc/ω. The loudspeaker cone, visible at the top of the enclosure, has an effective piston area, Apist

r, oscillates harmonically with a single frequency, ω ¼ 2πf.

As will be demonstrated, an ordinary loudspeaker enclosure, like the one shown in Fig. 12.2, can be treated as a "compact source," or monopole, if its characteristic physical dimensions are small compared to the wavelength of sound produced. If the loudspeaker enclosures were approximated by a rectangular parallelepiped with volume, V, then "compactness" would require that V1/3 λ.

The oscillatory motion of the loudspeaker cone will be the source of the oscillatory volume velocity in the surrounding fluid. If the cone's effective piston area is Apist, located at x ¼ 0, and its linear position as a function of time is x1ð Þ¼ 0, t ℜe bxejωt ½ , then Ubð Þ 0, t - - - - - -¼ ωj j bx Apist cosð Þ ωt þ ϕ .

As long as the compactness criterion is satisfied, it is theoretically impossible to remotely determine the physical shape of a compact source—if the radiation from a source is that of a monopole, then it is only the source's volume velocity, U1(a, t), that is related to the sound pressure, p1(R,t), detected at distances, R a, beyond the maximum physical extent of the source.1

Within the constraint of compactness, it does not matter if Ubð Þa -- - -- - ¼ 4πa2ω bξ -- - -- - or Ubð Þ0 -- - -- - ¼ Apistωj j bx , the sound radiated into the far field will be identical. Given that realization, then for a compact source, dimensional analysis guarantees that the solution of the steady-state radiation problem reduces to calculation of the acoustic transfer impedance, Ztr ¼ bpð Þ R =Ubð Þa , with units that are the same as those for the acoustic impedances we have studied in our investigation of lumped elements: |Ztr| / ρmc/A,

1 The inability to remotely determine the shape of a sound source that is smaller than the wavelength of the sound it is radiating is known in both acoustics and in optics as the Rayleigh resolution criterion. See Sect. 12.8.1.

where ρm is the mean density of the medium, c is its sound speed, and A is a constant with the dimensions of area [m2 ]. The transfer impedances for several systems were provided in Sect. 10.7.4 since they were required to make reciprocity calibrations of electroacoustic transducers.

#### 12.1 Sound Radiation and the "Causality Sphere"

Light travels at the speed of light; sound travels at the speed of sound. Although those statements seem simplistic, if not tautological, they have calculable consequences that are significant. If a disturbance has a duration, Δt, then those speeds guarantee that the consequences of that event can initially only have influence within a distance, d ¼ cΔt, from the source of that disturbance. The term causality sphere was introduced in Einstein's Theory of General Relativity as the boundary in spacetime beyond which events cannot affect an outside observer. Although most commonly associated with cosmological issues and black holes, the concept is directly relevant to every form of energy that can only move through space with a finite propagation speed.

Before attempting a direct calculation of the acoustic pressure radiated by a compact source that is pulsating in an unbounded fluid medium, it will be instructive to make a simple estimate of the acoustic transfer impedance between an oscillating source of fluid located at the origin of some coordinate system and the pressure at some remote location, a distance, R, from the origin, using only the adiabatic gas law derived in Sect. 7.1.3: pV γ ¼ constant. Before making such a calculation for a source of sound radiating spherically in three dimensions, it will be reassuring to use the same procedure to reproduce a result that we have derived earlier by other means.

If we consider a close-fitting piston at the end of an infinite tube of uniform cross-section,2 so that both the tube and the piston have a cross-sectional area, A, then that piston can launch a sound wave down the tube. If the piston produces a volume velocity, Uð Þ¼ 0, t ℜe Ubð Þ0 ejω t h i, then the acoustic transfer impedance, Ztr, of such a tube, provided in Eq. (10.85) or in Eq. (10.106), can be used to calculate the steady-state acoustic pressure of the traveling wave produced by the piston's oscillations.

$$
\hat{\mathbf{p}} = \mathbf{Z\_{tr}} \hat{\mathbf{U}}(0) = \frac{\rho\_m c}{A} \hat{\mathbf{U}}(0) \quad \text{and} \quad p\_1(\mathbf{x}, t) = \Re e \left[ \hat{\mathbf{p}} e^{j(\alpha t - k\mathbf{x})} \right] \tag{12.1}
$$

This result describes a plane wave that propagates down the tube with magnitude, j j bp , while assuming that thermoviscous dissipation on the tube walls is negligible.

Let's now calculate this same result in another way. If we consider an interval during which the piston moves between its extreme positions, the piston will sweep out a volume in one-half of a period, δV ¼ Ubð Þ0 - - - - - ð Þ¼ T=2 Ubð Þ0 - - - - - -=2f , where T ¼ f 1 ¼ 2π/ω is the period of the piston's oscillations. In that time, the piston can "influence" a volume of the fluid within the tube, V ¼ (Aλ/2) ¼ (AcT/2), since the sound could travel a distance equal to one-half wavelength. Substitution of δV and V into the adiabatic gas law produces a corresponding δp.

2 Making the tube infinitely long is just a computational convenience. Any tube long enough that any reflections from the end of the tube return to the region of interest long after the interval of interest would suffice. Alternatively, a tube that had an anechoic termination (e.g., absorbing wedge) or had a matching resistive termination, Rac ¼ ρmc/A, would also behave as though it were infinitely long.