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Appendix: Answers and Hints toProblems
1.1. d = 3; γ = 5/3; c = (γRoT /M)1/2.
1.2.d
dt
∫∫∫V ∗
dV =∫∫
S∗vn dS.
1.3.d
dt
∫∫∫V ∗
ρ dV =∫∫∫
V ∗∂ρ
∂tdV +
∫∫S∗
ρvn dS.
1.4. f B = −ρgez.1.5. a) T ds = cvdT − (RT/ρ)dρ; u = cvT .
b) p = Kργ ; K = poρ−γo e(s−so)/cv .
1.6. a) vo = 0; ∂po/∂t = 0; ∂ρo/∂t = 0.b) ∂p/∂t = c2[∂ρ′/∂t + v ·∇ρo].c) ∇ ·
e) All such fields can be expanded as a sum of a finite number of factoredterms, each factor being a function of only one of the three coordinates.
1.48. a) ∇ · (f ∇p) = −f k2p + ∇f ·∇p
b)∫∫∫
p∇2pdV =∫∫∫
∇ · (p∇p)dV −∫∫∫
∇p ·∇pdV
d) Let p + εf be the good guess, where εf is a priori unknown.
2.1. 0.0628, 0.0628, 0.1885, and 0.0628 W2.2. a) LA = 90 dB; b) 90 more looms2.3. a) 7.2 dB; b) 3 dB; c) 0.89 dB2.4. 69 dB (A-weighted)2.5. L = constant − 20 log r .
The sound level drops by 6 dB per doubling of distance.2.6. 73.2 dB2.7. For octave band, L ≈ 93.1 dB; for flat response, L ≈ 94 dB.2.8. Energy per unit area and frequency bandwidth (Hz) is
8P 2pk
ρc
1
(2πf )2 [cos(2πf T ) − (2πf T )−1 sin(2πf T )]2
2.9. a) p2f = 1.8 × 10−5(Pa)2/Hz
b) L = 4 dB
Appendix: Answers and Hints to Problems 715
2.10. a) L = 84.8 dB2.11. For two machines with background, L = 86.0 dB
2.12. C+(ΔL) ≈ 10
ln 1010−ΔL/10
2.13. Cbg(ΔL) ≈ 10
ln 1010−ΔL/10
2.14. Sound level repeats at intervals of 0.5 s, has minimum value 82 dB,maximum value 97 dB
2.15. a) K = 1.2337; b) Underestimates level by 0.74 dB2.16. A� (or B�) in the third octave above middle C2.17. Applicable intermediate result for cited special case is
L {p(t)} = Re
{Ae−iωt
2π
∫ t
o
h(ξ)eiωξ dξ
}
2.18. The decibel loss (with Q2 = 2βxf 2o ) is
−10 log
{√2
Q
∫ Q√
2
Q/√
2e−y2
dy
}
2.19. Dp(τ) = SoΔfsin(πτΔf )
πτΔfcos(π [f1 + f2]τ), where Δf = f2 − f1
2.20. Variance in (p2b)est is 〈p2
b〉2/[T Δf ] in both cases2.21. L125 = 10 log[10(LC+0.6)/10 − 10(LA+5.4)/10]2.22. a) p2
f (f ) = (2 × 10−3)(f/103)4e−2(f/103)2
b) LA ≈ 87.7 dB
2.23. Occasional pass-by’s of noisy vehicles, firing of different cylinders on sameengine, atmospheric turbulence, rush hour traffic, pavement roughness andirregularities, aerodynamic noise of flow around moving vehicles.
2.24. 0.63 m2.25. 3 dB per doubling of distance2.26. Ratio is 1/[T Δf ]2.27. N > 1002.28. I = π
12√
32.29. 15, 19, or 22 keys per octave2.30. To carry through heuristic derivation involving interchange of integration
order, insert convergence guarantor e−ετ and recognize a Dirac deltafunction in limit ε → 0
2.31. Proper assumptions imply n-th peak of running time average is 1/T timestotal time integral of p2
n,F where pn,F is acoustic pressure, after filtering, ofn-th pulse. Use Parseval’s theorem.
2.32. 1 − (6/π2) = 0.3922.33. Insert a factor of e−εt2
on left side before inserting Fourier transformrelations and interchanging order of integration.
716 Appendix: Answers and Hints to Problems
2.34. p2f (f ) ≈ 8π2
100
∫ 200/T
100/T
|g(2πf )|2df
2.35. p(ω) = ippk
2πω; pF (t) = ppk
π
∫ ωo
√2
ωo/√
2
sin[ω(t − τ)]ω
dω
fraction = 1 − 1
23/2π2 = 0.964
2.36. a) v2f (f ) = ω2F 2
f (f )
(k − mω2)2 + ω2b2
b) (v2)av = F 2f (fr)
4mb, where 2πfr = (k/m)1/2
2.37. If L is measured in nepers, then L1 ⊕ L2 = L1 + 1
2ln[1 + e−2(L1−L2)].
2.38. a) Admissible. b) Admissible. c) Admissible only if b < 2a.
2.39. LE = 10 log
{p2
pk/p2ref
2√
2π2fotref
}
which decreases by 3 dB when fo doubles.
2.40. a) Derive p(ω) = i
2πω
∫dp
dteiωtdt and let
dp
dtequal (Δp)δ(t − to) plus a
bounded quantity. The contribution from the latter goes to zero at largeω at least as fast as ω−2.
b) p(ω) = − 1
2πω2
∫d2p
dt2eiωtdt where
d2p
dt2is (Δp)δ(t − to)
plus a bounded quantity.
2.41. a) Integrate by parts and use (d/dτ)hF (t − τ) = −(d/dt)hF (t − τ) .b) Prove that filtering operation commutes with time and spatial differenti-
ations.
3.1. vr = −ωb sin θ sin(ωt − φ) at r = a
3.2. Applicable intermediate result is the ratio of the octave band contributionto the mean squared pressure, when reflection is included, to that due toincident wave alone, this ratio being
2 + 2
{sin(Ψf2) − sin(Ψf1)
(f2 − f1)Ψ
}
where Ψ = (4πy/c) cos θI . Required minimum distance for y is 1.62 m.3.3. Let η(t − [(x/c) sin θI ]) be the displacement of the interface, such that
(vy)(+)o =
(∂
∂t+ vo
∂
∂x
)η
3.4.Z
ρc= 3
√2
5− i
4√
2
5; α = 0.723
3.5. a) θI = 85.4◦
Appendix: Answers and Hints to Problems 717
3.6. With Z = ρc(ζR + iζI ), one finds
α(θI ) = 4ζR cos θI
(ζR cos θI + 1)2 + (ζI cos θI )2
Values of α for any two angles of incidence allow ζR to be uniquelydetermined, but one can only determine the absolute magnitude of ζI .
3.7. Z =(
ρc
cos θI
)Beiψ
2A − Beiψ
3.8. τ = (2ρcωπa2)2
(2ρcωπa2)2 + (keff − ω2meff)2
3.9.1
4k(x0
p)2; nonoscillatory if keffmeff < (ρcπa2)2
3.10. Δf = −3.8 Hz; Q = 39.33.11. Right side of equation for fraction reflected,
|RI,II|2 =[(ρc)II sec θII − (ρc)I sec θI
(ρc)II sec θII + (ρc)I sec θI
]2
and Snell’s law equation are unchanged when the wave comes from mediumII at angle of incidence θII
3.12. cII = 5596 m/s; L = 0.070 m; 100% transmitted
3.13. k ≈ ω
c
[1 + 2i
(500)2
]; α ≈ ω
c
[2+i500
]
Ix ≈ |p|22ρc
; Iy ≈ − |p|21000ρc
|p|2 = P 2 exp{
− 2(ω/c)y/500}
exp{
− 4(ω/c)x/(500)2}
3.14. Z = Rf + iρc cot kL; the fraction absorbed is
4Rf ρc
(Rf + ρc)2 + (ρc cot kL)2.
The maximum value 4Rf ρc/(Rf +ρc)2 occurs when L = (2n+1)π/2k,with n integer.
3.15. Energy at time 10L/c is 50(ρAL)V 2o .
3.16. No; Yes; No3.17. At the ground, vpk = 0.005 m/s; intensity was 0.005 W/m2; at the cited
ionospheric height, vpk = 50 m/s; intensity was 0.005 W/m2.3.18. Rf = 1.5ρc; α = 0.96; if wall not present, then RTL = 4.9 dB3.19. a) 4792 Hz; b) 2727 Hz; c) 5455 Hz; d) Ratio is always 2:1
718 Appendix: Answers and Hints to Problems
3.20.d
λ= 1
2− 1
2πtan−1(X/2); ΔRTL = 10 log
[1 + X4
4X2 + 4
];
X = ωmpl
ρc
3.21. Elliptical counterclockwise path:
(8/9) (δx)2 + (δy)2 = (Vo/ω)2 exp{−(32)1/2ωy/c
}
Lowest point in trajectory corresponds to surface wave trough.
3.22. a)RI
TIII= i
2
[(ρc)I
(ρc)II− (ρc)II
(ρc)I
]sin(ωd/cII)
3.23. kII determined from
{(B/A)eikΔL}b{(B/A)eikΔL}a = sin(kIIdb)
sin(kIIda)
Then ZII determined (data from either “a” or “b” experiment) from
σ 3/2ωwith Φ(η) determined from numerical solution of η = Φ−3 − Φ−1.
Appendix: Answers and Hints to Problems 719
3.28. Applicable intermediate results are
d
dyln p = iωρZ−1
local;d
dyln vy = iω
ρ(c−2 − v−2
tr )Zlocal
Mass law follows from zeroth order approximation to
Zlocal(0) − Zlocal(d) = −iω
∫ d
o
ρ dy + iω
∫ d
o
[c−2 − v−2tr ]ρ−1Z2
localdy
where ρ = mpl/d and d is plate thickness.
4.1. p = (ρcVo)(a
r
)e−(c/a)(t−c−1r) if t > r/c
Half of the energy stays in near field4.2. The quantity ect/aψ satisfies inhomogeneous ordinary differential equation
for a harmonic oscillator under influence of a transient force. Green’sfunction G(t |τ) is 0 if t < τ and is (a/c) sin[(c/a)(t − τ)] if t > τ .
4.3. p = (Δp)e−s[cos s − sin s]H(s); where s = c
a
(t − r
c+ a
c
)
and Δp = ρcvC
a
rcos θ is pressure jump at r � a
4.4. P = 4
3π
ρc3(Ωa/c)6b2
4 + (Ωa/c)4
4.5. EK − EP → 1
4md(v2
C)av1 + 1
2 (ka)2
1 + 14 (ka)4
4.6. Fn =1∑
u=2−n
Fn,u(a/r)u; Fn,u = in+1
(1 − u)!n−2+u∑
t=0
(−1)t
t !
Gn =∞∑
u=−1
Gn,u(kr)u; Gn,u = iu+1+n
(u + 1)!n−2∑t=0
(−1)t
t !
Method of matched asymptotic expansions requires Gn+u,−u = Fn,u
4.7. p2f (f ) = ρ2a4
r2[1 + (ka)2] ; Lb+1 − Lb ≈ 3 dB
4.8. Lb+1 − Lb ≈ 9 dB
4.9. P = 2P1
[1 − sin kd
kd
]
where P1 is the power when only one source is active4.10. Write Helmholtz equation, surface boundary condition, radiation condition,
and Eqs. (4.6.9) in dimensionless form using a as a length scale and vtyp asa velocity scale; conclude that p/ρcvtyp is function of ka and x/a.
4.11. Power proportional to pM7/2/γ 5/2, equal to 0.01Pav,0 and 7500Pav,0 forsecond and third cases.
720 Appendix: Answers and Hints to Problems
4.12. P = 2πρc
[k2a4
1 + (ka)2|v
S|2 + k4a6
4 + (ka)4
|vC |23
]
|vC |/|vS| = 34.5 for equal contributions when ka = 0.1
4.13. Ratio = 3[27 + 6(ka)2 + (ka)4] + i3(ka)5
81 + 9(ka)2 − 2(ka)4 + (ka)6 ;
The real part is less than 1.25 up to ka = 1.278; the imaginary part is lessthan one-fourth of the real part up to ka = 1.666.
4.14. An applicable intermediate result (kr � 1) is
p ≈ − iωvSρ
4πreikr
∫ π
o
e−ika cos θS
[1 + ka
i + kacos θS
]2πa2 sin θSdθS
4.15. In the limit of large r , the integral for p reduces to
− iωρvC
4π
eikr
r
∫∫eβnS ·er [1 + Der ·nS]nS · erdS
where D = β2 + β
2 + β2 + 2βwith β = −ika
4.16. |δφ| < 0.57 degrees if kr = 0.1; |δφ| < 5.7 degrees if kr = 1.0
4.17. Total power = 2π
ρcQ2
11k4; for one alone it is (1/5)-th of this value.
4.18. For r > a, t > 0, the acoustic pressure is nonzero only if r−a < ct < r+a,and then has value [(Δp)/2][1 − (ct/r)]
4.19. |p|2 ≈ 4p210106
[(97)2 + (30)2]1
k2r2sin4 θ cos2 φ sin2 φ
Pav = 163cp2
10
ω2ρ
p10 varies with ω as ω5, increases by factor of 32; Pav varies as ω8,increases by factor of 256.
4.20. Pav = 2π
3ρc
A2k2a4
1 + (ka)2
4.21. t = (ln 10)a
c
4 + (ka)4
(ka)4
M
md
; where k = 1c
(kspM
)1/2
4.22. If all four in phase, power increases by factor of 16. For the other statedphasing, one has two perpendicular dipoles, 90◦ out of phase; radiation ispredominantly horizontal with intensity proportional to sin2 θ .
4.23. LA = 106 dB4.24. P = 12.6 W
Appendix: Answers and Hints to Problems 721
4.25. Pav = 2πK2k2
3ρc
4.26. ppk = 21/210−3K1 cos θ
kr
[1 +
(1
kr
)2]1/2
; k = ω1
c1
4.27. prms = 0.5 Pa
4.28. Applicable intermediate results are Φin = 2Ωa2
3πcos η sin η sin φF 1
2 (ξ)
F 12 (ξ) → −16
5
( a
2r
)3 ; Φin → −4Ωa5
45π
∂2
∂y∂z
(1
r
)
4.29. Superimpose solution represented by Eqs. (4.8.8), (4.8.10), and (4.8.11) withresult from problem 4.28. Let vC = −ΩΔ and use p = −ρ∂Φ/∂t .
4.30. A simple example is two closed loops with a common segment. Each loopshould have a voltage source and other circuit elements. Let e1 be the voltageof the left loop’s voltage source and let i1 be the corresponding current. Youmust prove that i1/e2 when e1 = 0 equals i2/e1 when e2 = 0.
4.31. Start with Eq. (4.9.7) with surface S consisting of spheres S1 and S2enclosing points x1 and x2, respectively. When S1 and S2 become small,vb and pb are regarded as constant over S1, etc. One must also prove that
∫∫n1padS1 → 0 in the limit of vanishing sphere radius.
5.1. 103 dB
5.2. G = R−1eikR − R−1I eikRI → −2d
d
dz(r−1eikr )
where {R2, R2I } = x2 + y2 + (z ∓ d)2
5.3.P
Pav,ff= 1 + 3
sin 2kd
2kd+ 3
sin 2√
2kd
2√
2kd+ sin 2
√3kd
2√
3kdkd > 23 is necessary criterion
5.4. b) F(kx, ei ) = 8 cos(kxei · ex) cos(kyei · ey) cos(kzei · ez)
c) p = pi(0, 0, 0)F (kx, ei )
5.5. a) Method of images gives combination of four free-field Green’s functions,with appropriate signs.
b) r2|G2k| = 16 cos2(kxS sin θ cos φ) sin2(kzS cos θ)
c) P = Pff
[1 + sin 2kxS
2kxS
− sin 2kzS
2kzS
− sin 2k(x2S + z2
S)1/2
2k(x2S + z2
S)1/2
]
5.6. P = ρck2
2π|vn|2A2
5.7. a) Pav = 4ρc(ka)2πa2|vn|25.8. a) |vn| = 0.32m/s b) |F | = 15.5N
722 Appendix: Answers and Hints to Problems
5.9. a) I = ρc|vn|2k2a4
8π2r2
[sin((1/2)ka sin θ cos φ)
(1/2)ka sin θ cos φ
]2
×[
sin((1/2)ka sin θ sin φ](1/2)ka sin θ sin φ
]2
c) ka = 2π
5.10. η = 0.0035.11. a) ω = 6πc/a; b) |p| = 2ρc|vn|5.12. 6ρc|vn|5.13. a) P = 1
4ρc(ka)2πa2|vC |2
b)1
12ρc(ka)4πa2|vC |2
5.14. Single cycle of a sinusoidal signal, beginning at t = 3a/4c, ending at t =5a/4c, with peak amplitude ρc|vn|.
5.15. b) Images at φ = (m/n)2π ± φS where m = 0, 1, 2, . . . , n − 1.d) Power increases by factor of 6.
5.16. Applicable integral is∫ π/2
o
sin2q θ sin θ dθ = 22q(q!)2
(2q + 1)!5.17. a2ρc(v2
n)av5.18. With η abbreviating w/a, one should find
vw = 2vn
πη[K(η2) − E(η2)] for η < 1
= 2vn
π[K(η−2) − E(η−2)] for η > 1
vw ≈ vnη/2 if η � 1 and vw ≈ vn/2η2 if η � 1
vw/vn is always positive, but there is a logarithmic singularity at η = 1.5.19. a) Start with
p = − iωρvn
2π
∫ π/2
−π/2
∫ ∞
o
R−1eikRwSdwSdφS
where R = [z2 + w2S]1/2 when x = 0
b) Applicable intermediate result is
p = ρcvn
{eikzH(−x) + sign(x)
2π
∫ π/2
−π/2eikψdφ
}
where ψ = [z2 + x2 sec2 φ]1/2
5.20. z = π
4ka2
Appendix: Answers and Hints to Problems 723
5.21. Applicable intermediate results are
∫∫Ein(−ikR)dl · dls =
∫∫∫∫[∇ ·∇SEin(−ikR)]dASdA
∇ ·∇SEin(−ikR) = ikR−1eikR
5.22. a) Applicable approximation and intermediate results are
Ein(η) ≈ η − 1
4η2;
∫∫R2d l · d lS = −4(Area)2
b) Applicable intermediate result is
∫∫R dl · d lS = 2K(a) + 2K(b) − 2L(a, b) − 2L(b, a)
where K(a) =∫ a
o
∫ a
o
|x − xS |dxdxS ;
L(a, b) =∫ a
o
∫ a
o
[(x − xS)2 + b2]1/2dxdxS
5.23. a) Reflection through lower wall implies z → −z; reflection through upperwall implies z → 2h − z.
Given the abbreviations, R+,−n = [w2 + (z ∓ zS − 2nh)2]1/2, one starts
with
p = S
∞∑n=−∞
(R+n )−1eikR+
n + S
∞∑n=−∞
(R−n )−1eikR−
n
c) Applicable integrals are
∫ ∞+iπ/2
−∞−iπ/2eiα cosh νdν = πiH(1)
o (α);∫ ∞+iπ/2
−∞+iπ/2eiβ sinh νdν = πiH(1)
o (iβ)
5.24. Derive the intermediate result
p = −ρcvn
π
∫ 2a
o
cos−1(u/2a)d
du{eik[u2+z2]1/2}du,
integrate by parts, and change the integration variable to φ, whereu = 2a sin φ.
724 Appendix: Answers and Hints to Problems
5.25. a) With the abbreviation Φ = (k/2)[(z2 + a2)1/2 − z], one has
Iz,av = (constant)
(1 + z
(z2 + a2)1/2
)sin2 Φ
b) Iw,av ≈ −(constant)wa2
2(z2 + a2)3/2[sin2 Φ − k(z2 + a2)1/2 cos Φ sin Φ]
c) Note that Iz,av on z-axis goes to 0 when kz = 0, 2.5π , 8π , etc.5.26. ka > 166.1. αc = 0.396.2. 4.063 mV6.3. T60 = 13s6.4. α = 0.6256.5. T60 = 19.9s6.6. Pout = 51.5 dB6.7. T60,II = 1.6s6.8. T60 = 6(ln 10)L/cα
6.9. T60 = 6π(ln 10)AFloor
cαLP
6.10. T60 = 6(ln 10)(2L/c)
− ln(1 − α1) − ln(1 − α2)6.11. Start with
E′K = 2πP ff
cV
∑ ′ [k2(n)/k2]Ψ 2(x0, n)
[k2 − k2(n)]2 + k2/c2τ 2
Take km = (0.01)(π/2)cτk2 and assume cτk � 1.For the second part of the problem, an applicable intermediate result is
E′K = P
cπk2
∫ km
o
dk′
(1 − k′/k)2 + (cτk)−2
6.12. RTL = 40 dB6.13. L = 96 dB
6.14. RTL = Ro + 10 log[1 − 10−Ro/10] − 10 log
[ln 10
10R0
]
6.15. T60 = 14.2 s6.16. Sound level in room 2 is also 90 dB.6.17. 2.6 dB6.18. N = 256.19. L1 = 111.8 dB6.20. α = 0.0676.21. Lout = 52 dB
0.380, 0.400, 0.425. The calculated N is 10.37 when true N jumps from 9to 10. The corresponding leading term is 4.02.
6.25. Ψ = A cos
(nxπx
Lx
)cos
(nyπy
Ly
)cos
((nz + 1
2 )πz
Lz
)
f = c
2
⎡⎣
(nx
Lx
)2
+(
ny
Ly
)2
+(
(nz + 12 )2
Lz
)2⎤⎦
1/2
6.26. There are 9 possibilities for (nx, ny, nz): the 3 permutations of (0,5,0) andthe 6 permutations of (3,4,0).
6.27. p = −i4000LxS
2LyLz + LxLy + LxLz
6.28. N(ω) ≈ LxLy
4π
(ω
c
)2 + 2(Lx + Ly)
π
(ω
c
)+ 1
46.29. a) fSch = 46 Hz;
b) L − Lo = −12.5 dB, z = −2.88, Probability of 0.055;c) Probability of 0.0036
7.1. a) Only the plane wave mode (ny = 0, nz = 0); b) P = 0.092 W7.2. N = 637.3. a) Only ω = nΩ
b) p = ρcVo
∑q
KqnJn(ηqnr/a)einφeiβqnz
where J ′n(ηqn) = 0 and βqn = [k2 − (ηqn/a)2]1/2;
Kqn = k
βqn
∫ 1o
Jn(ηqnξ)ξdξ∫ 1o
J 2n (ηqnξ)ξdξ
c) For only one spinning mode, one should have cη1n/a < ω < cη2n/a.
7.4. a) P = 6π
k2APff; b) 0
7.5. a) p = − i
kA[2πρcPff]1/2eikx
b) Answer doubles when xo = λ/2. (Cancellation occurs if xo = λ/4.)
7.6. Start with Eq. (4.9.7) and use relations such as
(∫∫va ·nindS
)2
= −D21p1
where the indicated integral is over side 2.
726 Appendix: Answers and Hints to Problems
7.7. b) Continuous-pressure two-portc) Circuit should have capacitances C1 and C2 in series, and these should
be in parallel with capacitance C3.7.8. a) Zright = Zleft = i(ρc/A) cot(kL/2); Zmid = −i(ρc/A) sin(kL)
b) π -network, two acoustic compliances, CA = V/(2ρc2), and an acousticinertance, MA = ρL/A
c) Mass between two springs
7.9.4A1A2
(A1 + A2 + A3)2
7.10. |T |2 = (a2ωρc/4T )2
1 + (a2ωρc/4T )2
7.11. IL = 20 log
[1 + Ab
2A
]
7.12. The fraction into the branch is(4A/ρc)|ZL|2
|1 + (ZLA/ρc)|2 Re
(1
ZL
− 1
ZR
)
7.13. Equations imply (πx/a) + sign(x) ln[(α−1 + α)/2] → Φ/2B as |x| → ∞,so Φ has apparent discontinuity at x = 0 of 4B ln[csc(πb/2a)]. Criterionfor ignoring constriction is ka ln[csc(πb/2a)] � π/2.
7.14.4X
(2 + X)2absorbed;
X2
(2 + X)2reflected;
4
(2 + X)2transmitted;
where X = 0.01ρcA/b
7.15. ωr = (4c2a/V )1/2
7.16. a) ZA = −i(ωρl′/A)[1 − (ω2r /ω
2)]b) ωr = (ρl′/A)−1/2[(V/ρc2) + G]−1/2
c) G � V/ρc2
7.17. a) MA = 1
(V/ρc2)(2πfr)2
b) l′ = Ac2
(V )(2πfr)2
c) |pin/pext| = 2πc3
(V )(2πfr)3
7.18. a) RA = 1.18 × 104 kg/(s · m4); CA = 3.6 × 10−9 m4s2/kg;MA = 1.12 × 102 kg/m4
b) Q = 15
7.19. b) ZA = ω4C2AM2
A − 3ω2CAMA + 1
−iωCA[2 − ω2CAMA]c) (MACA)1/2ωr equal to 0.6180 or 1.6180d) 180◦ out of phase at higher resonance
7.20. ZHR = ± iρc
2A
αT
(1 − α2T )1/2
7.21. a) V = 0.0325 m3; b) Fraction is 0.9944; c) Fraction is 0.9946
Appendix: Answers and Hints to Problems 727
7.22. The excess kinetic energy is the limit as L− → ∞ and L+ → ∞ of
∫∫∫ ′ 1
2ρ(∇Φ)2dV − (ρU2
12L+/2A+) − (ρU212L−/2A−)
where the volume integration extends over the region −L− < x < L+. Theintegration is accomplished with aid of ∇·(Φ∇Φ) = (∇Φ)2 and with innerregion outer boundary conditions such as
Φ → Φ+∞(t) + (U/A+)x as x → ∞
7.23. a) 1.11 × 10−10 W; b) RTL = 64 dB
7.24. Psc = 32a2
πIav
7.25. ΔL = 10 log
[1 +
(ωw2
ac
)2]
7.26. Power dissipated ≈ (Rf /2πa2)|pext|2(ωMA)2 + (Rf /πa2)2
Power transmitted ≈ (ωa/2πc)(ωMA)|pext|2(ωMA)2 + (Rf /πa2)2
where MA = ρ/2a and |pext|2 = 8ρcIi,av7.27. Use a symmetrical conically converging-diverging flow over a region of
length L on each side of orifice. Then vary L. Principle of minimumacoustic inertance yields
MA � ρ23/2
πa[1 − (a/b)]3/2
If a/b � 1, actual MA should be ρ/(2a)
7.28. Result for b/a � 1 should be same as for open end of duct with infiniteflange. King’s exact answer is 0.261ρ/b. Karal’s approximate answer in theb/a � 1 limit is 0.270ρ/b.
7.29. Fraction of incident power that is radiated is approximately 2(ka)2
7.30. a) l = 0.310 m; b) P = 2.963 W; c) Q = 58.9; d) 750 Hz7.31. AM/A = 8.6 and L = 0.085 m
7.32. The fraction transmitted is4A1A4A
23
(A1A4 + A23)
2
7.33.∂
∂t
(ρU2
2A+ Ap2
2ρc2
)+ ∂
∂x(pU) = 0
(pU)av is independent of x.7.34. Intermediate result is Bessel’s equation (n = 0 and ξ = kx)
d2p
dξ2 + 1
ξ
dp
dξ+ p = 0
728 Appendix: Answers and Hints to Problems
7.35. See text’s discussion on horn design. Applicable intermediate result isUdia/Uth = 1 − ω2MACA − iωCAZth
7.36.d2p
dx2 + 1
c2 (ω2 − ω2c )p = 0, where ωc = c(2an/πb2)1/2 is the cutoff
frequency.If b = 0.05 m, a = 0.002 m, and n is such that 10% of the area is holes,
∂t/∂w1 = 0 implies sin θI = sin θR , where w1/h = tan θI .8.11. Equate 0 to the derivative with respect to w1 of
t = (h2 + w21)
1/2
cI+ [d2 + (w − w1)
2]1/2
cII
8.12. ct = 2(L2 − R2)1/2 + 2R sin−1(R/L)
8.13. Applicable approximations (when x/R � 1 and |R − ct | � R) are
x ≈ α[1 − (ct/R)] + 20.5ct (α/R)3
z − R ≈ (ct − R)[1 − (α2/2R2)] + (α/R)410.375R
8.14. a) Both∂
∂αx(α, t) = 0 and
∂
∂αz(α, t) = 0 yield the same equation for t in
terms of α.b) Substitute for t into equations for x(α, t) and z(α, t).c) Caustic begins with a cusp and asymptotically approaches the lines
±x/R = 0.1027(z/R) − 0.1826
8.15. R(θo) = 2H tan θo + 20H cot θo
Minimum Rmin = 12.65H obtained when tan2 θo = 108.16. a) With appropriate definition of angle φ, a ray has circle radius R = (H −
h)/ cos φ; the ray that grazes ground has radius R = H and touchesground at w = [2Hh − h2]1/2.
c) cot = H ln
(H + (2Hh − h2)1/2
H − h
)+ [w − (2Hh − h2)1/2]
8.17. sin(x/H) = (tan θo) sinh(z/H)
8.18. b) x = d/2; additional roots (possible when d > 2b) are
x = (d/2) ± [(d/2)2 − b2]1/2
8.19. a) Ray leaving surface at angle θo with respect to z-axis has circle radius(co/α) csc θo. Rotating radius vector makes angle θ with horizontal, suchthat θ = π/2 at trajectory’s lowest point.
b) Caustic condition is (2n + 1) cos θ = cos θo
730 Appendix: Answers and Hints to Problems
8.20. (p2)av = ρcP
4πx2
1
1 + (x/2H)2
8.21. No. The caustic condition (∂w/∂θo)(∂z/∂θ) − (∂w/∂θ)(∂z/∂θo) = 0 issatisfied only at the source point.
8.22. a) Applicable intermediate results are
tanh(coτ/2H) = cos θo + (z/w) sin θo;
cot θo = (w/2H) − (z/w) + (z2/2Hw)
b) w2 + {z + 2H sinh2(coτ/2H)}2 = H 2 sinh2(coτ/H)
8.23. a) v = 2e5f (t − c−1w d − 10c−1
a H)
[ρwcw + ρa,0ca,0][10(ca/cw)H + d]b) The ratio of intensities, source above ground and source below ground,
observed at height 10H is
2(ρwcw + ρa,0ca,0)2
ρwcωρa,0ca,0
(ca
cw
)2 [1 + (cw/ca)d
10H
]2
8.24. A ray initially making small angle ε (radians) with z-axis has path
w ≈ ε
co
∫ z
o
cdz. The ray tube area is πw2 and the power passing through
ray tube is (P/4π)πε2.8.25. One must prove that
kz,I
ρIω2 = R2kz,I
ρIω2 + T 2kz,II
ρII(ω − kxvII)2
8.26. a) Applicable intermediate results are ∇po = c2∇ρo and
b) Take dot product of first displayed equation with ρov′+vop
′/c2; multiplysecond displayed equation by v′ · vo + p′/ρo.
c) W = w + I · vo/c2 and I ≈ wcn yield W ≈ w/Ω where Ω = c/(c +
n · vo)
8.27. b)∂
∂t
{A
[1
2ρo(v
′)2 + (p′)2
2ρoc2 + p′v′vo
c2
]}
+ ∂
∂x
{A[p′ + ρovov
′][v′ + p′vo
ρoc2
]}= 0
d)P 2A(vo + c)2
ρoc3 = constant
Appendix: Answers and Hints to Problems 731
8.28. If the lens surface is taken as flat on the source side, with thickness ho atr = 0, and if d is distance from source side of lens to focal point, thenh(r) = ho + 0.634(d − ho) − {[0.634(d − ho)]2 − (1.224r)2}1/2 which isthe equation of an ellipse.
8.29. p = Pe−ikz{1 + (−2z)−1R1/2o (w − Ro)
1/2 exp[ik(w − Ro)2/(−2z)]}
8.30. a)Iwith
Iwithout= 8
5+ 2
√3√
5cos(4π cos θ)
b) Radiation pattern given in parametric form (θi ranging from 0 to π/2)by θ = 2θi − sin−1([3/4] sin θi);
9.1. a) TS = 10 log(σback/4πR2ref); σback = (25/9)πa2(ka)4
b) σback = πa2
c) Increases TS by 12 dB and 0 dB, respectively.
9.2. a)dσ
dΩ= (4/9π2)a2(ka)4 cos2 θ cos2 θk
b) σback = (16/9π)a2(ka)4 cos4 θk
c) TS = 10 log(σback/4πR2ref)
d) The flow velocity is parallel to the disk’s faces, so the disk does notdisturb the flow.
9.3. An intermediate result, obtained by the use of Gauss’s theorem, is
∫∫∫ [(Φν − xν)
∂Δ2
∂xμ
− (Φμ − xμ)∂Δ2
∂xν
]dV = 0
9.4. Applicable equations are Uinto = 4πS/iωρ;
pout = B + ikS; pout = ZHRUinto
732 Appendix: Answers and Hints to Problems
9.5. a) ω2r = ksp/(M + 1
2Md), where Md = 4
3πa3ρ is the displaced mass
c) D ≈ (i/4)kra3MdB
(M + 12Md)[1 − (ωr/ω)2] + (i/6)(kra)3Md
9.6. Relative phases, associated with travel time differences, must be randomlydistributed over a range of at least 2π for the assumption to hold. Dimen-sion of the scattering volume in the direction ei − esc must be at leastλ/[2 sin(θ/2)].
9.7. a) Energy scatter per unit time is approximately π2(a/c)f 2resp
2f (fres)/RA
b) Attenuation in nepers per unit propagation distance is
α(f ) = 4πa2N
[1 − (f/fres)2]2 + (2aRA/ρfres)2
With increasing x, the spectral density loses a narrow notch of frequen-cies centered at fres.
9.8. One must solve (numerically) the integral equation
(A1/2pecho)x=0 =∫ ∞
o
J (xo)f (t − 2xo/c)dxo
and then determine A(x) by solving the ordinary differential equation4A2dJ/dx = (A′)2 = 2AA′′
9.9. a) (p2sc,ap)av ≈
(ΔΩtr
4π
)2
(kh)2[δ(ρc)
ρc
]2
(p2i )av
b) σback = k2h4(ΔΩtr)2
4π
[δ(ρc)
ρc
]2
9.10. An approximate analysis suggests the replacement
ΔΩtr →∫ 2π
o
∫ π
o
e−αθ2exp{2ikh sin θ cos μ tan φ} sin θdθ dμ
which approximates to (π/α) exp[−(khφ/α)2].9.11. A = 1
9.12. a)ω − ωo
ωo
= 1
8− (3/8)(ct/r)
[8 + (ct/r)2]1/2
b) At time t = 0 one is still hearing sound that left the source when x wasnegative.
9.13. f/fo = 1.1526 if t < 0 and f/fo = 0.8676 if t > 0.
where the Airy functions are evaluated at η = −Mx/R.9.17. a) w1(τ − η) → (η − τ)−1/4eiπ/4 exp{i(2/3)(η − τ)3/2}
b) The problem reduces to proving that, up through first order in τ , thequantity Φ = kox + (koR/2)1/3xτ/R − (2/3)η3/2 − τη1/2 is a goodapproximate solution of
(∂Φ/∂x)2 + (∂Φ/∂y)2 = (1 + [2z/R])k2o
c)dx
dz= 1 + [τ/(2k2
oR2)1/3]
(2z/R)1/2 − (τ/2)(2R/z)1/2(2k2oR
2)−1/3
9.18. a) Start with general expression for a creeping wave,
p = F(x)Ai(b1 − yei2π/3)
where y = z/l and l = (R/2k2o)
1/3
b) eshead = (p2cw)av,0
4πρc
(2/kR)1/3
(0.536)2
c) eshed = ρc(v2z,cw)av,0
(kR/2)1/3
4π(0.701)2
9.19. Start with same expression as suggested for Problem 9.18 and derive
eabs = 1
2Re
[eiπ/6
ρωl|F(x)|2Ai′(b1)Ai∗(b1)
]
734 Appendix: Answers and Hints to Problems
For a nearly “rigid” surface, the ratio is 3.61 ρckol/ZS
For a nearly “soft” surface, the ratio is 3.10 ZS/ρckol
9.20. The wave speed is nearly c, and the creeping wave energy, E , per unit areasatisfies cdE /dx = −eav. Applicable equations are
eshed = (4πωρl)−1(p2cw)av,0/|Ai(a′
1)|2
α = (√
3/2)(−a′1)/2kl2
9.21. a) Appropriate substitution for path length is Rθ ; ray strip width isproportional to R sin θ ; replace w/2kl2 by Rθ/2kl2; replace 1/w by1/[Rθ1/2(sin θ)1/2]
b) Use Eq. (9.5.19a) with yo = 0 and f1(yo) = 1.d) Rp/S = −0.0397 + 0.008i
9.22. a) Ai(a′1 − yei2π/3) → exp(−ia′
1e−i2π/3y1/2)ei(2/3)y3/2
2π1/2eiπ/12y1/4
Applicable intermediate result is
2k2l2y1/2 + 2
3y3/2 = ωτTR
Ray tube area varies as cτTRr sin θ
b) Appropriate substitutions are cτTR → r and Δθ → θ − π/2.
9.23. Two rays arrive, with the one from the backside undergoing a phase shift.The superimposed wave, with the abbreviation ε = R/(−z), has a factorw−1/2e−ikz cos εe−ikw sin ε . According to Problem 9.15, this corresponds to(2πkR/r)1/2e−iπ/4Jo(kR sin θ)eikr . In the result for Problem 9.22, onemust replace (sin θ)−1/2 by e−iπ/4(2πkR)1/2Jo(kR sin θ).
9.24. a) p = pi
e−iπ/2 exp{i(ωR/vph)(θ − π/2)} exp{−(αR)(θ − π/2)}(2kl2 sin θ)1/2(2Rl)−1/4[−a′
1Ai(a′1)]
where pi is the acoustic pressure amplitude of the incident plane wave.9.25. a) Both eiξτw1(τ − η) and eiξτ v(τ − η) satisfy the parabolic equation.
b) Applicable intermediate results are
∂
∂x= ε−1R−1(1 + ε2h)−1
(∂
∂ξ+ 2ξ
∂
∂η
)
∂
∂y= ε−2R−1(1 + ε2h)−1
[−2
∂
∂η+ ε2
(2b
∂
∂ξ− 2a
∂
∂η
)]
where 2a = (η − ξ2) and 2b = (ξ − η1/2)
9.26. (8/7)P
9.27. pdiff = Se2ikr eiπ/4
(πkr)1/22r
Appendix: Answers and Hints to Problems 735
9.28. Approximate AD(X) to first order in X. The two results are consistent andthe fluid velocity (both radial and tangential components) is infinite at theedge. Flow locally resembles potential flow.
9.29. a) Draw a triangle, with sides r, rS, and R, and denote smaller interior angles
by α and β, such that their sum is φ. Then appropriate intermediate resultsare
L − R = rS(1 − cos α) + r(1 − cos β); h = r
Ssin α = r sin β
9.30. a) p = 2Sz−1eikz − 4S(πka)−1/2L−1eiπ/4eikL
where L = (z2 + a2)1/2 + a
b) Interference minima where kz + 2nπ = kL + (π/4)
c) p = (4S/L)(πka)−1/2ei(kL+π/4)
9.31. a) If one lets Δφ = φ − π/3 be angular deviation from the shadow zoneboundary, with φ reckoned from other wall, then the diffraction parameterX is −Γ Δφ and
Here Γ = (krrS/πL)1/2 and RI is distance from the image source(obtained by reflection through the φ = 0 plane).
b) NF = (L − RI )/(λ/2)
9.32. TS = 10 log
[k2a4
4π2R2ref
]
9.33. |p2|/|p1| = 0.247
9.34. p = iS
1282λ
10.1. Substitute κ ′o = κo
(T ′
o
To
)3/2To + TAe−TB/To
T ′o + TAe−TB/T ′
o
into κ = κ ′o
(T
T ′o
)3/2T ′
o + TAe−TB/T ′o
T + TAe−TB/T
10.2. Fractional error ≈ (3/4)(γ − 1)/Pr ≈ 0.00084.10.3. The following derivatives of unit vectors are applicable:
∂er
∂θ= eθ ; ∂eθ
∂θ= −er ; ∂er
∂φ= sin θeφ
∂eθ
∂φ= cos θeφ; ∂eφ
∂φ= −er sin θ − eθ cos θ
736 Appendix: Answers and Hints to Problems
10.4. Derivation from Eq. (10.1.15) starts with setting s = s+s′, where s is slowlyvarying. An applicable intermediate result is
ρT∂s
∂t− ∇ · (κ∇T ) = μ
2
∑ij
φ2ij + κ
T(∇T ′)2
For the example, the large t and large x/λ limit, one should obtain
T ≈ 2I
(t
πκρcp
)1/2
− I
κ
(e−2αx
2α+ x
)
10.5. a) Approximate dispersion relations:
k ≈ ω
c+ i
ω2
c3δcl; k ≈ ω
c+ i
k2
cδcl
b) The Green’s function satisfies 4Gν − Gμμ = 0 with G(μ, 0) = δ(μ)
c) One must numerically evaluate (with s = [x − ct]/L)
p = 2P
3√
πe−s2/9
∫ 1
o
e−2(s′)2/9 sinh( 29 ss′) sin(πs′)ds′
10.6. For the vorticity mode:
w ≈ (1/2)ρv2; I = −μ∑ij
ej viφij ; D = 1
2μ
∑ij
φ2ij
For the entropy mode:
w ≈ 1
2
(ρT
cp
)o
s2; I ≈ − κ
To
T ′∇T ′; D = κ
To
(ΔT ′)2
10.7. p = (κρc2ω/cp)1/2β(ΔT )s cos(ω[t − (x/c)] + π/4)
10.8. The absorption cross section is 6(ωμ/2ρc2)1/2πa2
10.9. The attenuation αwalls in nepers per meter is determined by
2i[(ω/c)2 − k2y − k2
z ]1/2αwalls = lvorΨvor + (γ − 1)lentΨent
where ky = nyπ/Ly
Ψvor = [(ω/c)2 − k2y]ε(ny)L
−1y + [(ω/c)2 − k2
z ]ε(nz)L−1z
Ψent = (ω/c)2[ε(ny)L−1y + ε(nz)L
−1z ]
with ε(n) = 1 if n = 0 and equal to 2 otherwise.
Appendix: Answers and Hints to Problems 737
10.10. RA = 24.9 × 103 Pa · s/m3, in contrast to a radiation resistance of
1.39 × 103 Pa · s/m3; Q = 56
10.11. α(θi) ≈ α(0)/ cos θi providing cos θi � α(0).
10.12. The approximation p = 0 at x = h leads to an additional factor1 − e2ikh
1 + e2ikh
in Eq. (10.5.23), where k = (1 + i)α and α =(
4μω
ρc2T a2
)1/2
. In the limit of
large |kh| the transmission loss is
RT L = 10 log
[8μe2αh
π2a6ωγN2ρ
]
10.13. d) power = ω5πμa4
6c3|ξ |2
10.14. (p2)av and the power both vary with U as U6.
10.15. p2f (f ) ≈ ρ2U5a3Q
c2r2 ,
where the dimensionless quantity Q is a function of the Strouhal number,the Reynolds number, and angular coordinates.
10.16. p = W
2πh= 0.47 Pa
10.17. If one takes NB = 6, the pm depend on m and θ through the factor
(RL/D cos θ − 6)m[(m/2) sin θ ]6m
(6m)!10.18. A marked increase is expected when ωRLeff/c goes from below unity
to above unity. If one requires the amplitude of the Airy function toexceed 1/2 of its peak value, then [ωRLeff/c]θ lies between the limits,1 − 0.28(ωR/ω)2/3 and 1 + 16(ωR/ω)2/3.
10.19. An applicable intermediate result is
((T ′ − Tν)2)av = (ωτν)
2
1 + (ωτν)2((T ′)2)av.
Use the approximation T ′ ≈ (Tβ/ρcp)op and the thermodynamic
relation β2 = (γ − 1)cp
c2T
10.20.(
1 + τν
∂
∂t
) (∂
∂x+ 1
co
∂
∂t
)p = (c−1
o − c−1∞ )τν
∂2p
∂t2
10.21. a) T60 = T60,n
[1 + 2cT60,nαpl
6 ln 10
]−1
where the nominal reverberation time T60,n corresponds to αpl = 0.
738 Appendix: Answers and Hints to Problems
b) αpl = 3.8 × 10−4 Np/m.c) It can occur at any frequency above 117 Hz if the humidity is right, and
at almost any frequency if the frequency is greater than 5000 Hz.
10.22. a > 2.3 m.10.23. Maximum of 0.0155 Np/m occurs when RH ≈10.5%.10.24. Expand the complex wave number k(ω,μ,μB, κ, cv1, cv2) in a power series
in μ, κ , etc., and keep only up through the first order terms. The coefficientof any such term should be independent of the parameters that are associatedwith dissipation.
10.25. Applicable first order intermediate result is
iωρ0seq/p = 2cp
πc2βTo
∑ν
(ανλ)mω2τν
1 − iωτν
+ k2κ(β/ρcp)o
10.26. At 50 Hz: αμ = 2.5 × 10−8, αμB= 1.1 × 10−8, ακ = 1.0 × 10−8, and
with relative humidities of 0, 50, and 100%, αν1 = 1.0 × 10−4, 7.2 × 10−7,and 3.0 × 10−7, while αν2 = 7.4 × 10−6, 9.3 × 10−6, and 4.9 × 10−6. At5000 Hz: αμ = 2.5 × 10−4, αμB
= 1.1 × 10−4, ακ = 1.0 × 10−4, and withrelative humidities of 0, 50, and 100%, αν1 = 1.2 × 10−4, 6.7 × 10−3, and3.0 × 10−3, while αν2 = 7.7 × 10−6, 1.8 × 10−4, and 3.5 × 10−4.
10.27. If the plane were flying at 3000m, the calculated upper limit would be 102.2dB; at 6000m, it would be 114.4 dB.
11.1. B/A = 2ρc
[(∂c
∂p
)T
+ βT
ρcp
(∂c
∂T
)p
]
yields 4.7 for fresh water and 5.0 for sea water.
11.2. c = co
(p
po
)(γ−1)/2γ
; v = 2co
γ − 1
[(p
po
)(γ−1)/2γ
− 1
]
11.3.∞∑
n=N
1
n4/3 converges, but∞∑
n=N
1
n1/3 diverges
11.4. Integral form of y-th component of Euler’s equation for a stationary controlvolume is
d
dt
∫∫∫ρvydV +
∫∫[ρvyv ·n + pny]dS = 0
A derivation similar to that of Sect. 11.3 yields
[ρvy(vx − vsh)]+ = [ρvy(vx − vsh)]−
11.5. vsh ≈ c + 1
2β
Δp
ρc− 1
8β2 (Δp)2
ρ2c3
11.6. a) xonset = ρc3To
2βPo
Appendix: Answers and Hints to Problems 739
b) N-wave of peak overpressure Po, positive phase duration To.
c) T/To = Po/P =[
1 −(
x − xonset
cτN
)]1/2
where τN = ρc2To
βPo
11.7. a) x = 2ρc3To
β(P1 + P2)b) After coalescence (at x of part (a)), there is a shock of overpressure (P1 +
P2) that moves with speed c + (β/2ρc)(P1 + P2).
11.8. A plausible assumption is that the fluid eventually returns to its original
pressure, so T δs = cpδT and one accordingly finds δT = βP 3
3ρ3c4cp
for net
temperature increase.11.9. b) K = c3/4δ; B = (c3/4πδ)1/2
c) Insert v(0, τ ) = sin ωτ . The integration, performed using technique ofEq. (2.8.6), yields e−αx sin ωt ′.
11.10. The equation vt + βvvx′ = δvx′x′ is satisfied by
v = a∂F (x′, t)/∂x′
F(x′, t)provided a = −2δ/β and Ft = δFx′x′
Initial value F(x′0) determined from setting
ln F(x′, 0) = −(β/2δ)
∫ x′
o
v(ξ, 0)dξ
Initial value problem for F has solution
F(x′, t) = 1
2(πδt)1/2
∫ ∞
−∞F(x′′, 0) exp{−(x′ − x′′)2/4δt}dx′′
11.11. b) δD = c
2√
π
(μ
ρc2
)1/2
[1 + (γ − 1)(Pr)−1/2]LP
Ac) In the coefficient of ∂p/∂t , one replaces c by c + βp/ρc.
11.12. a) Multiply vt + βvvx′ = δvx′x′ by ρv
b) After insertion of the expression from Eq. (11.6.23), one finds
ρδ
∫ ∞
−∞(∂v/∂ξ)2dξ = 1
6ρβv3
sh
c) With Δp = ρcvsh, result from (b) is that of Eq. (11.4.11).
11.13. T = 13.2μs; P = 203 Pa.11.14. a) ronset = [√
ro + (rP /2√
ro)]2, where rP = ρc3To/(2βPo)
740 Appendix: Answers and Hints to Problems
b) T/To = (ro/r)1/2(Po/P ) =[
1
2+ (1/rP )
√ro(
√r − √
ro)
]1/2
11.15. a) A = 2β
ρc3
√ro(
√r − √
ro)
b) T/To = (Po/P )(ro/r)1/2 = [1 + (Po/To)A ]1/2
11.16. a) pfs = ro
r[ln(r/ro)]−1/2
[4Poρc3
ωβro
]1/2
T =[
4Poβro
ωρc3
]1/2
[ln(r/ro)]1/2
b) r∗ = ro; K =[
4Po
ωρcβro
]1/2
c) K = 0.051; pfs = 15.5 Pa; T = 20.5 μs.
11.17. For very low amplitudes, the fraction approximates to1
4
(ωPoβ
ρc3m
)2
.
11.18. εcrit = Tρf c3 cos θ
4πβH11.19. Asymptotic waveform given parametrically, ψ ranging from 0 to 2π , by
2πt ′/T = ψ − (1/2) sin ψ; p = ε exp{(hf − z)/2H } sin ψ wherect ′ = ct − (hf − z).
11.20. The shock thickness due to classical absorption (including viscosity, bulkviscosity, and thermal conduction) is 1.04 × 10−3 m, but O2 relaxation hasthe strongest effect (φ = 1.58) and causes an increment (Δl)O2 = 8.56 ×10−3 m to be added to the shock thickness.
11.21. For x > 2ρc3Δ2/βK , the pulse is triangular with initial shock and positivephase duration given by
P =[
2ρc3K
βx
]1/2
; T =[
2βKx
ρc3
]1/2
11.22. p ≈ (M2 − 1)1/4Ψ (ξ)
2π(2r)1/2; Ψ (ξ) = ∂
∂ξ
∫ ξ
−∞fz(μ)dμ
(ξ − μ)1/2
where ξ = V t − x − (M2 − 1)1/2r; r = (−z)
b) FW(ξ) = (M2 − 1)1/2Ψ (ξ)
2πV 2ρ
pfs = 0.819(M2 − 1)3/8(ρc2FL)1/2
23/4β1/2r3/4L1/4M
T = 0.81921/4β1/2r1/4F
1/2L
ρ1/2c2(M2 − 1)1/8L1/4
Appendix: Answers and Hints to Problems 741
11.23. vr + v/r − (β/c2)vvt ′ = (δ/c3)vt ′t ′
11.24. a) FW(ξ) = 32R2max
5L3/2 [x1/2U(x)H(x) − (x − 1)1/2V (x − 1)H(x − 1)]where U(x) = 5 − 20x + 16x2 and V (y) = 5 + 20y + 16y2, with
x = ξ/L and y = x − 1.b) K = 0.679.
11.25. Asymptotic N-wave given by Eqs. (11.10.18) and (11.10.21) with K =(4/3π)1/2 = 0.651 and L replaced by LN .
Name Index
AAckeret, J., 699n
Adler, Laszlo, 469Airy, George Biddell, 262n, 530n, 662n
Akay, Adnan, 181n
Alembert, Jean le Rond d, 5n, 18n, 22n
Allen, Clayton H., 676n
Alsop, Leonard E., 579n
Ambaud, P., 562n
Ando, Yoichi, 400n
Andree, C.A., 306n
Andrejev, N., 40n
Antosiewi, Henry Albert, 533n
Arago, Dominique Francois Jean, 269n
Aristotle, 3n
Arons, Arnold Boris, 154n
Astrom, E.O., 597n
Atkinson, F.V., 205n
Atvars, J., 467n
BBaade, Peter K., 319n
Babinet, Jacques, 269n
Bach, Johann Sebastian, 67n
Backhaus, Hermann, 269n
Bagenal, Hope, 313n
Baker, Bevan Braithwaite, 201n
Baker, Donald W., 528n
Ballantine, Stuart, 232n
Ballot, see Buys BallotBarash, Robert M., 539n
Barnes, A., 443Barton, Edwin Henry, 427n, 430n
Bass, Henry Ellis, 637n
Batchelor, George Keith, 510n, 586n, 592n,623n, 641n, 678n
Bateman, Harry, 468, 678Bauer, H.-J., 637n
Bazley, E.N., 471n
Becker, R., 680n
Bell, Alexander Graham, 72n
Bender, Erich K., 403n
Beranek, Leo LeroyAcoustic Measurements, 619n
Acoustics, 127n, 147n, 310n, 384n, 412n
anechoic sound chambers, 132n
audience and seat absorption, 302n
impedance of commercial materials, 127n
Music, Acoustics, and Architecture, 313n
notebooks of W. C. Sabine, 292n
tiles and blankets, 167n
Berendt, Raymond D., 352Bergassoli, A., 562Bergmann, Peter Gabriel, 9, 481Bernoulli, Daniel, 30n, 133n, 164n, 328n,
401n
Bernoulli, James, 164n
Bethe, Hans Albrecht, 251n, 666n
Beyer, Robert Thomas, 463n, 631n, 656n
Bies, David Alan, 167n
Biot, Jean Baptiste, 12n
Biot, Maurice Anthony, 563n
Biquard, P., 596n
Blackman, Ralph Beebe, 100n
Blackstock, David Theobald, 51n, 660n, 668n,676n, 679n, 687n, 696n, 711n
Circular piston with baffle, 252–254far-field radiation, 261–263field on axis, 268–269pressure on surface, 252–254radiation impedance, 255–256radiation pattern, 262–263transient solution, 264–268transition to the far field, 271–283
Clamped electric impedance, 230Clebsch potentials, 463n
Coalescence of shocks, 675Cocktail party effect, 319–320, 359Coefficient of nonlinearity, 639Coincidence frequency, 145–146Complex elastic modulus, 165–166Complex number representation, 26–27Compliance, acoustic, 380Compressibility, 32, 643n
Equivalent area of open windows, 301, 324Ergodic process, 95Error function, 108Erythrocytes as scatterers, 529Euler-Bernoulli plate, 164Eulerian description, 6n
Euler-Lagrange equation, 433–434Euler-Mascheroni constant, 348Euler’s equation of motion for a fluid, 8–11Euler’s formula, 26
requirements for unique solution, 200–201solution for one-dimensional propagation,
54Inner expansion (see Matched asymptotic
expansions)Insertion loss
of barriers, 569–572of mufflers, 404–405
Instantaneous entropy function, 634–635Institute of Electric and Electronics Engineers
(IEEE), 138n, 476n, 494n
Integer-decibel approximation, 79, 80Integrodifferential equation for transient pulse
in absorbing duct, 710Intensity
acoustic (see Acoustic intensity)of radiation, 298n
Intensity level, 73Interface, 114, 123
between air and water, 153–154between different fluids, 148–153between fluid and elastic solid, 146between moving fluids, 114n, 160, 497point source above, 469–474(See also Boundary conditions; Reflection;
Transmission)Internal energy
of ideal gas, 31n
rotational, 631in second law of thermodynamics, 14translational, 631–634vibrational, 631
KKeller’s law of edge diffraction, 565Key note, 67Kinetic energy, 41
principle of minimum, 392–393Kinetic theory of gases, 31Kirchhoff approximation, 240–245
for orifice transmission, 378–379relation to rigorous diffraction theory, 244,
556Kirchhoff-Helmholtz integral theorem,
208–211in derivation of Rayleigh integral, 240–241extension to include viscosity, 590, 593integral equation for surface pressures, 213multipole expansion of, 213–214, 622
Kirchhoff’s dispersion relation, 599Kirchhoff’s laws of circuit analysis, 391
Pressure node in traveling wave, 24Pressure-release surface, 126, 133Principal value of integral, 155Probability density function, 343–344Propagation, 3Pulse-echo sounding, 507–509
Seismology of the atmosphere, 451Sensation unit, 72Separation constant, 329Separation of variables method, 362Shadow zone, 489, 540–550
behind curved body, 548caused by intervening wedge, 550–553external to main beam, 276–277limiting ray for, 540on nonilluminated side of caustic, 532in stratified medium, 541–544(See also Creeping waves; Diffraction)
Transfer functions, 77, 92–95, 104, 111Transient waves
diffracted by wedge, 489, 563Fourier integral representation, 91from piston in tube, 130–139, 555from piston in wall, 264–268reflection at interface, 154sound-exposure, 91–92from transversely oscillating sphere,