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Appendix III

Answers to problems

Problem 2.1.

From Eq. (2.112), we see that the eigenfrequencies of the pan flute are in multiples of 2n + 1. In other words, they are in odd multiples, and the even ones are missing. The fundamental resonance frequency is given by f = c/(4l) = 345/(4 × 0.294) = 293 Hz.

From Eq. (2.94), we see that the eigenfrequencies of the pan flute are in multiples of n. In other words, they are in odd and even multiples. The fundamental resonance frequency is given by f = c/(2l) = 345/(2 × 0.294) = 587 Hz.

Problem 2.2.

In the steady state, the homogenous wave equation in cylindrical coordinates from Eq. (2.23) becomes

$(\frac{\partial^{2}}{\partial w^{2}} + \frac{1}{w} \cdot \frac{\partial}{\partial w} + k^{2}) \tilde{p} (w) = 0.$

Although a solution is given by Eqs. (2.125) and (2.126), we omit the Y ₀ function because of continuity at the center so that the solution becomes

$\tilde{p} (w) = {\tilde{p}}_{-} J_{0} (k w),$

where the J ₀ function is analogous to the cosine function in the case of a closed tube and satisfies the boundary condition at the center

$\frac{\partial}{\partial w} {\tilde{p} (w) |}_{w = 0} = 0.$

At the cylinder wall, where w = a, we have

$\frac{1}{- j k ρ_{0} c} \cdot \frac{\partial}{\partial w} {\tilde{p} (w) |}_{w = a} = - {\tilde{u}}_{0},$

which leads to

${\tilde{p}}_{-} = \frac{j ρ_{0} c {\tilde{u}}_{0}}{J_{1} (k a)} .$

Hence,

$\tilde{p} (w) = j ρ_{0} c {\tilde{u}}_{0} \frac{J_{0} (k w)}{J_{1} (k a)} .$

(A3.1)

Next, we shall rewrite the inhomogeneous Eqs. (2.80) and (2.81) in cylindrical coordinates using the Laplace operator of Eq. (2.23)

$\begin{matrix} (\frac{\partial^{2}}{\partial w^{2}} + \frac{1}{w} \cdot \frac{\partial}{\partial w} + k^{2}) \tilde{p} (w) = - δ (w - a) \frac{\partial}{\partial w} {\tilde{p} (w) |}_{x = a} \\ = j k ρ_{0} c δ (w - a) {\tilde{u}}_{0}, \end{matrix}$

(A3.2)

which includes the sound source at w = a on the right-hand side, where δ is the Dirac delta function. Let the solution be in the form of an eigenfunction expansion

$\tilde{p} (w) = \sum_{n = 0}^{\infty} {\tilde{A}}_{n} J_{0} (β_{n} w / a),$

(A3.3)

where β _n is the solution to

$J_{1} (β_{n}) = 0.$

When the wall is stationary, this satisfies the boundary conditions

$\frac{\partial}{\partial w} {\tilde{p} (w) |}_{w = 0} = 0, \frac{\partial}{\partial w} {\tilde{p} (w) |}_{w = a} = 0.$

Inserting Eq. (A3.3) into Eq. (A3.2) and multiplying both sides by

$J_{0} (β_{m} w / a)$

while integrating over the radius of the cylinder gives

$\begin{matrix} \sum_{n = 0}^{\infty} {\tilde{A}}_{n} (k^{2} - \frac{β_{n}^{2}}{a^{2}}) \int_{0}^{a} J_{0} (β_{m} w / a) J_{0} (β_{n} w / a) w d w \\ = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{a} J_{0} (β_{m} w / a) δ (w - a) w d w . \end{matrix}$

The two integrals have the following identities:

$\int_{0}^{a} J_{0} (β_{m} w / a) J_{0} (β_{n} w / a) w d w = {\begin{matrix} 0, m \neq n \\ a^{2} J_{0}^{2} (β_{n}) / 2, m = n \end{matrix}$

$\int_{0}^{a} J_{0} (β_{m} w / a) δ (w - a) w d w = a J_{0} (β_{m}) .$

The first is the property of orthogonality from Eq. (A2.101a) of Appendix II, and the second is a property of the Dirac delta function from Eq. (A2.154) of Appendix II. Hence,

$J_{0} (β_{m} w / a)$

in this case is the orthogonal (or normalizing) function and we can eliminate the summation to yield

${\tilde{A}}_{n} = \frac{2 j k a ρ_{0} c {\tilde{u}}_{0}}{(k^{2} a^{2} - β_{n}^{2}) J_{0} (β_{n})}$

so that the solution of Eq. (2), given by Eq. (3), becomes

$\tilde{p} (w) = 2 j k a ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} \frac{J_{0} (β_{n} w / a)}{(k^{2} a^{2} - β_{n}^{2}) J_{0} (β_{n})} .$

(A3.4)

By comparing the pressures given by Eqs. (A3.1) and (A3.4) at w = a, we obtain

$\frac{J_{0} (k a)}{J_{1} (k a)} = \sum_{n = 0}^{\infty} \frac{2 k a}{k^{2} a^{2} - β_{n}^{2}} .$

Hence,

$Q (x) = \frac{2 J_{1} (x)}{x J_{0} (x)} = {(\sum_{n = 0}^{\infty} \frac{x^{2}}{x^{2} - β_{n}^{2}})}^{- 1} .$

For large n, we can substitute

$β_{n} \approx (n + 1 / 4) π, n \to \infty .$

Problem 7.1.

From Eq. (6.10), we obtain Q _TS = 0.46 × 7.58/(0.46 + 7.58) = 0.43. This is within 7% of the Q _TS value in Table 7.4 required for a 0.01 dB Chebyshev alignment. Multiplying the V _AB/V _AS figure from the table by V _AS of the drive unit gives the box volume V _AB = 1.5511 × 19.5 = 30.3 L. Also, from the table, we see that the required port resonance frequency is f _B = f _B/f _S × f _S = 0.8838 × 50 = 44.2 Hz. From the “Summary of bass-reflex design” on p. 334, we obtain the maximum peak pressure

$p_{\max} = 2 \sqrt{2} \times 10^{(\frac{S P L_{\max}}{20} - 5)} = 2.83 Pa$

and the maximum volume displacement require to produce that pressure

$V_{\max} = \frac{r p_{\max}}{2 π f_{B}^{2} ρ_{0}} = \frac{1 \times 2.83}{2 \times 3.14 \times {44.2}^{2} \times 1.18} = 195 {cm}^{3} .$

We let the port volume be V _P = 10V _max = 1.95 L. The approximate tube length t is given by Eq. (7.97)

$t \approx \frac{c}{ω_{B}} \sqrt{\frac{V_{P}}{V_{A B}}} = \frac{345}{2 \times 3.14 \times 44.2} \sqrt{\frac{1.95}{30.3}} = 31.5 cm$

where the cross-sectional area of the tube from Eq. (7.96) is S _P = V _P/t = 1.96 × 10⁻³/0.315 = 62.2 cm² and the diameter is

$d_{P} = 2 \sqrt{S_{P} / π} =$

8.9 cm. If we use a 9 cm diameter tube with a cross-sectional area S _P = ¼π × 9² = 63.6 cm², the exact length is given by Eq. (7.98)

$t = \frac{S_{P} c^{2}}{V_{A B} ω_{B}^{2}} - 0.84 \sqrt{S_{P}} = \frac{63.6 \times 10^{- 4} \times 345^{2}}{30.3 \times 10^{- 3} \times {(2 \times 3.14 \times 44.2)}^{2}} - 0.84 \sqrt{63.6 \times 10^{- 4}} = 25.7 cm .$

Also, from Table 7.4, we see that the cutoff frequency is f _3dB = f _3dB/f _S × f _S = 0.8143 × 50 = 40.7 Hz.

Problem 10.1.

The total volume is simply V = 12.5 × 10 × 8 = 1000 m³ and, from the table, the total absorptive area is S _tot = 100 + 125 + 32 = 257 m². The average Sabine absorption coefficient α _tot is given by Eq. (10.52)

$α_{t o t} = \frac{\sum α_{s, i} S_{i}}{S_{t o t}} = \frac{0.7 \times 100 + 0.6 \times 125 + 0.5 \times 32}{257} = \frac{161}{257} = 0.626.$

The reverberation time T is given by Eq. (10.50)

$T = \frac{0.161 V}{S_{t o t} α_{t o t}} = \frac{0.161 V}{\sum α_{s, i} S_{i}} = \frac{0.161 \times 1000}{257 \times 0.626} = 1 s,$

which according to Fig. 10.13 is fairly optimum for the size of auditorium. The sound strength G is given by Eq. (10.65)

$\begin{matrix} G = 10 \log_{10} (\frac{100}{r^{2}} + \frac{1600 π (1 - α_{t o t})}{S_{t o t} α_{t o t}}) \\ = 10 \log_{10} (\frac{100}{10^{2}} + \frac{1600 \times 3.14 \times (1 - 0.626)}{257 \times 0.626}) = 11 dB . \end{matrix}$

The distance r _rev from a source to equal direct and reverberant fields is given by Eq. (10.68)

$r_{e v} = \frac{1}{4} \sqrt{\frac{Q S_{t o t} α_{t o t}}{π (1 - α_{t o t})}} = \frac{1}{4} \sqrt{\frac{1 \times 257 \times 0.626}{3.14 \times (1 - 0.626)}} = 2.93 m .$

The average Eyring absorption coefficient α _ey is given by Eq. (10.70)

$α_{e y} = 1 - e^{- 0.161 V / (S_{t o t} T)} = 1 - {2.72}^{- 0.161 \times 1000 / (257 \times 1)} = 0.466.$

The required acoustic power is given by Eq. (10.71) using the maximum peak SPL of a large orchestra from column three of Table 10.2

$\begin{matrix} W = \frac{4 \times 10^{(S P L / 10) - 10}}{ρ_{0} c} {(\frac{Q}{4 π r_{r e f}^{2}} + \frac{4 (1 - α_{e y})}{S_{t o t} α_{e y}})}^{- 1} \\ = \frac{4 \times 10^{(102.5 / 10) - 10}}{1.18 \times 345} {(\frac{1}{4 π \times 10^{2}} + \frac{4 \times (1 - 0.466)}{257 \times 0.466})}^{- 1} = 0.938 W . \end{matrix}$

If the loudspeakers have an efficiency of 0.5%, the required amplifier power is 0.938 × 100/0.5 = 188 W.

Problem 12.1.

The specific radiation impedance is given by

$Z_{s} = \frac{\tilde{p} (a)}{{\tilde{u}}_{0}} = j ρ_{0} c \frac{H_{0}^{(2)} (k a)}{H_{1}^{(2)} (k a)},$

which is separated into real and imaginary parts as follows:

$\begin{matrix} Z_{s} = j ρ_{0} c \frac{H_{0}^{(2)} (k a) H_{1}^{(1)} (k a)}{H_{1}^{(2)} (k a) H_{1}^{(1)} (k a)} \\ = ρ_{0} c \frac{Y_{0} (k a) J_{1} (k a) - J_{0} (k a) Y_{1} (k a) + j (J_{0} (k a) J_{1} (k a) + Y_{0} (k a) Y_{1} (k a))}{J_{1}^{2} (k a) + Y_{1}^{2} (k a)} . \end{matrix}$

Applying the Wronskian yields,

$Z_{s} = ρ_{0} c (\frac{2}{π k a (J_{1}^{2} (k a) + Y_{1}^{2} (k a))} + j \frac{J_{0} (k a) J_{1} (k a) + Y_{0} (k a) Y_{1} (k a)}{J_{1}^{2} (k a) + Y_{1}^{2} (k a)}) .$

Problem 12.2.

The spherical cap is set in a rigid sphere of radius R and moves with a radial velocity

${\tilde{u}}_{0}$

such that the velocity distribution is described by

$\tilde{u} (R, θ) = {\begin{matrix} {\tilde{u}}_{0}, 0 \leq θ \leq α \\ 0, α < θ \leq π \end{matrix},$

(A3.1)

where α is the half angle of the arc formed by the cap. The total volume velocity is given by

${\tilde{U}}_{0} = {\tilde{u}}_{0} R^{2} \int_{0}^{2 π} \int_{0}^{α} sin θ d θ d φ = S {\tilde{u}}_{0},$

(A3.2)

where S is the effective surface area of the cap given by

$S = 2 π R^{2} (1 - \cos α) = 4 π R^{2} \sin^{2} (α / 2) .$

(A3.3)

Near-field Pressure. Again, assuming that the pressure field generated is a general axisymmetric solution to Eq. (2.180), the Helmholtz wave equation in spherical coordinates is

$\tilde{p} (r, θ) = ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} A_{n} h_{n}^{(2)} (k r) P_{n} (cos θ) .$

(A3.4)

Applying the velocity boundary condition gives

$\begin{matrix} \tilde{u} (R, θ) = \frac{1}{- j k ρ_{0} c} \frac{\partial}{\partial r} {p (r, θ) |}_{r = R} \\ = \frac{{\tilde{u}}_{0}}{- j k} \sum_{n = 0}^{\infty} A_{n} {h^{'}}_{n}^{(2)} (k R) P_{n} (\cos θ) = {\begin{matrix} {\tilde{u}}_{0}, 0 \leq θ \leq α \\ 0, α < θ \leq π, \end{matrix} \end{matrix}$

(A3.5)

where the derivative of the spherical Hankel function

${h^{'}}_{n}^{(2)} (k R)$

is given by Eq. (12.32). We now multiply both sides of Eq. (A3.5) with the orthogonal function P _m (cos θ) and integrate over the surface of the sphere, where the area of each surface element is given by δS = 2πR ² sin θ δθ, so that

$\frac{1}{- j k} \sum_{n = 0}^{\infty} A_{n} {h^{'}}_{n}^{(2)} (k R) \int_{0}^{π} P_{n} (cos θ) P_{m} (cos θ) sin θ d θ = \int_{0}^{α} P_{m} (cos θ) sin θ d θ,$

(A3.6)

from which we obtain the coefficients as follows:

$A_{n} = - j k (2 n + 1) \frac{\sin α P_{n}^{- 1} (cos α)}{2 {h^{'}}_{n}^{(2)} (k R)},$

(A3.7)

where we have used the integral identities from Eqs. (A2.66) and (A2.69) in Appendix II. Finally, by inserting Eq. (A3.7) in Eq. (A3.4), we can write the near-field pressure as

$\tilde{p} (r, θ) = - j k ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} (n + \frac{1}{2}) \sin α P_{n}^{- 1} (\cos α) P_{n} (\cos θ) \frac{h_{n}^{(2)} (k r)}{{h^{'}}_{n}^{(2)} (k R)} .$

(A3.8)

Far-field Pressure. In the far field, we can use the asymptotic expression for the spherical Hankel function from Eq. (12.18), which when inserted into Eq. (A3.8) gives

${\tilde{p} (r, θ) |}_{r \to \infty} = \tilde{p} (r, θ) = - j k ρ_{0} c S {\tilde{u}}_{0} \frac{e^{- j k r}}{4 π r} D (θ),$

(A3.9)

where S is the dome effective area given by Eq. (A3.3) and

$D (θ) = \frac{\sin α}{2 k^{2} R^{2} \sin^{2} (α / 2)} \sum_{n = 0}^{\infty} \frac{j^{n + 1} {(2 n + 1)}^{2} P_{n}^{- 1} (\cos α) P_{n} (\cos θ)}{n h_{n - 1}^{(2)} (k R) - (n - 1) h_{n + 1}^{(2)} (k R)} .$

(A3.10)

The far field on axis response is given by

$D (0) = \frac{\sin α}{2 k^{2} R^{2} \sin^{2} (α / 2)} \sum_{n = 0}^{\infty} \frac{j^{n + 1} {(2 n + 1)}^{2} P_{n}^{- 1} (\cos α)}{n h_{n - 1}^{(2)} (k R) - (n - 1) h_{n + 1}^{(2)} (k R)} .$

(A3.11)

Radiation Impedance. The total radiation force

$\tilde{F}$

is given by

$\begin{matrix} \tilde{F} = R^{2} \int_{0}^{2 π} \int_{0}^{α} \tilde{p} (R, θ) sin θ d θ d φ \\ = - 4 π R^{2} j ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} \frac{{(n + \frac{1}{2})}^{2} {(\sin α P_{n}^{- 1} (cos α))}^{2} h_{n}^{(2)} (k R)}{n h_{n - 1}^{(2)} (k R) - (n - 1) h_{n + 1}^{(2)} (k R)} \end{matrix}$

(A3.12)

using the identity of Eq. (A2.69). The specific impedance Z _s is then given by

$Z_{s} = \frac{\tilde{F}}{{\tilde{U}}_{0}} = - j ρ_{0} c \frac{\sin^{2} α}{\sin^{2} (α / 2)} \sum_{n = 0}^{\infty} \frac{{(n + \frac{1}{2})}^{2} {(P_{n}^{- 1} (\cos α))}^{2} h_{n}^{(2)} (k R)}{n h_{n - 1}^{(2)} (k R) - (n - 1) h_{n + 1}^{(2)} (k R)},$

(A3.13)

where we have used the expression for

${\tilde{U}}_{0}$

from Eq. (A3.2).

Problem 13.1.

The far-field pressure is given by Eq. (13.6)

$\tilde{p} (x, y, z) = 2 j k ρ_{0} c {\tilde{u}}_{0} \int_{- l_{y} / 2}^{l_{y} / 2} \int_{- l_{x} / 2}^{l_{x} / 2} {g (x, y, x | x_{0}, y_{0}, z_{0}) |}_{z_{0} = 0} d x_{0} d y_{0},$

where the Green's function is given by Eqs. (13.33) and (13.34)

$g (x, y, x | x_{0}, y_{0}, z_{0}) = \frac{e^{- j k (x (x - x_{0}) + y (y - y_{0}) + z (z - z_{0})) / R}}{4 π R},$

$R = \sqrt{x^{2} + y^{2} + z^{2}} .$

By combining the above equations, we get

$\begin{matrix} \tilde{p} (x, y, z) = 2 j k ρ_{0} c {\tilde{u}}_{0} \frac{e^{- j k R}}{4 π R} \int_{- l_{x} / 2}^{l_{x} / 2} e^{j k x x_{0} / R} d x_{0} \int_{- l_{y} / 2}^{l_{y} / 2} e^{j k y y_{0} / R} d y_{0} \\ = j k l_{x} l_{y} ρ_{0} c {\tilde{u}}_{0} \frac{e^{- j k R}}{2 π R} \cdot \frac{\sin (k l_{x} x / 2 R)}{k l_{x} x / 2 R} \cdot \frac{\sin (k l_{y} y / 2 R)}{k l_{y} y / 2 R} . \end{matrix}$

We note that sin θ _x = x/R and sin θ _y = y/R which yields

$\tilde{p} (x, y, z) = j k l_{x} l_{y} ρ_{0} c {\tilde{u}}_{0} \frac{e^{- j k R}}{2 π R} D (θ_{x}, θ_{y}),$

where

$D (θ_{x}, θ_{y}) = \frac{\sin (\frac{1}{2} k l_{x} \sin θ_{x})}{\frac{1}{2} k l_{x} \sin θ_{x}} \cdot \frac{\sin (\frac{1}{2} k l_{y} \sin θ_{y})}{\frac{1}{2} k l_{y} \sin θ_{y}} .$

Problem 13.2.

The near-field pressure is given by Eq. (13.6)

$\tilde{p} (x, y, z) = 2 j k ρ_{0} c {\tilde{u}}_{0} \int_{- l_{y} / 2}^{l_{y} / 2} \int_{- l_{x} / 2}^{l_{x} / 2} {g (x, y, x | x_{0}, y_{0}, z_{0}) |}_{z_{0} = 0} d x_{0} d y_{0},$

where the Green's function is given by Eqs. (13.33) and (13.34)

$g (x, y, x | x_{0}, y_{0}, z_{0}) = \frac{- j}{8 π^{2}} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \frac{e^{- j (k_{x} (x - x_{0}) + k_{y} (y - y_{0}) + k_{z} | z - z_{0} |)}}{k_{z}} d k_{x} d k_{y}$

$k_{z} = {\begin{matrix} \sqrt{k^{2} - k_{x}^{2} - k_{y}^{2}}, k_{x}^{2} + k_{y}^{2} \leq k^{2} \\ - j \sqrt{k_{x}^{2} + k_{y}^{2} - k^{2}}, k_{x}^{2} + k_{y}^{2} > k^{2} . \end{matrix}$

By combining the above equations, we get

$\tilde{p} (x, y, z) = \frac{k ρ_{0} c}{4 π^{2}} {\tilde{u}}_{0} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \frac{e^{- j (k_{x} x + k_{y} y + k_{z} | z |)}}{k_{z}} \int_{- l_{x} / 2}^{l_{x} / 2} e^{j k_{x} x_{0}} d x_{0} \int_{- l_{y} / 2}^{l_{y} / 2} e^{j k_{y} y_{0}} d y_{0} d k_{x} d k_{y} .$

The total radiation force is then given by

$\tilde{f} = \int_{- l_{y} / 2}^{l_{y} / 2} \int_{- l_{x} / 2}^{l_{x} / 2} {\tilde{p} (x, y, z) |}_{z = 0} d x d y .$

The imaginary parts of the complex exponents cancel over the negative and positive values of k _x and k _y so that this simplifies to

$\begin{matrix} \tilde{f} = \frac{k ρ_{0} c}{4 π^{2}} {\tilde{u}}_{0} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- l_{x} / 2}^{l_{x} / 2} \cos (k_{x} x) d x \int_{- l_{y} / 2}^{l_{y} / 2} \cos (k_{y} y) d y \\ \times \int_{- l_{x} / 2}^{l_{x} / 2} \cos (k_{x} x_{0}) d x_{0} \int_{- l_{y} / 2}^{l_{y} / 2} \cos (k_{y} y_{0}) d y_{0} \frac{1}{k_{z}} d k_{x} d k_{y} . \end{matrix}$

The specific radiation impedance is the ratio of the total radiation force to the total volume velocity. Hence, after integrating over the surface of the piston, we have

$Z_{s} = \frac{\tilde{f}}{{\tilde{U}}_{0}} = \frac{\tilde{f}}{l_{x} l_{y} {\tilde{u}}_{0}} = \frac{k l_{x} l_{y} ρ_{0} c}{4 π^{2}} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} {sinc}^{2} (k_{x} l_{x} / 2) {sinc}^{2} (k_{y} l_{y} / 2) \frac{1}{k_{z}} d k_{x} d k_{y},$

where sinc(z) = sin(z)/z. By using polar coordinates, where k _x = kt cosϕ, k _y = kt sinϕ, and dk _x dk _y = k ² tdtdϕ, we reduce the double infinite integral to a single infinite one over t and a single finite one over ϕ. Also, the infinite integral is split to produce the real resistance R _s and the imaginary reactance X _s so that

$Z_{s} = R_{s} - j X_{s},$

where

$R_{s} = \frac{k l_{x} l_{y} ρ_{0} c}{4 π^{2}} \int_{0}^{π / 2} \int_{0}^{1} {sinc}^{2} (\frac{k l_{x}}{2} t \cos ϕ) {sinc}^{2} (\frac{k l_{y}}{2} t \sin ϕ) \frac{t d t d ϕ}{\sqrt{1 - t^{2}}},$

$X_{s} = \frac{k l_{x} l_{y} ρ_{0} c}{4 π^{2}} \int_{0}^{π / 2} \int_{1}^{\infty} {sinc}^{2} (\frac{k l_{x}}{2} t \cos ϕ) {sinc}^{2} (\frac{k l_{y}}{2} t \sin ϕ) \frac{t d t d ϕ}{\sqrt{t^{2} - 1}} .$

Substituting t = sin θ in the above equation yields

$R_{s} = \frac{k l_{x} l_{y} ρ_{0} c}{4 π^{2}} \int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} {sinc}^{2} (\frac{k l_{x}}{2} \sin θ \cos ϕ) {sinc}^{2} (\frac{k l_{y}}{2} \sin θ \sin ϕ) \sin θ d θ d ϕ, X_{s} = \frac{k l_{x} l_{y} ρ_{0} c}{4 π^{2}} \int_{0}^{\frac{π}{2}} \int_{\frac{π}{2} + j 0}^{\frac{π}{2} + j \infty} {sinc}^{2} (\frac{k l_{x}}{2} \sin θ \cos ϕ) {sinc}^{2} (\frac{k l_{y}}{2} \sin θ \sin ϕ) \sin θ d θ d ϕ,$

where we note that sin(π/2 + j∞) = cos j∞ = cosh ∞ = ∞.

Problem 13.3.

$\begin{matrix} D (θ, ϕ) = \frac{2 J_{1} (k a \sin θ)}{k a \sin θ} \cdot \frac{\sin (2 k d \sin θ \sin ϕ)}{4 \sin (\frac{1}{2} k d \sin θ \sin ϕ)} \\ = \frac{2 J_{1} (k a \sin θ)}{k a \sin θ} \cdot \cos (\frac{1}{2} k d \sin θ \sin ϕ) \cos (k d \sin θ \sin ϕ) \end{matrix}$

Problem 14.1.

Let the Green's function be of the form

$G (x, y | x_{0}, y_{0}) = \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} A_{m n} \sin (m π x / l_{x}) \sin (n π y / l_{y}) .$

Inserting this into Eq. (14.24), multiplying both sides by sin(pπx/l _x) sin(qπy/l _y), and integrating over the surface of the membrane gives

$\begin{matrix} \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} A_{m n} (k_{D}^{2} - \frac{m^{2} π^{2}}{l_{x}^{2}} - \frac{n^{2} π^{2}}{l_{y}^{2}}) \int_{0}^{l_{x}} \sin (\frac{m π x}{l_{x}}) \sin (\frac{p π x}{l_{x}}) d x \int_{0}^{l_{y}} \sin (\frac{n π y}{l_{y}}) \sin (\frac{q π y}{l_{y}}) d y \\ = - \int_{0}^{l_{x}} δ (x - x_{0}) \sin (\frac{p π x}{l_{x}}) d x \int_{0}^{l_{y}} δ (y - y_{0}) \sin (\frac{q π y}{l_{y}}) d y . \end{matrix}$

Applying the integral of Eq. (14.18) together with the property of the Dirac delta function of Eq. (A2.154) from Appendix II yields

$A_{m n} = \frac{4}{l_{x} l_{y} (\frac{m^{2} π^{2}}{l_{x}^{2}} + \frac{n^{2} π^{2}}{l_{y}^{2}} - k_{D}^{2})} \sin (\frac{m π x_{0}}{l_{x}}) \sin (\frac{n π y_{0}}{l_{y}}) .$

Hence,

$G (x, y | x_{0}, y_{0}) = 4 \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{\sin (m π x / l_{x}) \sin (n π y / l_{y}) \sin (m π x_{0} / l_{x}) \sin (n π y_{0} / l_{y})}{(k_{m n}^{2} - k_{D}^{2}) l_{x} l_{y}},$

where

$k_{m n} = π \sqrt{\frac{m^{2}}{l_{x}^{2}} + \frac{n^{2}}{l_{y}^{2}}} .$

Problem 14.2.

Let the Green's function be of the form

$G (w, ϕ | w_{0}, ϕ_{0}) = \sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} A_{m n} \cos (m ϕ) J_{m} (α_{m n} w / a) .$

Inserting this into Eq. (14.44), multiplying both sides by cos(pϕ)J _p(α _pq w/a), and integrating over the surface of the membrane gives

$\begin{matrix} \sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} A_{m n} (k_{D}^{2} - \frac{α_{m n}^{2}}{a^{2}}) \int_{0}^{2 π} \cos (m ϕ) \cos (p ϕ) d φ \int_{0}^{a} J_{m} (α_{m n} w / a) J_{p} (α_{p q} w / a) w d w \\ = - \int_{0}^{2 π} δ (ϕ - ϕ_{0}) \cos (p ϕ) d ϕ \int_{0}^{a} δ (w - w_{0}) J_{p} (α_{p q} w / a) d w . \end{matrix}$

Applying the integrals of Eqs. (14.38) and (14.39) together with the property of the Dirac delta function of Eq. (A2.154) from Appendix II yields

$A_{m n} = \frac{2 \cos (m ϕ_{0}) J_{m} (α_{m n} w_{0} / a)}{π (1 + δ_{m 0}) (α_{m n}^{2} - k_{D}^{2} a^{2}) J_{m + 1}^{2} (α_{m n})} .$

Hence,

$G (w, ϕ | w_{0}, ϕ_{0}) = \frac{2}{π} \sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} \frac{\cos (m ϕ) J_{m} (α_{m n} w / a) \cos (m ϕ_{0}) J_{m} (α_{m n} w_{0} / a)}{(1 + δ_{m 0}) (α_{m n}^{2} - k_{D}^{2} a^{2}) J_{m + 1}^{2} (α_{m n})} .$

Problem 14.3.

$\begin{matrix} f_{01} = \frac{α_{01}^{2} h}{4 π a^{2}} \sqrt{\frac{Y}{3 (1 - ν^{2}) ρ_{D}}} = \frac{{2.2215}^{2} \times 0.2 \times 10^{- 3}}{4 π \times {(12.5 \times 10^{- 3})}^{2}} \sqrt{\frac{69 \times 10^{9}}{3 \times (1 - {0.3}^{2}) \times 2700}} = 1.54 kHz, \\ f_{02} = {(\frac{α_{02}}{α_{01}})}^{2} f_{01} = {(\frac{5.4516}{2.2215})}^{2} 668 = 9.26 kHz, \\ f_{03} = {(\frac{α_{03}}{α_{01}})}^{2} f_{01} = {(\frac{8.6114}{2.2215})}^{2} 668 = 23.1 kHz, \end{matrix}$

Problem 14.4.

Let the Green's function be of the form

$G (w, φ | w_{0}, φ_{0}) = \sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} A_{m n} η_{m n} (w, φ),$

where η _mn(w, ϕ) = cos(mϕ) (J _m(α _mn w/a) − B _mn I _m(α _mn w/a) + C _mn). Inserting this into Eq. (14.149), multiplying both sides by

$η_{p q}^{*} (w, ϕ)$

, and integrating over the surface of the shell gives

$\begin{matrix} \sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} A_{m n} (\frac{α_{m n}^{4}}{a^{4}} - k_{D}^{4}) \int_{0}^{2 π} \int_{0}^{a} η_{m n} (w, φ) η_{p q}^{*} (w, φ) w d w d ϕ \\ = - \int_{0}^{2 π} \int_{0}^{a} δ (w - w_{0}) δ (φ - φ_{0}) η_{p q}^{*} (w, φ) d w d ϕ . \end{matrix}$

Applying the integrals of Eq. (14.142) together with the property of the Dirac delta function of Eq. (A2.154) from Appendix II yields

$A_{m n} = \frac{a^{2} η_{m n}^{*} (w_{0}, φ_{0})}{π Δ_{m n} (k_{D}^{4} a^{4} - α_{m n}^{4})} .$

Hence,

$G (w, φ | w_{0}, φ_{0}) = \frac{a^{2}}{π} \sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} \frac{η_{m n} (w, φ) η_{m n}^{*} (w_{0}, φ_{0})}{Δ_{m n} (k_{D}^{4} a^{4} - α_{m n}^{4})} .$

Problem 14.5.

$\begin{matrix} f_{01} = \frac{h}{4 π a^{2}} \sqrt{\frac{Y}{3 ρ_{D}} (\frac{β_{1}^{4}}{1 - ν^{2}} + 48 \frac{H^{2}}{h^{2}})} \\ = \frac{40 \times 10^{- 6}}{4 π \times {(12.5 \times 10^{- 3})}^{2}} \sqrt{\frac{69 \times 10^{9}}{3 \times 2700} (\frac{- 109.2 \times 10^{3}}{1 - {0.3}^{2}} + 48 \times {(\frac{2 \times 10^{- 3}}{40 \times 10^{- 6}})}^{2})} = 0 Hz, \\ f_{02} = \frac{40 \times 10^{- 6}}{4 π \times {(12.5 \times 10^{- 3})}^{2}} \sqrt{\frac{69 \times 10^{9}}{3 \times 2700} (\frac{{5.9034}^{4}}{1 - {0.3}^{2}} + 48 \times {(\frac{2 \times 10^{- 3}}{40 \times 10^{- 6}})}^{2})} = 20.7 kHz, \\ f_{03} = \frac{40 \times 10^{- 6}}{4 π \times {(12.5 \times 10^{- 3})}^{2}} \sqrt{\frac{69 \times 10^{9}}{3 \times 2700} (\frac{{9.1898}^{4}}{1 - {0.3}^{2}} + 48 \times {(\frac{2 \times 10^{- 3}}{40 \times 10^{- 6}})}^{2})} = 21.3 kHz, \\ f_{04} = \frac{0.2 \times 10^{- 3}}{4 π \times {(12.5 \times 10^{- 3})}^{2}} \sqrt{\frac{69 \times 10^{9}}{3 \times 2700} (\frac{{12.386}^{4}}{1 - {0.3}^{2}} + 48 \times {(\frac{2 \times 10^{- 3}}{40 \times 10^{- 6}})}^{2})} = 22.7 kHz . \end{matrix}$

Problem 15.1.

$\begin{matrix} p_{r m s} (r, 0) = 2 π f ε_{0} a^{2} {(\frac{E_{P}}{d})}^{2} \frac{1}{2 \sqrt{2} r c} \\ = \frac{2 \times 3.14 \times 10^{3} \times 8.85 \times 10^{- 12} \times {0.05}^{2} \times {({2000 / 10}^{- 3})}^{2}}{2 \times \sqrt{2} \times 1 \times 345} = 0.570 Pa \end{matrix}$

Hence, the sound pressure level is 20 log₁₀(0.57/(20 × 10⁻⁶)) = 89.1 dB SPL.

Problem 15.2.

From Eq. (15.119), the polarization voltage is

$E_{P} = (\frac{E_{P}}{d}) d = \frac{2000}{10^{- 3}} \times 0.224 \times 10^{- 3} = 448 V .$

From Eq. (15.86), the minimum tension is

$T = \frac{ε_{0} a^{2}}{d} {(\frac{E_{P}}{d})}^{2} = \frac{8.85 \times 10^{- 12} \times {0.05}^{2}}{0.224 \times 10^{- 3}} {(\frac{2000}{10^{- 3}})}^{2} = 395 N/m .$

From Eq. (15.114), the resonance frequency is

$f_{0} = \frac{1}{3.26 a} \sqrt{\frac{T}{a ρ_{0}}} = \frac{1}{3.26 \times 0.05} \sqrt{\frac{395}{0.05 \times 1.18}} = 502 Hz .$

From Eq. (15.117), the resistance is

$R_{S} = \frac{2}{Q} \sqrt{\frac{ρ_{0} T}{a}} = \frac{2}{1} \sqrt{\frac{1.18 \times 395}{0.05}} = 193 rayls .$

From Eq. (15.113), the maximum tension is

$T = 0.6 σ_{\max} h = 0.6 \times 55 \times 10^{6} \times 12 \times 10^{- 6} = 396 N/m .$

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Table of Contents for Appendix III. Answers to problems

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Table of Contents for
Appendix III. Answers to problems