Acoustics

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

13.14. Radiation from an infinitely long oscillating strip in an infinite baffle [37,38]

Boundary conditions

Essentially this is the limiting case of a rectangular piston as one of its dimensions tends to infinity. The infinitely long strip of width d shown in Fig. 13.40 is mounted in an infinite baffle in the xy plane and oscillates in the z direction with a harmonically time-dependent velocity

{\tilde{u}}_{0}

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

. As with the circular piston in an infinite baffle, the monopole source elements, together with their images, coalesce to form elements of double strength. Because of the symmetry of the pressure fields on either side of the baffle, there is the following Neumann boundary condition on its surface:

Figure 13.40 Geometry of infinitely long rigid strip in infinite baffle. The point of observation P is located at position (x,y,z) in rectangular coordinates.

\frac{\partial}{\partial z} \tilde{p} {(x, z) |}_{z = 0 +} = 0, {\begin{matrix} - \infty \leq x < - \frac{d}{2}, \\ - \frac{d}{2} < x \leq \infty \end{matrix}

$\frac{\partial}{\partial z} \tilde{p} {(x, z) |}_{z = 0 +} = 0, {\begin{matrix} - \infty \leq x < - \frac{d}{2}, \\ - \frac{d}{2} < x \leq \infty \end{matrix}$

(13.273)

which is satisfied automatically. On the surface of the strip there is the coupling condition

\frac{\partial}{\partial z} \tilde{p} {(x, y) |}_{z = 0 +} = - j k ρ_{0} c {\tilde{u}}_{0}, - \frac{d}{2} \leq x \leq \frac{d}{2},

$\frac{\partial}{\partial z} \tilde{p} {(x, y) |}_{z = 0 +} = - j k ρ_{0} c {\tilde{u}}_{0}, - \frac{d}{2} \leq x \leq \frac{d}{2},$

(13.274)

where ρ ₀ is the density of air or any other surrounding medium, c is the speed of sound in that medium, and k = 2π/λ = ω/c is the wave number.

Far-field pressure

The pressure at point P due to a single line source at x ₀ is obtained from Eq. (12.6) to give

\tilde{p} (r, θ) = \frac{ρ_{0} c ({\tilde{U}}_{0} / l)}{2} \sqrt{\frac{k}{2 π r_{1}}} e^{- j (k r_{1} - \frac{π}{4})},

$\tilde{p} (r, θ) = \frac{ρ_{0} c ({\tilde{U}}_{0} / l)}{2} \sqrt{\frac{k}{2 π r_{1}}} e^{- j (k r_{1} - \frac{π}{4})},$

(13.275)

where

({\tilde{U}}_{0} / l)

$({\tilde{U}}_{0} / l)$

is the volume velocity per unit length, k = ω/c = 2π/λ is the wave number, and

r_{1}^{2} = r^{2} \cos^{2} θ + {(r \sin θ - x_{0})}^{2} .

$r_{1}^{2} = r^{2} \cos^{2} θ + {(r \sin θ - x_{0})}^{2} .$

(13.276)

Hence, the pressure due to the strip is the integral across its width of Eq. (13.275) for a single line source taking into account the double strength sources and letting

{\tilde{U}}_{0} = l d {\tilde{u}}_{0}

${\tilde{U}}_{0} = l d {\tilde{u}}_{0}$

as follows:

\tilde{p} (r, θ) = ρ_{0} c {\tilde{u}}_{0} \int_{- \frac{d}{2}}^{\frac{d}{2}} \sqrt{\frac{k}{2 π r_{1}}} e^{- j (k r_{1} - \frac{π}{4}) d x_{0}} .

$\tilde{p} (r, θ) = ρ_{0} c {\tilde{u}}_{0} \int_{- \frac{d}{2}}^{\frac{d}{2}} \sqrt{\frac{k}{2 π r_{1}}} e^{- j (k r_{1} - \frac{π}{4}) d x_{0}} .$

(13.277)

At a large distance r, the terms containing r in Eq. (13.276) dominate. Hence the remaining terms can be replaced with ones that enable r ₁ to be factorized as follows

\begin{matrix} r_{1}^{2} = r^{2} + x_{0}^{2} - 2 r x_{0} \sin θ \\ \approx r^{2} + x_{0}^{2} \sin^{2} θ - 2 r x_{0} \sin θ \\ = {(r - x_{0} \sin θ)}^{2}, \end{matrix}

$\begin{matrix} r_{1}^{2} = r^{2} + x_{0}^{2} - 2 r x_{0} \sin θ \\ \approx r^{2} + x_{0}^{2} \sin^{2} θ - 2 r x_{0} \sin θ \\ = {(r - x_{0} \sin θ)}^{2}, \end{matrix}$

(13.278)

which after inserting into Eq. (13.277) gives

\begin{matrix} \tilde{p} (r, θ) = ρ_{0} c {\tilde{u}}_{0} \sqrt{\frac{k}{2 π r}} e^{- j (k r - \frac{π}{4})} \int_{- \frac{d}{2}}^{\frac{d}{2}} e^{j k x_{0} \sin θ} d x_{0} \\ = \frac{k d ρ_{0} c {\tilde{u}}_{0}}{2} \sqrt{\frac{d}{π r}} e^{- j (k r - \frac{π}{4})} D (θ), \end{matrix}

$\begin{matrix} \tilde{p} (r, θ) = ρ_{0} c {\tilde{u}}_{0} \sqrt{\frac{k}{2 π r}} e^{- j (k r - \frac{π}{4})} \int_{- \frac{d}{2}}^{\frac{d}{2}} e^{j k x_{0} \sin θ} d x_{0} \\ = \frac{k d ρ_{0} c {\tilde{u}}_{0}}{2} \sqrt{\frac{d}{π r}} e^{- j (k r - \frac{π}{4})} D (θ), \end{matrix}$

(13.279)

where the directivity function D(θ) is given by

D (θ) = \frac{\sin (\frac{1}{2} k d \sin θ)}{{(\frac{1}{2} k d)}^{3 / 2} \sin θ},

$D (θ) = \frac{\sin (\frac{1}{2} k d \sin θ)}{{(\frac{1}{2} k d)}^{3 / 2} \sin θ},$

(13.280)

which is the same as that of a finite line source of length d in its plane, as given by Eq. (4.89). The directivity pattern is shown in Fig. 4.18. The on-axis pressure is given by

D (0) = 1 / \sqrt{\frac{1}{2} k d} .

$D (0) = 1 / \sqrt{\frac{1}{2} k d} .$

In Section 2.1, we saw that in the case of a piston radiating into an infinitely long tube, the pressure

\tilde{p}

$\tilde{p}$

along the tube is directly proportional to the piston velocity

{\tilde{u}}_{0}

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

. Assuming the tube is much narrower than the wavelength, this represents a one-dimensional system. In three-dimensional space, such as that of a piston in an infinite baffle radiating into free space, the radiated pressure is proportional to the acceleration of the piston

j ω {\tilde{u}}_{0}

$j ω {\tilde{u}}_{0}$

. Not surprisingly, in the two-dimensional space of an infinite strip we find that the pressure is proportional to

\sqrt{j ω} {\tilde{u}}_{0}

$\sqrt{j ω} {\tilde{u}}_{0}$

Radiation impedance

Using the Bouwkamp impedance theorem (see Section 13.13), the radiated power per unit length l is given by

W = {| \frac{{\tilde{U}}_{0}}{\sqrt{2}} |}^{2} R_{A R} = \frac{l}{ρ_{0} c} \int_{- \frac{π}{2}}^{\frac{π}{2}} {| \frac{\tilde{p} (r, θ)}{\sqrt{2}} |}^{2} {r d θ |}_{r \to \infty},

$W = {| \frac{{\tilde{U}}_{0}}{\sqrt{2}} |}^{2} R_{A R} = \frac{l}{ρ_{0} c} \int_{- \frac{π}{2}}^{\frac{π}{2}} {| \frac{\tilde{p} (r, θ)}{\sqrt{2}} |}^{2} {r d θ |}_{r \to \infty},$

(13.281)

where the integration is taken over a half-cylindrical surface in the extreme far field. Using the pressure from Eq. (13.279), we obtain the real and imaginary parts of the impedance as follows:

R_{s} = l d R_{A R} = ρ_{0} c \frac{k d}{π} \int_{0}^{\frac{π}{2}} D^{2} (θ) d θ,

$R_{s} = l d R_{A R} = ρ_{0} c \frac{k d}{π} \int_{0}^{\frac{π}{2}} D^{2} (θ) d θ,$

(13.282)

X_{s} = l d X_{A R} = ρ_{0} c \frac{k d}{π} \int_{\frac{π}{2} + j 0}^{\frac{π}{2} + j \infty} D^{2} (θ) d θ,

$X_{s} = l d X_{A R} = ρ_{0} c \frac{k d}{π} \int_{\frac{π}{2} + j 0}^{\frac{π}{2} + j \infty} D^{2} (θ) d θ,$

(13.283)

where R _s is the specific radiation resistance in N·s/m³ (rayl), where the bold R indicates that the quantity varies with frequency, and X _s is the specific radiation reactance in N·s/m³ (rayl). Substituting t = sin θ yields

\begin{matrix} R_{s} = l d R_{A R} = ρ_{0} c \frac{k d}{π} \int_{0}^{1} {(\frac{\sin (\frac{k d}{2} t)}{\frac{k d}{2} t})}^{2} \frac{d t}{\sqrt{1 - t^{2}}} \\ = ρ_{0} c {\frac{k d}{2}}_{1} F_{2} (\frac{1}{2}; \frac{3}{2}, 2; - \frac{k^{2} d^{2}}{4}) \\ \approx ρ_{0} c \frac{k d}{2}, k d < 0.5, \end{matrix}

$\begin{matrix} R_{s} = l d R_{A R} = ρ_{0} c \frac{k d}{π} \int_{0}^{1} {(\frac{\sin (\frac{k d}{2} t)}{\frac{k d}{2} t})}^{2} \frac{d t}{\sqrt{1 - t^{2}}} \\ = ρ_{0} c {\frac{k d}{2}}_{1} F_{2} (\frac{1}{2}; \frac{3}{2}, 2; - \frac{k^{2} d^{2}}{4}) \\ \approx ρ_{0} c \frac{k d}{2}, k d < 0.5, \end{matrix}$

(13.284)

\begin{matrix} X_{s} = l d X_{A R} = ρ_{0} c \frac{k d}{π} \int_{0}^{\infty} {(\frac{\sin (\frac{k d}{2} t)}{\frac{k d}{2} t})}^{2} \frac{d t}{\sqrt{t^{2} - 1}} \\ = ρ_{0} c \frac{1}{k d} G_{2,4}^{2,1} ({\frac{k^{2} d^{2}}{4} |}_{1,1,0, \frac{1}{2}}^{1, \frac{3}{2}}) \\ \approx ρ_{0} c \frac{k d}{π} (\frac{3}{2} - γ - \ln (\frac{k d}{2})), k d < 0.5, \end{matrix}

$\begin{matrix} X_{s} = l d X_{A R} = ρ_{0} c \frac{k d}{π} \int_{0}^{\infty} {(\frac{\sin (\frac{k d}{2} t)}{\frac{k d}{2} t})}^{2} \frac{d t}{\sqrt{t^{2} - 1}} \\ = ρ_{0} c \frac{1}{k d} G_{2,4}^{2,1} ({\frac{k^{2} d^{2}}{4} |}_{1,1,0, \frac{1}{2}}^{1, \frac{3}{2}}) \\ \approx ρ_{0} c \frac{k d}{π} (\frac{3}{2} - γ - \ln (\frac{k d}{2})), k d < 0.5, \end{matrix}$

(13.285)

where F is the hypergeometric function, G is the Meijer G function, and γ = 0.5772 is Euler's constant. Separate plots of R _s/ρ ₀ c and X _s/ρ ₀ c are shown in Figs. 13.43 and 13.44, respectively, as a function of kd.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for Acoustics

Create new playlist

Sign In

Sign Up

13.14. Radiation from an infinitely long oscillating strip in an infinite baffle [37,38]

Boundary conditions

Far-field pressure

Radiation impedance

Table of Contents for
Acoustics