Vibroacoustic Simulation. Alexander PeifferЧитать онлайн книгу.

target="_blank" rel="nofollow" href="#fb3_img_img_165892ac-ca2f-5a48-bdcf-dde5a6d6b4ce.png" alt="normal upper Pi equals one-half upper R e left-parenthesis bold-italic p bold-italic upper Q Superscript asterisk Baseline right-parenthesis equals one-half upper R e left-parenthesis bold-italic upper Z Subscript a Baseline bold-italic upper Q bold-italic upper Q Superscript asterisk Baseline right-parenthesis equals upper R e left-parenthesis bold-italic upper Z Subscript a Baseline right-parenthesis upper Q Subscript r m s Superscript 2"/> (2.87)

or for constant source pressure

(2.88)

It is instructive to see the mechanical properties considering the limit cases from above and extract the mass that is moved by the surface. From Newtons’s law a force given by F=4πR2p leads to an in-phase acceleration of jωvr of a mass m

bold-italic upper F equals bold-italic p Baseline 4 pi upper R squared equals m j omega bold-italic v Subscript r

hence

m equals upper R e left-parenthesis StartFraction 1 Over j omega EndFraction StartFraction bold-italic p Over bold-italic upper Q EndFraction right-parenthesis equals StartFraction 1 Over omega EndFraction upper I m left-parenthesis bold-italic upper Z Subscript a Baseline right-parenthesis (2.89)

For kR≪1 we get: m=4πR3ρ0=3Vsphρ0. Thus, at low frequencies the source surface motion carries three times the fluid volume of the sphere. This motion near the source is called an evanescent wave, because it is oscillatory motion of fluid that does not radiate.

2.4.1.4 Point Sources

A point source is a spherical source with an infinitely small radius. Performing the limit kR→0 for Equation (2.73) leads to the velocity potential for point sources of strength Q

normal upper Phi left-parenthesis r comma omega right-parenthesis equals minus StartFraction bold-italic upper Q left-parenthesis omega right-parenthesis Over 4 pi r EndFraction e Superscript minus j k r (2.90)

The pressure and velocity field of such a source is given by

bold-italic p left-parenthesis r comma omega right-parenthesis equals minus j k rho 0 c 0 normal upper Phi equals StartFraction j k rho 0 c 0 bold-italic upper Q left-parenthesis omega right-parenthesis Over 4 pi r EndFraction e Superscript minus j k r (2.91)

and

(2.92)

All other relations regarding power and intensity expressions remain. We see that the limit is expressed for kR and not for the wavelength. The reason is that it is the ratio of a characteristic length (in this case the sphere radius) to the wavelength that determines if the geometrical details must be considered or not. In other words, a wave of a certain wavelength doesn’t care about details that are much smaller.

With the D’Alambert solution for spherical waves (2.66) we can also derive a point source in time domain

normal upper Psi left-parenthesis r comma t right-parenthesis equals minus StartFraction upper Q left-parenthesis t minus c 0 slash r right-parenthesis Over 4 pi r EndFraction (2.93)

The point source is of great importance for the solution of the inhomogeneous wave equation in combination with complex boundary conditions. Any source can be reconstructed by a superposition of point sources as shown in Section 2.7.

Performing the limit process with kR→0 and taking the power from equation 2.86 we get the intensity of the point source:

mathematical left-angle upper I left-parenthesis r right-parenthesis mathematical right-angle Subscript upper T Baseline equals StartFraction upper Q Subscript rms Superscript 2 Baseline k squared rho 0 c Over 16 pi squared r squared EndFraction (2.94)

and the total radiated power

mathematical left-angle normal upper Pi mathematical right-angle Subscript upper T Baseline equals StartFraction upper Q Subscript rms Superscript 2 Baseline k squared rho 0 c 0 Over 4 pi EndFraction (2.95)

with radiation impedance following from this

upper Z Subscript a Baseline equals StartFraction k squared rho 0 c 0 Over 4 pi EndFraction (2.96)

2.5 Reflection of Plane Waves

A plane wave striking a plane surface is a first example of interaction with obstacles. Imagine a configuration as shown in Figure 2.7. The impedance of the surface is z2, and it is given by using the velocity vz perpendicular to the plane.

Figure 2.7 Reflection of a plane wave at an infinite surface with impedance Z2. Source: Alexander Peiffer.

Without loss of generality the wave front is parallel to the y-axis and all properties are functions of x and z. The solution in the half space of z>0 is the superposition of two plane waves.

normal upper Phi 1 left-parenthesis r right-parenthesis equals normal upper Phi e Superscript minus j bold k bold r Baseline plus normal upper Phi Superscript left-parenthesis upper R right-parenthesis Baseline e Superscript minus j bold k Super Superscript left-parenthesis upper R right-parenthesis Superscript bold r (2.97)

With the following arguments of the exponential function

StartLayout 1st Row 1st Column bold k bold r 2nd Column equals k left-parenthesis sine theta x minus cosine theta z right-parenthesis EndLayout (2.98)

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