Therefore, it requires only half of the source strength to produce the same pressure field as a full spherical pressure field. In other words, if the same source strength is put on the half space, half of the source strength will be reflected to the half space and produce twice the pressure field in the full spherical space created by the same source strength.

6.5Acoustic Intensity and Sound Power

Sound intensity I is the acoustic intensity, formulated in the previous section as:

where A was calculated for a point source as:

A ¼ Qs ρoc k

4π

For a spherical source with radius r ¼ a , acoustic intensity I is constant on the surface of the sphere. Therefore, sound power can be formulated without integration:

Z

w IðrÞ ds

s

¼ 4πa2Iðr ¼ aÞ

Even though the above formula for sound power is formulated at r ¼ a, this formula is valid for any other choice of surface for integration. It makes sense because the sound power is the total radiation energy of the source and will not vary with the choices of the surface for integration. This can be checked by comparing the above sound power, at r ¼ a, to the sound power at r ¼ 2a as shown below:

Example 6.2 (A Point Source)

An acoustic pressure p(r, t) is created by a surface vibration of a spherical source with a radius a ¼ 1001 ½m&. Given the surface velocity of the sphere as:

where the surface velocity is Ua ¼ 20 [m/s] and frequency of radiation is f ¼ 680 [Hz].

Assuming that this spherical source can be treated as a point source, calculate

(a) the source strength, (b) flow velocity, (c) acoustic pressure, (d) intensity, and

(e) power radiating from the point source.

Use 415 [rayls] for characteristic impedance (ρ0c) of air and 340 [m/s] for the speed of sound in air. Show units in the Meter-Kilogram-Second (MKS) system.

Example 6.2 Solution

Similar to Example 6.1, the angular frequency can be calculated as:

ω ¼ 2πf ¼ 2π 680 h1si ¼ 1360π h1si

And the wave number can be calculated as:

Compare the given surface velocity of the sphere:

to the boundary condition of a small pulsating spherical source in the summary table below:

ϕ ¼ tan 1 1 kr

Part (c)

The formula for the acoustic pressure from a point source is shown in the summary table as:

pðr, tÞ ¼ Ar cos ðωt kr þ θÞ

¼ 10r:4 cos ð1360πt 4πrÞ

Part (d)

Acoustic intensity:

Part (e)

Sound power:

Remarks

The results show that the acoustic quantities calculated from a small spherical formulation (Example 6.1) and a point source formulation (Example 6.2) are identical, as expected.

For a small sphere, the constant A is calculated by comparing the magnitudes of the velocity:

to arrive at:

This shows that the constant A calculated from a point source is the same or close to the value calculated from a small spherical source. Therefore, we can conclude that this spherical source can be modeled as a point source.

Example 6.3

The rectangular plate has the width Lx and the height Ly with the following given dimensions:

Lx ¼ 0:8 ½m&

Ly ¼ 0:6 ½m&

= cos 2 +

Use m × n point sources

m=2 (x-direction) n=2 (y-direction)

at (0.1, 0, 0.1) [m]

Assume that this plate vibrates at frequency f and the amplitude U of surface velocity is:

f ¼ 340 ½Hz&

hmi

U ¼ 20 s

θo ¼ 0 ½rad&

Therefore, the surface velocity is:

hmi uðtÞ ¼ U cos ð2πftÞ ¼ 20 cos ð2π 340 tÞ s

The function POINTSOURCE.m and the main program VibrationPlate.m of this example can be found on Moodle.

The outputs of the first 11 time steps are:

The output of the first 51 time steps can be plotted as: