Pressure zone and boundary microphones 1

[otaio]

Dear all,

I would like to understand the physics of why sound pressure doubles at a surface and how the thickness of the this boundary can be calculated.

B

 

[GregLocock]

It doesn't.

The reflected wave is of the same magnitude as the incident wave.

If you are referring to the +6 dB seen when you place a speaker near a wall, compared with the same speaker in open air, then it is due to constructive interference between the reflected wave and the direct wave. The beat way to understand this is to draw some sine waves on a diagram showing you, the speaker and the wall. You should see that there is a maximum reinforcement at 0 Hz, of a factor of 2, followed by a cancellation at a frequency closely related to the number of wavelengths you can fit between the wall and the speaker, and so on. The minima will be at those frequencies where the reflected wave arrives at the speaker 180 degrees out of phase with the direct sound.

Reality is more complex than that, but it is a good place to start for low frequency noise. Cheers

Greg Locock

 

[MikeyP]

I beg to differ, Greg. The pressure near a surface due to a wave incident on the surface can be up to double that of the wave precisely because "The reflected wave is of the same magnitude as the incident wave", if these waves interfere constructively.

See for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/sound/reflec.html

The diagram here is a little confusing as it shows the wave as tranverse (rather than as a longitudinal sound wave).

The phenomenon is analogous (I think) to what happens in transmission line theory, where a pulse travels down an unterminated (open circuit) wire with a characteristic impedance. In this case, there is a doubling of voltage at the discontinuity. There is a nice animation here:

http://othello.mech.nwu.edu/ea3/book99/transline/text.html

The distance where pressure doubling occurs depends on 3 main parameters: The angle of incidence of the wave, the absorpative characteristics of the surface (both attenuation and phase shift) and the wave length. The first link above shows the simplest case of a normally incident wave on a perfectly reflecting surface.

Be aware that this simple case is the only situation where the pressure is actually doubled.

M

 

[GregLocock]

AAGH. I knew I'd get it wrong. Why couldn't our ears work on intensity rather than pressure, then it would be easy to keep the signs straight?




Cheers

Greg Locock