For normal air at STP, what is the linear amplitude of air
displacement as sound propagates at different Sound Pressure Levels
(SPL) and frequencies.

Now, how does it shift with temperature

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Request for Question Clarification by hedgie-ga on 05 Jun 2004 20:16 PDT

gaussfan

by STP do you mean Standart pressure, i.e. 1 atmosphere?

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Clarification of Question by gaussfan-ga on 05 Jun 2004 22:51 PDT

STP= standard tempeature and pressure ie. 1atm, 25deg C.

To further clarify, a sinusoidal spherical or plane wave of single
frequency F propagates in an isotropic air medium. By "amplitude of
displacement" I mean the magnitude in say, microns, of the excursion
of air particles about their equilibrium position.

Thank you for the clarification.

HERE IS THE ANSWER:

  Sound is a compression wave in a medium (air) which can be
represented as an Ideal Gas. Here is a animated picture of that:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/sound/u11l1c.html

   Variables density and compression modulus (or bulk modulus)
depend on temmperature and determine  "speed of sound" v ,
and "impedance of the medium Z". Basic material constant is gamma,
about 1.44 for air (a  mixture of diatomic gases), which is defined here:
http://hypertextbook.com/physics/waves/sound/

Quick look on the Efffect of T on speed v (same as c here) is here:
http://encyclopedia.thefreedictionary.com/Speed of sound

Detailed, 35 pages long, technical description is here:
http://www.mech.soton.ac.uk/mh/ME318/Compressed_notes_mh.pdf

Salient points are this: The compression is adiabatic (i.e.so fast
that there is no thermal equilibration), 
(this seems to be  different then view expressed in the comment)
and  quantities  in the Hooks law:  strain * K = stress  are

strain .. dimensionless (=change of density / density)
stress .. pressure (Pascals - N /m. m )
K = bulk modulus is in Pa and equals gamma*P 
This is consequence of Ideal Gas equation od state for adiabatic compression
as explained on page 11 here:

http://fisicanet.terra.com.br/cursos/msu/m202.pdf
 

So , combining all that: to get your answer you
first get P amplitude of the pressure vawe,  from SPL (which is Lp
here)  using this equation
                  Lp= 20 * log(P/P0),
which explained here:
http://physics.mtsu.edu/~wmr/log_3.htm

From that and K you get you get your strain amplitude. K is
proportional to P which is proportional to absolute temperature T.
Strain is a dimensionless number.(A surface wave measures displacement
in microns, but compression wave masures strain in percents. Even the
loudest sound (meaning at treshold of pain) is less then 3%.)
So, the same pressure amplitude, at lower temperature, will produce
slightly larger strain (=cold air is stiffer).
Let's say, in summer T = 273K and in winter T=253K. Then strain for
very loud sound, (a dimensionless number) would be 9/273  and 9/253
respectively.

So, what is the displacement of particles in microns? During load
sound three percent of particles must escape the volume defined by the
wavelength. The actual speed of the particles is comparable to the
speed of sound. The sound is just changing counts of particles
arriving and leaving the volumes of compressions and rarefications.

hedgie 

hedgie

I was fascinated to read your question. Usually we measure the
fluctuations of pressure and density.  To find the true answers to
your question I would recommend a numerical investigation using the
Euler equations normalized about a mean value in one dimension.  (That
is the conservation, momentum, and energy equation)  The temperature
and pressure are both represented in the energy equation - you can
convert between the two once you have solved the full set of
equations.  If the acoustic pressure fluctuates then the acoustic
temperature will also fluctuate proportionally.

Go to this website, just past question No. 8 is a working example
which tells you everything you need to know.

http://www.phys.unsw.edu.au/~kryw/clectures/tutorials/Tutorial4/Tutorial4.html

Euler equations normalized about a mean value in one dimension,
saem_aero?!  Who has the patience for that?!  :-)

The question is very clearly stated, and it is not answered in the "Answer".