From: davidc@usthk.ust.hk
Subject: Re2: Acoustics Problem for Swimmers: EXPLANATION-2
Organization: The Hong Kong University of Science & Technology (HKUST)
Date: Tue, 15 Dec 1992 14:01:24 GMT

In article <1992Dec14.182239.5789@novell.com>, dseeman@novell.com (Daniel Seeman) writes:
> In article <1gde3lINNfj5@crcnis1.unl.edu> e_p@unlinfo.unl.edu (edgar pearlstein) writes:
>>
>>A few days ago I posted the following:
>>
>>
>>>                    AN ACOUSTICS PROBLEM FOR SWIMMERS
>>>      When under water in a swimming pool one hears general noise - a 
>>> hissing sound, something like white noise.  I assume that this is due 
>>> to splashing and circulating water, as well as noise entering the water 
>>> from the air above.  A pool has very hard walls and bottom, so one 
>>> might expect a fairly uniform distribution of the sound. 
>>>      Now here's the problem:  I have noticed that when my head is only 
>>> a little below the surface, the hiss has a higher average pitch than 
>>> when it is farther down.  Why is this?
>>>      I think I know the answer, but I won't announce it yet.
>>
>>       Here is the explanation that I favor:
>>            First consider standing waves within the water, in the ideal 
>>       case of nobody in the pool and the surface perfectly still.  The 
>>       wave pattern, for any frequency of sound that happens to be 
>>       present in the water, will have pressure ANTINODES at the bottom 
>>       and sides, but pressure NODES where the water meets air.  This 
>>       latter is because the speed of sound in water is much greater - 
>>       about a factor of 4.2 - than in air.  (Note that the ear responds 
>>       to pressure, not displacement.)  How deep does the node go?  
>>       Well, a quarter-wavelength below the surface there will be a 
>>       pressure ANTInode, for waves travelling perpendicular to the 
>>       surface. 
>>            Now let's put a swimmer in the water.  For sufficiently long 
>>       wavelengths, the pattern will be pretty much as described in the 
>>       paragraph above.  But for those frequencies whose wavelength is 
>>       smaller than, say, a few feet, the swimmer's body will greatly 
>>       change the standing wave pattern.  Also, ripples on the surface 
>>       will randomize things for wavelengths smaller than, say, four 
>>       times the ripple amplitude.  
>>            Thus we see that for high frequencies (short wavelength), 
>>       the standing wave pattern will be essentially random, while for 
>>       low frequencies there will be a pressure node close to the 
>>       surface.  So, since all frequencies are present in the noise,  
>>       the amount of high frequency sound perceived will be pretty much 
>>       independent of depth, while the amount of low frequency sound 
>>       will decrease as the ear gets close to the surface (pressure 
>>       antinode for low frequencies).  So the apparent pitch will rise 
>>       as the ear gets closer to the surface. 
> 
> I understand your arguement and it seems to make sense.  But if these standing
> waves and their ANTI-nodes are the explanation, then there must be other areas
> in the pool where only the higher frequencies can be heard.  Or, does this mean
> the standing waves are shaped EXACTLY like the pool?  Certainly for the case 
> with only ONE node this may be the case.  But what about
> higher order standing waves?  There must be other areas in the pool where
> only the higher frequencies can be heard (not just areas near the surface...
> ). I say this because of the distribution of the nodes/ANTI-nodes in the
> 3-dimensional volume of the pool when considering higher order standing waves.
> 
> Your microphone experiment would be able to map this out though.  It sounds
> like a good senior project for someone.  Try it out, write your paper and
> let us all know!  This is a nice project!
> 
> dks.
>>            It remains to discuss what we mean by "high" and "low" 
>>       frequencies.  From the considerations of the second paragraph, it 
>>       appears that we mean wavelengths that are small and large 
>>       compared with a few feet.  In water, a wavelength of four feet 
>>       corresponds to a frequency of about 1200 Hz. 
>>                       ---------------------------------
>>            A few people suggested an explanation in terms of how one's 
>>       hearing is affected by the pressure of water, since as one goes 
>>       deeper, the pressure is greater.  For persons whose eustachian 
>>       tubes get blocked easily, there might be something to this, 
>>       although I doubt that it would explain the magnitude of the 
>>       effect.  
>>            For someone who has both the ambition and the equipment, I 
>>       can suggest an experiment to settle the question of whether the 
>>       effect is objective or subjective:  Use an underwater microphone, 
>>       connected to an amplifier and earphones above water.  Then listen 
>>       on the earphones as the mike's depth is varied.  To be fancier, 
>>       one could even Fourier analyze the signal from the mike. 
>>
>>Edgar Pearlstein, University of Nebraska, Lincoln
>>
>>

>>
	If your standing wave model is correct,  you cannot hear any "high"
frequency sound.  It is because standing wave cannot exist in the situation
that the "length" of the system is not well defined.  In fact,  if the case
is consistent with what you said,  high frequency wave will average out to 
zero.

	I think the explanation to this phenomena is that the high frequency
wave is absorbed by water as it travels deeper.  Since matter is dispersive,
a square wave (containing all kinds of frequency) will change to a round 
peak when it passes through the matter.