Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

According to my understanding, at the smallest level sound is just a collection of particles colliding on a wave.

So, assuming we can arbitrarily reduce the size of any human being, how small can we reduce someone while this someone can still "hear" normal size human beings talk coherently.

EDIT: In other words, how small do we have to be in order for it to become impossible to differentiate random collisions with other particles from collisions that result from the wave?

EDIT2: To avoid the shrinking human problem, how about thinking about small recorder? How small can the recorder be in order to record (and play back) frequencies heard by the human ear and for it to NOT play random noise generated by random collisions?

share|improve this question
2  
An interesting related puzzle is "Why can we use low frequency sound for localization?" When the frequency drops below about 1000 Hz the size of the sound wave gets to be several times the separation between our ears, yet we can still detect the direction from which sounds in the few hundred Hertz range are coming. What's up with that? (I can think of several ways to do this in electronics, but they all require sophisticated and/or precision manipulation of the signal.) –  dmckee Jul 30 '13 at 13:33
    
How about this as a recorder pimall.com/nais/microrecorder.html . And nanotechnology is getting more and more nano, so I might guess nano meters. –  anna v Jul 30 '13 at 14:21
    
@dmckee Is this with a pure sine wave, or are the harmonics responsible for enabling directional detection? –  Michael Mar 20 at 18:58
    
@Michael That is a very good question, and I don't know the answer. // Adds it to my list of projects for seniors –  dmckee Mar 20 at 19:16

2 Answers 2

up vote 11 down vote accepted

If you want to hear frequencies up to 20,000 Hz, it means that you must be able to perceive the changes of the pressure that are changing each 50 microseconds. But in 50 microseconds, even an atomic-sized recorder is hit by a large yet finite number of air molecules. How do we calculate their number?

Well, the background pressure is 100,000 pascals which means that each second, the air molecules deposit 100,000 newton-seconds of momentum per squared meter. $10^5$ per squared meter is $10^{-13}$ per square nanometer. The momentum of a typical air molecule is about $10^{-24}$ newton-seconds so the atmospheric pressure corresponds to $10^{11}$ (a ratio) collisions with a squared nanometer per second or a million of collisions per several microseconds.

This number 1 million has statistical noise of order "the square root of this million" i.e. one thousand. So if you will have a small animal or recorder, it will indeed be losing the sound in the noise. Only sufficiently loud sound – the loudness is an important quantity you haven't considered – will be audible. The calculation above suggests that a nanometer-sized animal could only distinguish sounds with pressure variations at least 1/1,000 of the atmosphere i.e. 100 pascals which is an extremely loud sound.

To compare, humans hear 1,000-Hz sound that is as silent as 20 micropascals in the pressure variations.

share|improve this answer
    
I think this is the answer but, wouldn't we need to perceive pressure changes every 25 microseconds? (This is from the Nyquist theorem.) Also thanks! I didn't know how to approach this problem, but it seems logical now. Kudos! –  user27653 Jul 30 '13 at 15:00
    
Yup, 25 microseconds is more accurate. ;-) –  Luboš Motl Jul 31 '13 at 8:47
    
So as diaphragm size increases, signal-to-noise ratio improves? –  endolith Sep 13 '13 at 3:34
    
Also, you can hear tones even if their RMS value is lower than the RMS value of white noise. –  endolith Sep 13 '13 at 3:39

For loud enough sound, a microphone could in principle be as small as a few molecules. This can be inferred as follows.

Air has a finite number density $n$ measured as number of molecules per unit volume. These molecules move at a thermal velocity $v$. So per unit of time, an area $A$ gets bombarded by $nvA$ molecules. And within a sound cycle at frequency $f$, the number of hits will be $nvA/f$. The square root of this number gives us the signal-to-noise for very loud sound.

With typical values $n = 0.25\ 10^{26}\ m^{-3}$, $v =5\ 10^2\ m/s$, $f = 10^4\ s^{-1}$, and a molecular sized microphone ($A \approx 10^{-19}\ m^2$) it follows that $S/N \approx \sqrt{nvA/f} \approx 300$.

So if one could design and construct a molecular size device capable of measuring fast pressure variations, there is no reason to believe that it would not be able to record loud sounds up to the high end of the audible frequencies. How to engineer and produce such a minuscule microphone is an entirely different story.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.