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Re: [Phys-l] frequency and wavelength of sound in air



On 04/04/2009 06:37 PM, Julie Quah wrote:

Why does sound frequency not change when travelling from one
medium (eg air) to another (eg water)? Is it because the sound
frequency is like the finger print for each wave ? Then why do we hear
sound pitch increase when it travels through helium, which is another
different medium as compared to air?

1) First of all, sound pitch does *not* increase when it crosses
the boundary between helium and air.

The resonant frequency of a resonator (e.g. organ pipe or voice
box) depends on the speed of sound of the medium, but that is not
a boundary-crossing issue.



2) Returning to the air/water boundary, let's assume for simplicity
it is flat, or locally approximated as flat. Assume without further
loss of generality the surface is in the XY plane.

To an excellent approximation, when a sound wave comes to the air/water
boundary, it doesn't go across at all. Mostly it just reflects off
the humongous difference in index.

To the extent that there is any transmission at all, when a plane
wave crosses the boundary,
-- frequency stays the same
-- kx stays the same
-- ky stays the same
-- kz does not stay the same.

There's nothing special about frequency. It is not the "fingerprint".
It is not any more special than kx or ky. The physics is simple: you
have a wavelike solution on one side of the boundary and another
wavelike solution on the other side, and they need to make the same
prediction *at* the boundary. That's all there is to it.

The change in kz with no change in kx and ky causes the direction of
propagation to change. This is called refraction.

There will be a huge range of incident angles for which it is not possible
to simultaneously match frequency, kx, and ky. This corresponds to total
internal reflection. (Internal to the air, external to the water.)

The physics of all this is essentially the same as the physics for light
at a change of index of refraction ... which you can look up in almost
any physics text. Actually the theory is easier for sound, because you
don't need to worry about polarization.