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Although most commonly associated with a moving object emitting a tone (like a car or train), a sound may also be doppler shifted if the listener is moving or (even less obvious) if the air medium is moving between a stationary tone and stationary listener.

My question is: If the only way allowed to observe a tone generating object is by listening once, can a distinction be made whether the object, the listener, or the air is moving?

Additional constraints are:

  1. The tone generator frequency at its source is fixed and known
  2. Only one listener observation may be made yielding a single frequency from which (taken with constraint #1) the amount of doppler shift can be determined.

I'm not sure I'm allowed to say this (since this is my first question ever to post here), but I think the answer is no, but wanted to make sure I considered everything. (Apologies for my wordiness & thanks for your time)

Hal McKinney
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2 Answers2

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I believe you are correct. Given known atmospheric conditions you could determine the amount of the shift, but motions being relative, you could not know whether it was from motion of the emitter, the wind, the observer, or a combination of their movements.

Adrian Howard
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  • Luckily though, we can usually measure if we (the receiver) is moving or not through other means. As well as wind speed and direction. – DKNguyen May 09 '21 at 02:22
  • Thanks. I'd vote this up but I don't have enough creds to vote just yet since I'm a newby here. :-) – Hal McKinney May 09 '21 at 02:29
  • I believe if you are allowed to use two microphones and hence get phase information you can tell at least if there is listener or air motion. Because if you know the speed of sound in the current conditions, then there can only be one wavelength for any given frequency. If the phase implies a different wavelength then it is because the microphones or the air are moving. – geshel May 09 '21 at 03:37
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If the frequency generated by the source is $f_s$ and the frequency heard at the receiver is $f_r$ then

$(c-v_s)f_r = (c+v_r)f_s$

where $c$ is the speed of sound and $v_s$ and $v_r$ are the speeds of the source and receiver relative to the air (the signs assume that source and receiver are approaching each other).

If the speed with which the source approaches the receiver is $v$ then $v=v_r +v_s$, so

$(c-v+v_r)f_r=(c+v_r)f_s$

If you know $f_s, f_r, v$ and $c$ then you can solve this equation to find $v_r$, then $v_s=v-v_r$, and you now know the speeds of the receiver and the source relative to the air.

What this doesn’t tell you is whether the air itself is moving relative to some third object.

gandalf61
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  • Great detail in this answer @gandalf61, but since (as your last sentence points out) distinguishing the speed of the air is needed as well, your good work here seems to further confirm my expectations. Thanks! – Hal McKinney May 10 '21 at 13:38