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For instance: In the diagram below suppose A shouts at B who is at a distance of S apart while the platform moves in a uniform velocity V1. If the speed of sound in air at *rest is V, wouldn't the time it takes for the sound to travel from A to B is, t = S/(V-V1).

Intuitively, I suppose an observer on earth would see sound from A has travelled further than S distance forward to reach B since B is also moving forward, so that isn't it logical to think it takes more time for sound to reach B than it should be when the platform is at rest.

I guess that, in relative to B the speed of sound is V-V1 and in relative to the earth the speed of sound is V. Am I correct?

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    $\begingroup$ Ever noticed that an emergency vehicle's siren has a different pitch when the vehicle is moving toward you and when it is moving away from you? Now imagine that while the vehicle is approaching you, there is somebody else standing behind it as it moves away from them. What does each of you hear? $\endgroup$
    – The Photon
    Sep 1, 2021 at 5:08
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    $\begingroup$ It makes all the difference if this vehicle is open, like in the drawing, or closed, like a car. Sound waves travel on a medium, and it is relative to the average speed of that medium. If the vehicle is closed, the air moves with it, so the speed of sound is relative to the vehicle. If the vehicle is open, the speed of sound is relative to the wind felt by the vehicle (which itself depends on the speed of the vehicle). $\endgroup$
    – lvella
    Sep 1, 2021 at 9:42
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    $\begingroup$ Speed of sound is constant to the air it moves through, and to nothing else. If the wind is gently blowing, the speed of sound is gently sped up/slowed down relative to the ground. Sound is stupid, it has not heard of Einstein, much less heard of relativity. $\endgroup$
    – PcMan
    Sep 1, 2021 at 11:13
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    $\begingroup$ If sound itself hasn't heard of relativity, how can anybody else have heard of it then?... @Ivella yes. In fact, even if the vehicle is open, there will probably a lot of turbulence etc. that skews the results. $\endgroup$
    – leftaroundabout
    Sep 1, 2021 at 15:12
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    $\begingroup$ @PcMan: It's not stupid so much as slow. If you somehow had a weird medium with a very fast (i.e. significant fraction of c) speed of sound, you would need to start doing Lorentz transformations to calculate the Doppler shift, and this all gets a lot more exciting. $\endgroup$
    – Kevin
    Sep 1, 2021 at 19:41

4 Answers 4

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No, the speed of sound is not the same everywhere. There is a special frame, and that is the special frame in which the air itself is at rest.

This has rather profound results when the speed of the platform V1 exceeds the speed of sound V. The term V-V1 becomes negative, because the observers are now moving at supersonic speeds.

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  • $\begingroup$ what if the vehicle is closed. wouldn't it still take t = S/V-V1 for the sound to reach from A to B since sound speed only depends on properties of the medium such as temperature, stiffness, density ...not the velocity of the medium itself. $\endgroup$
    – madu
    Sep 20, 2021 at 10:50
  • $\begingroup$ @madu: The speed of sound does depend on the velocity of the medium, even when that velocity is a velocity field. E.g. things get a more complex inside a rotating cylinder where the air rotates with the cylinder. But for a closed vehicle moving in a straight line, the air inside is presumably entirely independent from the air outside, so there is no V1 term anymore. $\endgroup$
    – MSalters
    Sep 20, 2021 at 11:05
  • $\begingroup$ thank you very much. $\endgroup$
    – madu
    Sep 20, 2021 at 17:52
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You are correct. Since A and B are at rest with respect to eachother, they will hear the same frequency. But an observer on the ground (say C) will hear a frequency which is either greater or less depending on whether she is in front of the cart or behind. In the former case, we would get \begin{align} & \lambda_C = \lambda_B \\ & f_C^{-1} = \frac{v_B}{v_C} f_B^{-1} \\ & f_C = \frac{v}{v - v_1} f_B. \end{align} This checks out because if $v_1 = v$, the sound speed and therefore the frequency observed by B is zero. This makes the finite frequency heard by the stationary observer infinitely larger.

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    $\begingroup$ Oops. I referred to an observer on the ground in front of the cart but forgot to say this was C. Fixed. $\endgroup$
    – Connor Behan
    Aug 31, 2021 at 13:37
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    $\begingroup$ "Since A and B are at rest with respect to each other, they will hear the same frequency." If the cart is travelling faster than the speed of sound relative to the air, when A shouts B will hear nothing. $\endgroup$
    – alephzero
    Aug 31, 2021 at 20:00
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    $\begingroup$ @madu note that when light moves in a dense medium, it's phase velocity is lower, and then it also depends on the reference frame and there is even a kind of ‘luminal boom’, the Cherenkov radiation. It is the speed of light in vacuum that is special, not the light itself. $\endgroup$
    – Jan Hudec
    Aug 31, 2021 at 20:56
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    $\begingroup$ Supersonic ears cannot hear supersonic mouths. If B hears nothing then A also hears nothing and the point stands... $\endgroup$
    – Connor Behan
    Aug 31, 2021 at 21:05
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    $\begingroup$ @madu, if you replace sound for light, you'll involve relativistic speeds and therefore have to switch from Galilean transformation to Lorentz transformation under which the time in the reference frame of the cart is different from reference frame of the ground. So while it will still take $t = \frac{S}{c-v_1}$ in the ground frame, it will take $t' = \frac{S}{c}$ in the cart (if it's in vacuum), because the speed of light is invariant. I'm not trying to derive the more complex dense medium case for a comment. $\endgroup$
    – Jan Hudec
    Sep 2, 2021 at 19:01
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The velocity of the sound relative to the observer depends on the observer's velocity, but "the speed of sound" is generally understood to mean the speed of sound relative to the medium.

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You're right, the speed is not the same. Extending your thought experiment with measurement equipment may help to understand in more detail.

Consider a sound receiver which activates a light that the sender can see. The local propagation speed of light in air is near $c$ (and $>> V$) so we can consider the light speed signal instantaneous for the purposes of measuring the speed of sound between the two points on the platform. This avoids a need to invoke outside observers, letting these observers directly measure the speed of sound.

(The signal could be a momentary chirp, with the receiver lighting up an LED after detecting a few oscillations at that frequency. Or it could be a modulation (such as AM or FM) of a carrier audio frequency with the receiver demodulating, and encoding that info onto an LED in some form. By sending a continuously varying signal, the sender can detect how far "away" in milliseconds of propagation time the receiver is.)

Note that unlike an ultrasound ranging device (e.g. with a piezo to chirp and then detect the return), the sound is only propagating one way, with the other way carried over light.

At 90% of the speed of sound in the forward direction, the sound waves aren't going much faster through the air than the platform (and sender / receiver), and sound will take about 10x longer to get there. (Tending towards infinity in the limit as you get closer to the singularity in your formula.)

In the other direction, at close to Mach 1.0 sound will take close to half the time. In theory you can make the propagation time arbitrarily short by going many times the speed of sound, but hearing anything over the sonic boom will be problematic. And if you get into relativistic speeds in air, you will have bigger problems (xkcd).

Connor's answer shows that the speed of the source and receiver relative to the air can cancel out for subsonic cases, so you hear the same frequency. But that's not at all the same thing as a constant speed of sound.

Unprotected humans will have a Bad Time at transonic speeds (just below Mach 1), so let's assume sender and receiver are androids, or that the humans left a computerized experiment set up and wisely got off. We're also neglecting that air having to go around things will have to speed up a bit, potentially going supersonic.

This is why planes not designed to break the sound barrier will also have a Bad Time as flows around parts of the aircraft become supersonic. One of the speed limits for normal passenger jets is a maximum Mach number of typically 0.85 or so for the plane as a whole, which keeps airflow over control surfaces and wing roots comfortably below Mach 1.0.

All these speeds are of course relative to the air, which might be moving relative to the ground (aka non-zero wind speed). Mach numbers are implicitly air speed, so yes, the air itself is a special frame of reference, relative to which true air speeds are measured.

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    $\begingroup$ I wonder if you could make a practical physics demonstration out of this. Ultrasound in a wind tunnel or something. $\endgroup$
    – TimWescott
    Aug 31, 2021 at 23:57
  • $\begingroup$ Thank you very much @Peter Cordes and yes TimeWescott I would need to do a small demo right after lock down is over. waiting eagerly for that. $\endgroup$
    – madu
    Sep 1, 2021 at 15:47
  • $\begingroup$ It would be very interesting for me to know what would happen if it was just an EM wave (i.e. light) that sent across instead of sound wave. wonder if there would be any phase shift at the receiver or a frequency change or none... $\endgroup$
    – madu
    Jul 13 at 15:36
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    $\begingroup$ @madu: Light behaves the same in all inertial frames (for two observers at rest wrt. each other). That's the premise from which special relativity derives. But if they're moving through some medium at near light speeds, and the local speed of light in the medium is lower than c (as in most media), you could get something noticeable happening, I guess. $\endgroup$
    – Peter Cordes
    Jul 13 at 17:36

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