I have a question about the interpretation of the Doppler effect, when you look at the results as a change in octaves. Nothing actually changes when you look at the result in octaves instead of frequency shift, obviously, but it suddenly seems a whole lot less intuitive. This makes me wonder whether I understand things correctly.

If we assume a stationary listener and medium and moving sound source ($-c < v_s < c$, where $c \approx 343 \frac{m}{s}$, speed of sound), the observed frequency $f_L$ is:

$$f_L = \left(\frac{c}{c - v_s} \right) f_0$$

where $f_0$ is the emitted frequency. For an object moving away from you (negative $v_s$), the maximum observed pitch is $\frac{1}{2} f_0$, which is exactly one octave down. Intuitively, I find that this makes sense: all frequencies are pitched down, but not by a crazy amount.

When the source moves away ($v > 0$), however, one octave upis observed at half the speed of sound: $v = \frac{1}{2} c \Rightarrow f_L = 2 f_0$. And at $v = \frac{3}{4}c$ , it's two octaves up, and you can go all the way up to infinity. So in this case, all frequencies are suddenly scaled to a larger and larger range of frequencies, which seems odd.

Is this actually a correct way of looking at it? And what happens for the limit of $v_s \rightarrow c$ (just before the sonic boom)?

  • $\begingroup$ I think your doppler equation is a bit off. For your setup, I'd use $f_L=(\frac{c}{c-v_s})f_0$ $\endgroup$ – Jim Jan 28 '14 at 15:41
  • $\begingroup$ You seem to be right, thanks. I changed it and the implications accordingly. But the question remains... $\endgroup$ – Yellow Jan 28 '14 at 15:59

You are visualizing fine, the Doppler effect is usually experienced at ground speeds (much less than speed of sound). As you approach the sound barrier, waves will compress so much that instead of hearing a high pitch, you will experience a sudden shock wave (known as "sonic boom"). And at higher speeds than sound, the Doppler effect for the incoming object will not work practically, because you will not hear anything (object is moving faster than sound). But once it passes you, and you receive the shock wave, you will hear a much lower pitch of the source sound.

  • $\begingroup$ Thanks for the sound explanation (no pun intended). So let me just confirm that the observed frequency in terms of octaves is therefore much higher when the object approaches you than when it's gone past you, right? At least at say $|v| > 300 \frac{m}{s}$. $\endgroup$ – Yellow Jan 28 '14 at 16:46
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    $\begingroup$ @Yellow, Well, in term of octaves, for sounds above 300m/s but below 343 m/s, lets say, the source $A_4=400 Hz$, traveling at you at $V_s=321.5625 m/s$, you will hear $A_8=7040 Hz$ or 4 octaves higher. Once it passes you, going away this turns out to be 227 Hz, almost near $A_3$ or just an octave lower. $\endgroup$ – CAGT Jan 28 '14 at 17:10

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