Doppler effect observed in octaves

Question

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)?

Answer 1

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.