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A detector recedes from a stationary source of sound with a speed that increases continuously and without any limit. Describe qualitatively, the frequency of sound detected by the detector. Assume that the sound waves propagate indefinitely without attenuation and the detector does not create any air drag.

My answer: decreases continuously till no sound is detected.

Official answer: First decreases, becomes zero and then increases and eventually no sound is heard.

I used the Doppler formula : $\displaystyle f=f_0\left(\frac{v+v_{observer}}{v-v_{source}}\right)$ where $v_{observer}$ is negative because the detector is moving away. But when the speed of the detector exceeds that of sound, the sound waves which are subsequently emitted by the source will never be able to reach the detector. This was my reasoning, and I understand the "becomes 0" part, but where does the frequency increase?

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    $\begingroup$ When dealing with Doppler effect, remembrer that it isn't symmetrical. Are you sure that the formula you're using is correct for this specific scenario? Reversing the sign of velocity usually doesn't work due to this asymmetry. $\endgroup$
    – Miyase
    Jun 25, 2022 at 13:34
  • $\begingroup$ @Miyase AFAIK this is the only formula for Doppler Effect. $\endgroup$
    – insipidintegrator
    Jun 25, 2022 at 14:07
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    $\begingroup$ Have you considered the sound waves that passed you that you are now catching up with and passing? $\endgroup$
    – DJohnM
    Jun 25, 2022 at 14:09
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    $\begingroup$ Consider the same example with cars... If a bunch of cars run with constant speed and you accelerate then they'll initially pass you with decreasing frequency (because you accelerate) and after your velocity gets higher than theirs you'll start catching them up with increasing frequency (again because you accelerate). Hope this makes sense. $\endgroup$
    – ZaellixA
    Jun 25, 2022 at 16:35
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    $\begingroup$ Yes @ZaellixA. Yukterez ‘s answer clears everything in the first two lines itself. $\endgroup$
    – insipidintegrator
    Jun 25, 2022 at 16:36

3 Answers 3

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Consider, by analogy, a small, fast boat sitting almost motionless in the open ocean.

It's moving just fast enough to maintain steerage way, and keeps moving slowly with a following sea. The waves approach from behind, lift the stern and then the bow; the boat moves up and down with a frequency very close to the stationary frequency of the waves.

Now you advance the throttle and start moving away from the moving waves more quickly. The waves catch up, but at a lower frequency than before. Just what you would expect from the Doppler effect.

Eventually, your boat is travelling at the same speed as the waves in the water. The boat is suspended, motionless in terms of moving relative to the waves. Visualize a surfer, keeping just ahead of the peak of the breaking wave. A somewhat precautious situation.

Now, if the boat speeds up even more, it is now not just fleeing the waves. It is catching up with and passing the waves ahead of it. It experiences waves coming from bow to stern with increasing frequency as it goes faster and faster.

Because surface water waves move so slowly, most boats today operate in this "supersonic" regime. No matter how the boat twists and turns, the waves always seem to be coming from ahead, and the boat is designed to handle this one situation.

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insipidintegrator wrote: "But when the speed of the detector exceeds that of sound, the sound waves which are subsequently emitted by the source will never be able to reach the detector. "

This example assumes that the sound was continuous and began long before the observer started to accelerate, so the whole air around him is already filled with sound waves. Then the observer will hear the sound backwards when his velocity is greater than the speed of sound, since now he overtakes the soundwaves instead the other way around.

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    $\begingroup$ The first two lines of your answer solve my concerns. +1 $\endgroup$
    – insipidintegrator
    Jun 25, 2022 at 15:46
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There are four distinct phases for the situation in question:

  1. Subsonic movement
    As the detector moves ever faster away from the source, the frequency of the waves drops because the waves move slower and slower in relation to the detector.

  2. Sonic movement
    When the detector hits the speed of sound, it's moving along with the wave, and thus the wave seems to be standing still from its viewpoint. It does not register any sound at all.

  3. Supersonic movement
    The detector catches up with the waves from behind. Note that these are the waves that have already passed the detector in the first phase, and that the wave crests are now encountered in reverse order! As the detector accelerates further, the frequency of the perceived sound goes up.

  4. Waves left behind
    The detector has passed the first wave that was ever emitted. It is now further away than the sound from the source has traveled. As such, it cannot detect the signal anymore.

These four phases are what the official answer alludes to.

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    $\begingroup$ Thank you for a clear and concise 4th phase!+1 $\endgroup$
    – insipidintegrator
    Jun 26, 2022 at 7:55

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