Can anyone explain why they say Doppler effect does not depend on acceleration?

Would having acceleration not affect the frequency?

If the source emits the first circular wave, moving at 50 m/s towards you and speeding up towards you, by the time it sends out the second circular wave, the velocity will have increased which will affect the displacement of the source.

So is f5 really equal to f6 and f7?

  • $\begingroup$ In your text, the reception of the sound is supposed instantaneous. However, if you want to study the pheonomena along some amount of time, you might be interested by a previous answer to a similar question. $\endgroup$ – Trimok Sep 13 '14 at 11:41

How long do you have to listen to a sound to know its frequency? An entire period? Only half a period? This answer can become somewhat complicated, and I doubt that Fourier transforms were discussed in the class this quiz is from.

So let's just say you don't have to wait for an entire period in order to know the frequency. Actually, let's just say you know determine the frequency instantaneously. That could conceptually do away with accelerated sources having an effect on observed frequencies.

  • $\begingroup$ you edited your post, but I didn't have the chance to read it all. Can you post first post? $\endgroup$ – yolo123 Sep 13 '14 at 1:27
  • $\begingroup$ Ah! The frustrating wait and see. Although I would greatly want more out of my questioning, I guess this is as far as it goes. :( $\endgroup$ – yolo123 Sep 13 '14 at 1:28
  • 1
    $\begingroup$ It's gone forever because I edited it less than 5 minutes after being created; sorry. I didn't like that argument, but here it is again briefly (remember I'm not a fan of it). If the emitter frequency is high, the period is short; it could be so short that the extra distance traveled due to an accelerating source during this period is negligible. $\endgroup$ – BMS Sep 13 '14 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.