Doppler effect formula signs: when is it plus and when is it minus for source and observer?

6 answers & replies

• [+][–]Aman YadavDropperSample· 1mo ago

  Rule. Approaching = frequency up. Receding = frequency down.

  Numerator, observer: moving towards the source gives $v + v_o$, away gives $v - v_o$. Denominator, source: moving towards the observer gives $v - v_s$, away gives $v + v_s$.

  Your case: source approaches, so denominator is $v - v_s = 320$. $f' = 500\times\frac{340}{320} \approx 531$ Hz. You used the wrong sign in the denominator.

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  • [+][–]Farhan ShaikhDropperSample· 1mo agoAuthor

    got it, I was copying the sign from the other case. Approaching = bigger, receding = smaller, I'll check that first. thanks everyone!

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• [+][–]Pooja Iyer12thSample· 1mo ago

  More numbers for practice!! Say $v = 340$ m/s.

  • Stationary source 500 Hz, observer runs towards it at 20 m/s: $f' = 500\times\frac{360}{340} \approx 529$ Hz.
  • Source moves away at 20 m/s from a stationary observer: $f' = 500\times\frac{340}{360} \approx 472$ Hz. (That's your 472, it was the receding answer!)

  Notice that towards-source and towards-observer give slightly different values even though the relative speed is the same. It's because the medium (air) is the reference, not either of them.

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• [+][–]Tushar PawarDropperSample· 1mo ago

  For (a): let the source be behind the observer, both moving forward at speed $u$. The observer is moving away from the source: numerator $v - u$. The source is moving towards the observer: denominator $v - u$. So $f' = f$. No change. Makes sense, the gap between them doesn't change, so the crests arrive at the same rate as they leave. If they move in opposite directions the answer changes of course.

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• [+][–]Rohan Sharma11thSample· 1mo ago

  for (b) when the source moves it literally changes the frequency it produces, the siren makes higher frequency sound as it speeds up, because the engine is working harder. the train horn gets higher pitch 🚂 and then when it passes you the pitch drops because the horn relaxes? that's how I explain it to myself at least

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  • [+][–]Zoya Khan12thSample· 1mo ago

    Rohan, I understand why it feels that way but it's not the siren itself. The source keeps emitting waves at the same frequency $f$ no matter how fast it moves. What changes is where those waves are in space. If the source moves towards you, each new crest is emitted from a spot closer to you, so the crests are bunched up: the wavelength reaching you is shorter, $\lambda' = (v - v_s)/f$, and the sound speed is unchanged, so you receive a higher frequency. If instead the observer moves, the wavelength doesn't change, the observer simply meets the crests faster. So 'apparent' frequency is the right word.

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