Will the frequency observed by a stationary observer will remain same if only the medium between the source and observer is moving?(ie. both source and observer are at rest and wind is blowing from source to the observer)

My textbook states that frequncy will not change and movembnt of medium will not haave any effect on frequency.

I think it is wrong. I think that initially it should change after that it would remain same.

My reason is that initially the travelling air (considered as medium) toward observer from source will increase the speed of waves in between source and observer after that the new waves will have changed wavelength ofsetting the change in velocity so the frequency will again become equal to the original.

Correct me if i am wrong, please.

$Any \ help\ is\ greatly\ appreciated$

  • $\begingroup$ Have you tried searching for an answer? eg A Doppler Effect problem with a moving medium : physics.stackexchange.com/q/66353 $\endgroup$ May 20, 2016 at 13:00
  • $\begingroup$ @sammygerbil Yes , but iam confused, according to that answer , as per my view, it says that frequency would change in my question , but my textbook says that frequency will not change and wind causes no effect. Who is correct? Please help! $\endgroup$
    – user102917
    May 20, 2016 at 13:34

2 Answers 2


No, the frequency will not change. If the wind is blowing at constant speed and the distance between source and observer remains constant, then the time it takes for a sound wave to get from source to observer will be constant. So the time interval between wave peaks (period T) when they are detected by the observer remains equal to the interval between the wave peaks when they are emitted. It is only when the distance between the source and observer is changing that the observed frequency will be different from the emitted frequency.

The effect of the wind is to make the wave peaks arrive earlier or later than they would in still air (ie they take less or more time to get from source to observer). But the time interval between the arrival of wave peaks is the same, and therefore the observed frequency is the same.

I do not understand what you mean about "initially it should change, after that it would remain same."

If the wind blows towards the observer, then the wave speed $c$ relative to the ground is higher, and the wavelength $\lambda$ (distance between peaks) is longer, but the frequency $f$ remains the same :
$c = f\lambda$.
Increased speed is compensated by longer wavelength.

  • 2
    $\begingroup$ I think in this way: first i assume that in between the source and the medium there are a number of waves present and many new are being generated by the source, now all of a sudden if wind starts to blow (entire medium moves uniformly like a block) then the crest present between source and observer will be driven faster and the observer will count more crests in less time as medium is driven towards it intact, as wind blows new waves now form from source which will have longer wavelengths and would then cancel the effect of increase in velocity . $\endgroup$
    – user102917
    May 20, 2016 at 14:11
  • $\begingroup$ Yes, if the wind speed is changing (accelerating), the situation is more complicated. Like your textbook, I assume that the wind speed is constant, since the question does not say otherwise. $\endgroup$ May 20, 2016 at 14:40
  • $\begingroup$ No, i didnt assume the medium is accelerating, the assumption i made is purely for constant velocity, hope you understand. Please have a look at it again. $\endgroup$
    – user102917
    May 20, 2016 at 15:34
  • $\begingroup$ Your comment says "if the wind starts to blow..." This means the wind speed is changing - eg from 0 to 20mph in 2mins. I think the book is assuming that we have reached a 'steady state' - the wind has been blowing with constant speed for some time, so that any initial effects are in the past. $\endgroup$ May 20, 2016 at 16:35

If you are not understanding still you can apply Doppler's effect formula here i.e. $$f'=f((v+w)+0)/((v+w)+0)=f$$ where $v+w$ is speed of sound now in wind.


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