Effect of motion of medium on frequency 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.
$Any \ help\ is\ greatly\ appreciated$
 A: 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.
A: 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.
