Why Does Moving Medium Not Change Detected Frequency?

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SUMMARY

The discussion centers on the phenomenon where the frequency of sound waves emitted by a stationary source remains unchanged despite the movement of the medium (air) between the source and the observer. The key point is that while the speed of sound waves increases with the medium's velocity, the time interval between wavefronts, defined as the period T, remains constant. This is mathematically demonstrated by analyzing the arrival times of successive wavefronts, confirming that the frequency detected by the observer does not vary with the medium's motion.

PREREQUISITES
  • Understanding of wave mechanics and sound propagation
  • Familiarity with the concepts of frequency and period in wave physics
  • Basic knowledge of relative motion and reference frames
  • Mathematical skills to analyze wavefront arrival times
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This discussion is beneficial for physics students, acoustics researchers, and anyone interested in the principles of sound wave behavior in moving media.

mathman44
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Hi all. I am asking for some help with getting a physical intuition for the following situation:

Consider an observer and source, at rest. The source emits sound waves of a particular frequency, 'f'. Letting the medium (air for example) between the source and the observer move with a particular velocity, why is it that the frequency of the emitted waves, as detected by the observer, remains unchanged?

It is intuitively clear that the velocity of the emitted wave increases as the medium gains speed, but it is less clear why the frequency detected by the observer remains unchanged.

We all know from experience that the sounds in the distance, traveling through moving air, don't get modified in frequency. But why must this be so, physically?

Any insight would be greatly appreciated!
 
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Say the speed of sound is v, and the flow rate of air is u. The distance between source and observer is d, which is fixed. The source generates a sound with frequency f. The period is T = 1/f.

The wavefronts leaving the source travel at speed v relative to the air, and the air travels at speed u relative to the source and the observer. So the wavefront travels at speed v + u between the source and observer. That means that wavefront 1, leaving at t = 0, arrives at time t1 = d/(v+u) later.

If wavefront 2 leaves at time t = T, it arrives a time t2 = T + d/(v+u) later.

t2 - t1 = T + d/(v+u) - d/(v+u) = T

So... the spacing at which the observer receives the wavefronts is still T and is unaffected by the travel time, it would seem.
 

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