Why Does Moving Medium Not Change Detected Frequency?

AI Thread Summary
The discussion explores why the frequency of sound waves remains unchanged for an observer when the medium, such as air, moves with a certain velocity. Despite the medium's movement increasing the speed of the emitted sound waves, the time interval between wavefronts, or the period, remains constant. This is illustrated by calculating the arrival times of successive wavefronts, which shows that the spacing between them is unaffected by the medium's velocity. Consequently, the observer detects the same frequency as emitted by the source. The conclusion emphasizes that the frequency detected by the observer is determined by the source's emission rate, not the medium's motion.
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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