Example of Doppler Effect With Two Shifts

A stationary source emits a sound wave of 5000 Hz. An object approaches to this source with a velocity of 3.5 m/s. What is the frequency of the wave reflected on the object?

Solution:
In this problem there are two Doppler shifts.

The first one, because the object acts like an observer in movement an "detects" a wave sound of frequency
f' = [(v + v0)/v]·f = [(343 m/s+ 3.5 m/s)/(343 m/s)]·5000 Hz = 5051 Hz.

The second one, because the object acts like a source in movement that "remits" (reflects) the sound, and hence the reflected frequency is
= v/(v - vs) = [(343 m/s)/(343 m/s - 3.5 m/s)]·5051 Hz = 5103 Hz

Sound Waves Theory and Examples, Visit:
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