**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 - v_{s})
= [(343 m/s)/(343 m/s - 3.5 m/s)]·5051 Hz = 5103 Hz

**Sound Waves Theory and Examples, Visit:**

**
Sound Waves Concepts and Examples
Sound Waves: Standing Wave, Interference, Doppler Effect - Examples **

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