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Prove. Let *f* be the emitted frequency, *f'* be the observed frequency, v be the velocity of sound,
v_{o} be the velocity of the observer and v_{s} be the velocity of the source where v_{o} and v_{s} can be
either positive or negative. Let the positive directions of v_{o} and v_{s} correspond to that of a vector
extending from observer to source. Then

Derivation. In Fig. 1, the observer is at the left and the source is at the right and the positive
direction is from left to right. The sound source is at point b at time t = 0 and at point c at time t.
The outer circle represents the wave front of the wave emitted at time t = 0. It corresponds to a
sphere in three dimensional space and moves outward with a velocity v (The fact that the wave
originated at a moving source does not affect its velocity after leaving the source.) The radius ba
of this sphere is thus vt. The distance bc equals v_{s}t. Thus

distance ac = vt + v_{s}t = (v + v_{s})t

distance cd = vt - v_{s}t = (v - v_{s})t

In the time interval from t = 0 to t = t, the number of waves emitted by the source is ft. In front of the source the waves are crowded into the distance cd and behind the source they are spread out over the distance ac. The wavelength in front of the source is thus

and the wavelength behind the source is

The wave fronts are approaching the observer at an effective speed of v + v_{o} . The time interval
between the time when wave front i passes the observer and when wave front i+1 passes the
observer is given by

See Fig. 2. The frequency he hears is then

If we consider the case where the observer is stationed to the right of the source, we get the same formula.

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