Standard 5.2 – The Doppler Effect
Background: The Doppler effect occurs whenever the source of a sound has a different velocity from
the detector of the same sound. If the source and detector are converging (getting closer together), the
detected frequency will be greater than the actual frequency. If the source and the detector are diverging
(getting farther apart), the detected frequency will be less than the actual frequency.
Doppler effect problems can be solved using the following equation:
fd = fs(v – vd) / (v – vs)
fd = frequency detected
fs = frequency produced by the source
v = velocity of the wave
vd = velocity of the detector
vs = velocity of the source
Pay particular attention to direction. The direction the wave moves to get from the source to the detector
is always the positive direction.
Example 1: A police siren produces a sound wave with a frequency of 550 Hz. If the police car
approaches a stationary bystander at a speed of 23.0 m/s, what frequency is heard by the bystander?
Solution: First, determine whether the detected frequency will be greater than or less than the actual
frequency. In this case, the source and detector are converging. The detected frequency should be greater
than 550 Hz. fs = 550 Hz
v = 343 m/s
vd = 0 m/s
vs = +23.0 m/s
fd = (550)(343 - 0) / (343 – 23) = 589 Hz
Always check to see that your answer matched your initial prediction. The 589 Hz detected is indeed
greater than the 550 Hz actually produced by the siren.
Example 2: Suppose the police car approaches a jogger headed toward the moving police car at a speed
of 8.5 m/s. What frequency does the jogger hear?
Solution: Again, we must first make a general prediction about the frequency detected. Because the
jogger is headed toward the police car, we expect the detected frequency to be even greater than the
frequency detected by the stationary bystander.
fs = 550 Hz
v = 343 m/s
vd = -8.5 m/s vs = +23.0 m/s
Notice that the velocity of the jogger has to be negative since the jogger is headed the opposite direction
as the sound.
fd = (550)(343 + 8.5) / (343 – 23) = 604 Hz
Example 3: A train whistle emits a frequency of 256 Hz. A stationary observer detects a frequency of
247 Hz after the train passes. How fast is the train moving?
Solution:
fd = 247 Hz
fs = 256 Hz
v = 343 m/s
vd = 0 m/s
247 = (256)(343 - 0) / (343 – vs)
We need to multiply both sides by (343 – vs) to get the variable out of the denominator.
247(343 – vs) = (256)(343)
84721 – 247vs = 87,808
-247vs = 3087
vs = -12.5 m/s (the negative velocity implies that the train is moving away from the detector)
Applications of the Doppler Effect
1) Doppler radar sends out waves that are reflected by precipitation particles (rain, snow, sleet) and
measures the amplitude and frequency of the reflected waves. The more wave signals that get
returned (higher amplitude), the heavier the precipitation. The frequency of the returning signals
accounts for both the direction and speed of the storm.
2) Red shift / Blue shift: Every star contains hydrogen and helium. Those elements, like all
elements, emit a unique set of colors that show up as narrow bands of color when viewed through
a spectrometer. These lines represent very specific frequencies of light. When a star is moving
relative to us, the Doppler Effect causes the bands to show up as different colors, depending on
the direction and speed of movement. Red light is the lowest frequency of visible light while blue
light is one of the highest frequencies. If a star is moving away from us, we see the bands of
color shifted toward the red side of the spectrum because the detected frequency is lower than the
actual frequency. Stars moving toward us have lines that are shifted toward the blue end as the
detected frequency is greater than the actual frequency. Every star outside our galaxy
produces a red shift. This tells us that the entire universe is expanding.