Doppler Effect • Frequency shift when sound source and/or listener

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Doppler Effect •
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Frequency shift when sound source and/or listener is moving When source is moving towards listener pitch appears higher When source is moving away from listener, pitch appears lower Example: Plane flying overhead, https://www.youtube.com/watch?v=eo_owZ2UK7E Applications: Weather radar: Developed in 1988, the Doppler shift is used to determine the radial velocity component of a target. It is also used to determine rotations in wind, which predicts severe weather. The frequency shift is very small, so the Doppler radar actually uses a phase shift. Echocardiography: Ultrasound and Doppler technology are used to visualize the structure of the heart. Areas of the heart that have a high and low blood velocity can be seen in different hues of the resulting picture. The Doppler shift can be used to determine how fast an object is moving, and is the basis for radar guns. Most radar guns use microwaves, but we will look at a simple application using sound waves. The shift in frequency is determined as:  v  vs
f   f 0 
 v  vs

 
where f0 is the initial frequncy, f′ is the shifted frequency, v is the speed of sound (usually taken as 767 mph), and vs is the speed of the object. Using some fun algebra, the speed can be written in terms of f0 and the frequency shift, (f′‐ f0). vs 
 f   f0 
v 2 f0   f   f0 
The instrument measures the beat frequency, (f′‐ f0), and from that calculates the speed of the object. We have a rational function, where the independent variable is (f′‐ f0) and the dependent variable is vs. The function is ploteed on the next page for a frequency of f0 = 5000 Hz and v= 767 mph. 140
120
100
vs (mph)
Speed vs. Frequency Shift
80
60
40
20
0
vs 
0
500
1000
(f'‐f0), Hz
 f   f0  v
2 f0   f   f0 
1500
2000
This graph looks approximately linear. Can you explain why? What do you think will happen as the frequency shift continues to increase? 700
600
vs (mph)
500
400
Speed vs. Frequency Shift
300
200
vs 
100
 f   f0  v
2 f0   f   f0 
0
0
10000
20000
30000
(f'‐f0), Hz
40000
50000
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