sound

1. Doppler Effect/Shift
• By the end of the lecture students will be
able to calculate frequencies of sound as
they apply to the Doppler effect/shift.

2. Doppler Effect
A tone is not always heard at the same frequency at which it is
emitted. When a train sounds its horn as it passes by, the pitch of
the horn changes from high to low. Any time there is relative
motion between the source of a sound and the receiver of it, there is
a difference between the actual frequency and the observed
frequency. This is called the Doppler effect.
The Doppler effect applied to electromagnetic waves helps
meteorologists to predict weather, allows astronomers to estimate
distances to remote galaxies, and aids police officers catch you
speeding.
The Doppler effect applied to ultrasound is used by doctors to
measure the speed of blood in blood vessels, just like a cop’s radar
gun. The faster the blood cell are moving toward the doc, the greater
the reflected frequency.

3. Doppler
Equation
fD = fS
v ± vD
v ± vS
)(
fD = frequency as heard by a listener
fS = frequency produced by the source
v = speed of sound in the medium
vD = speed of the listener
vS = speed of the source
This equation takes into account the speed of the source of the
sound, as well as the listener’s speed, relative to the air (or
whatever the medium happens to be). The only tricky part is the
signs. First decide whether the motion will make the observed
frequency higher or lower. (If the source is moving toward the
listener, this will increase fD, but if the listener is moving away
from the source, this will decrease fD.) Then choose the plus or
minus as appropriate. A plus sign in the numerator will make fD
bigger, but a plus in the denominator will make fD smaller.
Examples are on the next slide.

4. Which way + or - ?
D S
Negative Direction Positive Direction

5. Doppler Set-ups fD = fS
v ± vD
v ± vS
)(
still10 m/s
f D = 1000
343
343 -10 )(
= 1030 Hz
The horn is producing a pure 1000 Hz tone. Let’s find the frequency as
heard by the listener in various motion scenarios. The speed of sound
in air at 20 °C is 343 m/s.
still
10 m/s
fD = 1000
343 + 10
343 )(
= 1029 Hz
Note that these situation are not exactly symmetric.
Also, in real life a horn does not produce a single tone.
More examples on the next slide.

6. 3 m/s
Doppler Set-ups
(cont.)
fD = fS
v ± vD
v ± vS
)(
10 m/s
fD = 1000
343 - 3
343 -10 )(
= 1021 Hz
The horn is still producing a pure 1000 Hz tone. This time both the
source and the listener are moving with respect to the air.
fD = 1000
343 + 3
343 - 10 )(
= 1039 Hz
10 m/s 3 m/s
Note the when they’re moving toward each other, the highest
frequency possible for the given speeds is heard. Continued . . .

7. 10 m/s
10 m/s
3 m/s
Doppler Set-ups
(cont.)
fL = fS
v ± vD
v ± vS
)(
fD = 1000
343 - 3
343 + 10 )(
= 963 Hz
The horn is still producing a pure 1000 Hz tone. Here are the final
two motion scenarios.
fD = 1000
343 + 3
343 + 10 )(
= 980 Hz
3 m/s
Note the when they’re moving toward each other, the highest
frequency possible for the given speeds is heard. Continued . . .

8. Doppler Problem
Mr. Magoo & Betty Boop are heading toward each other. Mr. Magoo
drives at 21 m/s and toots his horn (just for fun; he doesn’t actually see
her). His horn sounds at 650 Hz. How fast should Betty drive so that
she hears the horn at 750 Hz? Assume the speed of sound is 343 m/s.
21 m/s vD
fD = fS
v ± vD
v ± vS
)( 750 = 650
343 + v D
343 - 21 )(
vD = 28.5 m/s