Section 9.5: The Doppler Effect
Tutorial 1 Practice, page 435
1. Given: vsource = 20.0 m/s; f0 = 1.0 kHz;
vdetector = 0 m/s; vsound = 330 m/s
Required: fobs
Analysis:
!v
+ vdetector $
f obs = # sound
& f0
" vsound + vsource %
Solution:
!v
+ vdetector $
f obs = # sound
& f0
" vsound + vsource %
! 330 m/s + 0 m/s
=#
#" 330 m/s + '20.0 m/s
(
! 330 m/s $
=#
& 1.0 kHz
" 310 m/s %
(
)
$
& 1.0 kHz
&%
(
)
!v
+ vdetector $
f obs = # sound
& f0
" vsound + vsource %
)
Solution: Determine the speed of sound at 15 °C:
(
(
f obs = 1.1 kHz
Statement: The detected frequency of the
approaching police car is 1100 Hz, or 1.1 kHz.
2. Given: fobs = 900.0 Hz; vdetector = 0 m/s;
f0 = 950.0 Hz; vsound = 335 m/s
Required: vsource
Analysis:
!v
+ vdetector $
f obs = # sound
& f0
v
" sound + vsource %
f0
v
+ vdetector
f obs sound
(
vsound + vsource =
)
(
(
vsource
) (
950.0 Hz
335 m/s + 0 m/s ! 335 m/s
900.0 Hz
95 Hz
=
335 m/s ! 335 m/s
90 Hz
= 18.6 m/s
)
$
& 300.0 Hz
&%
(
)
)
f obs = 320 Hz
)
=
!v
+ vdetector $
f obs = # sound
& f0
v
" sound + vsource %
(
f0
v
+ vobserver ! vsound
f obs sound
(
Determine the frequency detected by the observer:
! 340.5 m/s $
=#
& 300.0 Hz
" 315.5 m/s %
)
)
vsound = 340.5 m/s
(
f
= 0 vsound + vdetector ' vsound
f obs
(
!
m/s $
= 331.4 m/s + # 0.606
& 15 °C
"
°C %
! 340.5 m/s + 0 m/s
=#
#" 340.5 m/s + '25 m/s
Solution:
vsource =
)
vsound = 331.4 m/s + 0.606 m/s/°C T
= 1100 Hz
vsource
(b) Answers may vary. Sample answer:
Two examples of the Doppler effect are the noise
of a jet at an air show and the sound of a racecar to
someone near the track.
2. A sound wave has a higher frequency when the
source is approaching a stationary observer
because the sound waves are compressed as the
source gets closer to the observer. Compressed
sound waves mean a higher frequency.
3. Given: f0 = 300.0 Hz; T = 15 °C;
vdetector = 0 m/s; vsource = 25 m/s
Required: fobs
Analysis: vsound = 331.4 m/s + ( 0.606 m/s/°C )T ;
)
)
Statement: The speed of the ambulance is
18.6 m/s.
Section 9.5 Questions, page 435
Statement: The detected frequency of the object is
320 Hz.
4. Given: f0 = 850 Hz; ∆f = 58 Hz; vdetector = 0 m/s;
vsound = 345 m/s
Required: vsource
Analysis: f obs = f0 + ! f ;
!v
+ vdetector $
f obs = # sound
& f0
" vsound + vsource %
vsound + vsource =
vsource =
f0
v
+ vdetector
f obs sound
(
)
f0
v
+ vdetector ' vsound
f obs sound
(
)
1. (a) The Doppler effect describes the changing
frequency of sound as the source is in motion
relative to an observer.
Solution: Determine the observed frequency:
Copyright © 2011 Nelson Education Ltd.
Chapter 9: Wave Interactions
f obs = f0 + ! f
= 850 Hz + 58 Hz
f obs = 908 Hz
9.5-1
Determine the speed of the fire truck:
vsource
f
= 0 vsound + vdetector ! vsound
f obs
(
=
vsource
850 Hz
908 Hz
= !22 m/s
)
(345 m/s + 0 m/s) ! (345 m/s)
Statement: The speed of the fire truck is 22 m/s.
5. Given: vsource = 0 m/s; f0 = 440.0 Hz;
vdetector = 90 km/h; T = 0 °C
Required: fobs
Analysis:
f obs
!v
+ vdetector $
= # sound
& f0
v
" sound + vsource %
6. Given: vdetector = 0 m/s; fobs = 560 Hz;
vsound = 345 m/s; f0 = 480 Hz
Required: vsource
Analysis:
!v
+ vdetector $
f obs = # sound
& f0
" vsound + vsource %
vsound + vsource =
vsource =
vsource =
vsource = 25 m/s
Determine the observed frequency of the horn as I
approach the observer:
)
f0
v
+ vdetector ' vsound
f obs sound
(
)
f0
v
+ vobserver ! vsound
f obs sound
(
)
(
(
vsource
) (
560 Hz
345 m/s + 0 m/s ! 345 m/s
480 Hz
56 Hz
=
345 m/s ! 345 m/s
48 Hz
= 58 m/s
=
vsource = 90 km/h
km ! 1000 m $ ! 1 h $ ! 1 min $
#
&
h #" 1 km &% #" 60 min &% " 60 s %
(
Solution:
Solution: Since the temperature is 0 °C, the speed
of sound is 331.4 m/s.
Convert vsource to metres per second:
= 90
f0
v
+ vdetector
f obs sound
)
)
Statement: The speed of the source is 58 m/s.
7. The frequency reduces. The effect is not
instantaneous as it depends on the speed of the
source and how far the source is from the observer.
!v
+ vdetector $
f obs = # sound
& f0
" vsound + vsource %
! 331.4 m/s + 0 m/s
=#
#" 331.4 m/s + '25 m/s
(
! 331.4 m/s $
=#
& 440 Hz
" 306.4 m/s %
(
)
$
& 440 Hz
&%
(
)
)
f obs = 480 Hz
As I pass the observer, the person will detect the
exact frequency of the horn.
Statement: The person will detect a frequency of
480 Hz as I approach, and a frequency of 440 Hz
as I pass.
Copyright © 2011 Nelson Education Ltd.
Chapter 9: Wave Interactions
9.5-2