Lecture 5
Ch17. Longitudinal Waves
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2013
Homework 4: Two Speakers
Two speakers separated by distance d1
= 2 m are in phase. A listener observes
at distance d2 = 3.75 m directly in front
of one speaker. Consider the full audible
range for normal human hearing, 20 Hz
to 20 kHz. Sound velocity is 343 m/s.
(a) What is the lowest frequency fmin,1 that gives minimum
signal (destructive interference) at the listener’s ear?
(b) What is the second lowest frequency fmin,2 that gives
minimum signal?
(c) What is the lowest frequency fmax,1 that gives maximum
signal (constructive interference) at the listener’s ear?
(d) What is the highest frequency fmax,n that gives maximum
signal?
Erwin Sitompul
University Physics: Wave and Electricity
5/2
Solution of Homework 4: Two Speakers
(d3 ) 2  (d1 ) 2  ( d 2 ) 2
d3
d3  (2)2  (3.75) 2  4.25 m
L  d3  d 2  4.25  3.75  0.5 m
vsound  f  , vsound  343 m s
Erwin Sitompul
University Physics: Wave and Electricity
5/3
Solution of Homework 4: Two Speakers
(a) What is the lowest frequency fmin,1 that gives minimum
signal (destructive interference) at the listener’s ear?
 Fully destructive
interference
L


L
vsound f min
 0.5,1.5, 2.5,
 0.5,1.5, 2.5,
vsound
 0.5,1.5, 2.5,
L
343

 0.5,1.5, 2.5,
0.5
 686 Hz  0.5,1.5, 2.5,
f min 
f min,1  686 Hz  0.5  343 Hz
(b) What is the second lowest frequency fmin,2 that gives
minimum signal?
f min  686 Hz  0.5,1.5, 2.5,
f min,2  686 Hz 1.5  1029 Hz
Erwin Sitompul
University Physics: Wave and Electricity
5/4
Solution of Homework 4: Two Speakers
(c) What is the lowest frequency fmax,1 that gives maximum
signal (constructive interference) at the listener’s ear?
 Fully constructive
interference
L


 0,1, 2,
L
 0,1, 2,
vsound f max
vsound
f max 
 0,1, 2,
L
343

 0,1, 2,
0.5
 686 Hz  0,1, 2,
f max,1  686 Hz 1  686 Hz
(d) What is the lowest frequency fmax,1 that gives maximum
signal (constructive interference) at the listener’s ear?
f max  686 Hz  0,1, 2,
f max,n  686 Hz  29  19894 Hz  Highest constructive
frequency that still can be
listened by human, < 20 kHz
Erwin Sitompul
University Physics: Wave and Electricity
5/5
Beats
 If two sounds whose
frequencies are nearly equal
reach our ears simultaneously, what we hear is a
sound whose frequency is
the average of the two
combining frequencies.
 We also hear a striking
variation in the intensity of
this sound –it increases and
decreases in slow, wavering
beats that repeat at a
frequency equal to the
difference between the two
combining frequencies.
Erwin Sitompul
University Physics: Wave and Electricity
5/6
Beats
 Let the time-dependent variations of the displacements due to
two sound waves of equal amplitude sm be
s1 ( x, t )  sm cos(k1 x  1t )
s2 ( x, t )  sm cos(k2 x  2t )
 From superposition principle, the resultant displacement is:
s( x, t )  sm cos(k1 x  1t )  sm cos(k2 x  2t )
k

 2 sm cos(kx  t )  cos(
x
t)
2
2
k
 

  2 sm cos(
x
t )   cos(kx  t )
2
2 

Amplitude
modulation,
depends on
Δk/2 and Δω/2
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Oscillating term,
a traveling wave,
depends on k and ω
k  k1  k2
k  12 (k1  k2 )  k
  1  2
  12 (1  2 )  
cos   cos   2cos 12 (   ) cos 12 (   )
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Beats
k
 

s ( x, t )   2 sm cos(
x
t )   cos(kx  t )
2
2 

ampl
f ampl
 1  2


2
2
f1  f 2
f


2
2
f beat  2 f ampl  f1  f 2
 In 1 amplitude cycle, we
will hear 2 beats
(maximum amplitude
magnitude)
 The "beat" wave
oscillates with the
frequency average, and
its amplitude varies
according to the
frequency difference
cos(
Erwin Sitompul
k

x
t)
2
2
University Physics: Wave and Electricity
5/8
Example: Beats
The A string of a violin is not properly tuned. Beats at 4 per
second are heard when the string is sounded together with a
tuning fork that is oscillating accurately at concert A (440 Hz).
(a) What are the possible frequencies produced by the string?
f beat  f1  f 2
f beat  fstring  f fork
4  fstring  440  fstring  436 Hz or 444 Hz
(b) If the string is stretched a little bit more, beats at 5 per
second are heard. Which of the possible frequencies are
the the frequency of the string?
 A string is stretched tighter  The frequency will be higher
 The frequency of beats increases  The frequency difference
increases
 If the string frequency becomes higher and its difference to
440 Hz increases  The frequency of the string is 444 Hz.
Erwin Sitompul
University Physics: Wave and Electricity
5/9
The Doppler Effect
 The Doppler Effect deals with the relation between motion
and frequency.
 The body of air is taken as the reference frame.
 We measure the speeds of a sound source S and a sound
detector D relatif to that body of air.
 We shall assume that S and D move either directly toward or
directly away from each other, at speeds less than the speed
of sound (vsound = 343 m/s).
Erwin Sitompul
University Physics: Wave and Electricity
5/10
The Doppler Effect: D Moving S Stationary
 If the detector moves toward the source, the number of
wavefronts received by the detector increased.
 The motion increases the detected frequency.
 If the detector moves away from the source, the number of
wavefronts received by the detector decreased.
 The motion decreases the detected frequency.
Erwin Sitompul
University Physics: Wave and Electricity
5/11
The Doppler Effect: S Moving D Stationary
 If the source moves toward the detector, the wavefronts is
compressed. The number of wavefronts received by the
detector increased.
 The motion increases the detected frequency.
 If the source moves away from the detector, the distance
between wavefronts increases. The number of wavefronts
received by the detector decreased
 The motion decreases the detected frequency.
Erwin Sitompul
University Physics: Wave and Electricity
5/12
The Doppler Effect
 The emitted frequency f and the detected frequency f’ are
related by:
v  vD
f f
v vS
where v is the speed of sound through the air, vD is the
detector’s speed relative to the air, and vS is the source’s
speed relative to the air.
+ The detector moves
v  vD
f f
v vS
toward the source
– The detector moves away
from the source
– The source moves
toward the detector
+ The source moves away
from the detector
Erwin Sitompul
University Physics: Wave and Electricity
5/13
Checkpoint
The figure indicates the directions of motion of a sound source
and a detector for six situations in stationary air. For each
situation, is the detected frequency greater than or less than
the emitted frequency, or can’t we tell without more information
about the actual speeds?
Source
(a)
Greater
(b)
Less
Detector
zero speed
zero speed
Source
(d)
Need more
information
(e)
Greater
(c)
(f)
Need more
information
Less
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Detector
University Physics: Wave and Electricity
5/14
Example: The Doppler Effect
A toy rocket flies with a velocity of 242 m/s toward a
mast while emitting a roaring sound with frequency
1250 Hz. The sound velocity is 343 m/s.
(a) What is the frequency heard by an observer who
is standing at the mast?
f  1250 Hz
vS  242 m s,
toward D
vD  0
v  vD
343  0
 4245 Hz
f f
 1250
v vS
343  242
(b) A fraction of the soundwaves is reflected by the mast and
propagates back to the rocket. What is the frequency
detected by a detector mounted on the head of the rocket?
f  4245 Hz
vS  0
vD  242 m s,
toward S
Erwin Sitompul
343  242
v  vD
 4245
 7240 Hz
f f
v vS
343 0
University Physics: Wave and Electricity
5/15
Supersonic Speeds
Mach Number 
vsource = vsound
(Mach 1 - sound barrier)
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v
vsound
vsource > vsound
(Mach 1.4 - supersonic)
University Physics: Wave and Electricity
5/16
Homework 5: Ambulance Siren
An ambulance with a siren emitting a whine at 1600 Hz
overtakes and passes a cyclist pedaling a bike at 8 m/s. After
being passed, the cyclist hears a frequency of 1590 Hz.
(a) How fast is the ambulance moving?
(b) What frequency did the cyclist hear before being overtaken
by the ambulance?
Illustration only
• Concorde, the supersonic
turbojet-powered supersonic
passenger airliner
• Average cruise speed
Mach 2.02 or about 2495 kmh
Erwin Sitompul
University Physics: Wave and Electricity
5/17
Homework 5A: Bat and Insect
1. An iron bar produces sound with a frequency of 335 Hz when struck.
When the iron bar is struck together with a steel bar, beats with a
frequency of 5.5 Hz will be heard. A piece of thread is then tied to the iron
bar and its frequency is lowered slightly. When struck at the same time
again, both the iron and steel bars now produce a beat with frequency of
8.2 Hz.
(a) What is the frequency of the iron bar after the thread is tied to it?
(b) What is the frequency of the steel bar?
2. (a) A stationary observer hears a frequency of
560 Hz from an approaching car. After the car
passes, the observed frequency is 460 Hz.
What is the speed of the car? (speed of sound
is air is 343 m/s.)
(b) A bat, moving at 5 m/s, is chasing a flying
insect. If the bat emits a 40 kHz chirp and
receives back an echo at 40.4 kHz, at what
speed is the insect moving toward or away
from the bat?
Erwin Sitompul
University Physics: Wave and Electricity
5/18