University Physics - Erwin Sitompul

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Lecture 5
Ch17. Longitudinal Waves
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
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  (d2 )2
d3
d3  (2) 2  (3.75) 2  4.25 m
L  d3  d2  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 
fmin,1  686 Hz  0.5  343 Hz
(b) What is the second lowest frequency fmin,2 that gives
minimum signal?
fmin  686 Hz  0.5,1.5, 2.5,
fmin,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,
fmax,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?
fmax  686 Hz  0,1, 2,
fmax,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
Intensity and Sound Level
 There is more to sound than frequency, wavelength, and
speed. We are well with something called intensity.
 The intensity I of a sound wave at a surface is the average
rate per unit area at which enery is transferred by the wave
through or onto the surface.
P
I
A
Erwin Sitompul
University Physics: Wave and Electricity
5/6
The Decibel Scale
 Human ear can bear the displacement amplitude that ranges
from about 10–5 m for the loudest torelable sound to about
10–11 m for the faintest detectable sound.
 The ratio between the highest and the lowest amplitude is
106.
 To deal with such an enormous range of values, people use
logarithmic scale instead of linear scale.
I
  (10 dB) log
I0
 β is called the sound level.
 dB is the abbreviation for decibel, the unit of sound level.
 I0 is a standard reference intensity 10–12 W/m2, chosen
because it is near the lower limit of human range of hearing.
Erwin Sitompul
University Physics: Wave and Electricity
5/7
Intensity and Sound Level
 β increases by 3 dB every time
the sound intensity is doubled
(increases by a factor of 2).
 β increases by 10 dB every time
the sound intensity increases by
an order of magnitude
(increases by a factor of 10).
Erwin Sitompul
University Physics: Wave and Electricity
5/8
Traveling Sound Waves
 Here we examine the displacements and pressure variations
associated with a sinusoidal sound wave traveling through air.
 The figure below displays such a wave traveling rightward
through a long air-filled tube.
 For a thin element of air
of thickness Δx, as the
wave travels through this
portion of the tube, the
element of air oscillates
left and right in a simple
harmonic motion about
its equilibrium position.
Erwin Sitompul
University Physics: Wave and Electricity
5/9
Traveling Sound Waves
 We choose to use a cosine
function to show the
displacements s(x,t):
s( x, t )  sm cos(kx  t )
Erwin Sitompul
University Physics: Wave and Electricity
5/10
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
5/11
Beats
 Let the time-dependent variations of the displacements due to
two sound waves of equal amplitude sm be
s1 ( x, t )  sm cos(k1x  1t )
s2 ( x, t )  sm cos(k2 x  2t )
 From superposition principle, the resultant displacement is:
s( x, t )  sm cos(k1x  1t )  sm cos(k2 x  2t )
k

 2sm cos(kx  t )  cos(
x
t)
2
2
k
 

  2sm cos( x 
t )   cos(kx  t )
2
2 

Amplitude
modulation,
depends on
Δk/2 and Δω/2
Erwin Sitompul
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 (   )
University Physics: Wave and Electricity
5/12
Beats
k
 

s( x, t )  2sm cos( x 
t )   cos(kx  t )
2
2 

 1  2

2
2
f f
f

 1 2
2
2
ampl 
f ampl
fbeat  2 fampl  f1  f 2
 In 1 amplitude cycle,
we will hear 2 beats
(maximum
amplitude
magnitude)
cos(
Erwin Sitompul
k

x
t)
2
2
University Physics: Wave and Electricity
5/13
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?
fbeat  f1  f2
f beat  fstring  ffork
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
5/14
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
5/15
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
5/16
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
5/17
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
5/18
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
343  0
v  vD
 1250
f f
 4245 Hz
343  242
v vS
(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
f f
 7240 Hz
343 0
v vS
University Physics: Wave and Electricity
5/19
Supersonic Speeds
vsource = vsound
(Mach 1 - sound barrier)
Erwin Sitompul
vsource > vsound
(Mach 1.4 - supersonic)
University Physics: Wave and Electricity
5/20
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
5/21
Homework 5: Ambulance Siren
New
(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?
(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
5/22

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