waves_04

Report
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waves_04
SOUND WAVES
flute
clarinet
click for sounds
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waves_04: MINDMAP SUMMARY - SOUND WAVES
Sound waves, ultrasound, compressional (longitudinal) waves, pressure,
particle displacement, medical imaging, superposition principle, stadning
waves in air columns, constructive interference, destructive interference,
boundary conditions, pipe – open and closed ends, nodes, antinodes, speed
of sound in air, period, frequency, wavelength, propagation constant (wave
number), angular frequency, normal modes of vibrations, natural
frequencies of vibration, fundamental, harmonics, overtones, harmonic
series, frequency spectrum, radian, phase, sinusoidal functions, wind
musical instruments, beats, beat frequency, Doppler Effect, Doppler radar,
shock waves
v f 

T

fbeat  f1  f2

k
2
  2 f 
T
v  vo
f o  fs
v  vs
1
T
f
k
2

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SOUND WAVES IN AIR
• Longitudinal wave through any medium which can be
compressed: gas, liquid, solid
• Frequency range 20 Hz - 20 kHz (human hearing range)
• Ultrasound: f > 20 kHz
For medical imaging f = 1-10 MHz
Why so large?
• Infrasound: f < 20 Hz
• Atoms/molecules are displaced in the direction of
propagation about there equilibrium positions
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SOUND WAVES IN AIR
0
max
0
max
0
displacement
min
0
max
0
min
pressure
Displacement of molecules
Pressure variation
ULTRASOUND
Ultrasonic sound waves have frequencies greater than 20 kHz and,
as the speed of sound is constant for given temperature and medium,
they have shorter wavelength. Shorter wavelengths allow them to
image smaller objects and ultrasonic waves are, therefore, used as a diagnostic tool and
in certain treatments.
Internal organs can be examined via the images produced by the reflection and
absorption of ultrasonic waves. Use of ultrasonic waves is safer than x-rays but
images show less details. Certain organs such as the liver and the spleen are invisible to
x-rays but visible to ultrasonic waves.
Physicians commonly use ultrasonic waves to
observe fetuses. This technique presents far less
risk than do x-rays, which deposit more energy in
cells and can produce birth defects.
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What is the physics of this image?
CP 495
ULTRASOUND
Flow of blood through the placenta
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 Speed
of sound wave in a fluid
The speed of a sound wave in a fluid depends
on the fluid’s compressibility and inertia.
v
B

B : bulk modulus of the fluid
 : equilibrium density of the fluid
B
 Speed
v
Y

 Speed
v  331
P
elasticproperty
v
V / V
inertialproperty
of sound wave in a solid rod
Y : Young’s modulus of the rod
 : density of the fluid
of sound wave in air
T
273
v = 343 m.s-1 at T = 20oC
Above formulae not examinable
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Energy and Intensity of Sound waves
 Intensity level in decibel
• The loudest tolerable sounds have intensities about 1.0x1012 times
greater than the faintest detectable sounds.
• The sensation of loudness is approximately logarithmic in the human
ear. Because of that the relative intensity of a sound is called the
intensity level or decibel level, defined by:
 I 
  10 log  
 I0 
I0 = 1.0x10-12 W.m-2 : the reference intensity
the sound intensity at the threshold of hearing
 1.0 1012
  10log
12
 1.0 10
 1.0 1011
  10 log
12
 1.0 10
W/m2 
  10log(1)  0 dB
2 
W/m 
Threshold of hearing
W/m2 
  10 log(10)  10 dB
2 
W/m 
 1.0 1010 W/m2 
  10 log(100)  20 dB
  10 log
12
2 
 1.0 10 W/m 
 1.0 W/m2


  10log(1012 )  120dB
  10log
12
2 
 1.0 10 W/m 
Threshold of pain
Not examinable
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Problem 1
A noisy grinding machine in a factory
produces a sound intensity of 1.00x10-5 W.m-2.
(a) Calculate the intensity level of the single grinder.
 1.00105 W/m2 
7

  10log

10
log(
10
)  70.0 dB
12
2 
1
.
00

10
W/m


(b) If a second machine is added, then:
 2.00105 W/m2 
  10log(2 107 )  73.0 dB
  10log
12
2 
 1.0010 W/m 
(c) Find the intensity corresponding to an
intensity level of 77.0 dB.
I 
I
7.70
  10 
1.00 1012
 I0 
  77.0 dB  10log 
I  5.01105 W.m-2
Not examinable
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Problem 2
A point source of sound waves emits a disturbance with a
power of 50 W into a surrounding homogeneous medium.
Determine the intensity of the radiation at a distance of 10 m
from the source. How much energy arrives on a little detector
with an area of 1.0 cm2 held perpendicular to the flow each
second? Assume no losses.
Solution
P = 50 W r = 10 m I = ? W.m-2 A = 1.0 cm2 = 1.0×10-4 m2 W = ? J
I = P / (4 r2) = 4.0×10-2 W.m-2
W = I A = 4.0×10-6 J
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Problem 3
A small source emits sound waves with a power output of 80.0 W.
(a) Find the intensity 3.00 m from the source.
I
Pav
-2

0.707
W.m
4 r 2
(b) At what distance would the intensity be one-fourth as much as it
is at r = 3.00 m?
r
Pav
80.0

m  6.00 m
4 I
4 (0.707) / 4.0
(c) Find the distance at which the sound level is 40.0 dB?
I 
 I 
40.0  10log    4.00  log    I  104.00 I 0  1.00 108 W.m-2
 I0 
 I0 
I1 r22
I
I
 2  r22  r12 1  r2  r1 1  2.52104 m
I 2 r1
I2
I2
BEATS
• Loud-soft-loud modulations
of intensity are produced
when waves of slightly
different frequencies are
superimposed.
• The beat frequency is equal
to the difference frequency
fbeat = | f1 - f2|
1 beat
Used to tune musical instruments to same pitch
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CP
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BEATS two interfering sound waves can make beat
Two waves with different frequency
create a beat because of interference
between them. The beat frequency is
the difference of the two frequencies.
fbeat  f2  f1
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BEATS
Superimpose oscillations of equal amplitude, but different
frequencies
A sin(2 f1t )  A sin(2 f 2t )
( f1  f 2 )
( f1  f 2 )
 2 A sin(2
t ) cos(2
t)
2
2
( f1  f 2 ) 
( f1  f 2 )

  2 A cos(2
t )  sin(2
t)
2
2


Modulation of amplitude
Oscillation at the average
frequency
frequency of pulses is | f1-f2 |
CP 527
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BEATS – interference in time
Consider two sound sources producing audible sinusoidal waves at slightly different
frequencies f1 and f2. What will a person hear? How can a piano tuner use beats in
tuning a piano? If the two waves at first are in phase they will interfere constructively
and a large amplitude resultant wave occurs which will give a loud sound. As time
passes, the two waves become progressively out of phase until they interfere
destructively and it will be very quite. The waves then gradually become in phase again
and the pattern repeats itself. The resultant waveform shows rapid fluctuations but with
an envelope that various slowly.
The frequency of the rapid fluctuations is the average frequencies =
The frequency of the slowly varying envelope =
f1  f 2
2
f1  f 2
2
Since the envelope has two extreme values in a cycle, we hear a loud sound twice
in one cycle since the ear is sensitive to the square of the wave amplitude.
The beat frequency is
fbeat  f1  f2
CP 527
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click for sound
60
50
40
f =100
f = 104
beats
30
20
10
0
0
0.05
0.1
0.15
0.2
0.25
time
f1 = 100 Hz f2 = 104 Hz frapid = 102 Hz Trapid = 9.8 ms
fbeat = 4 Hz Tbeat = 0.25 s (loud pulsation every 0.25 s)
CP 527
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60
click for sound
50
40
30
20
10
0
0
0.05
0.1
0.15
0.2
0.25
time
f1 = 100 Hz f2 = 110 Hz frapid = 105 Hz Trapid = 9.5 ms
fbeat = 10 Hz Tbeat = 0.1 s (loud pulsation every 0.1 s)
CP 527
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24 25
click for sound
60
50
40
f =100
f = 120
beats
30
20
10
0
0
0.05
0.1
0.15
0.2
0.25
time
f1 = 100 Hz f2 = 120 Hz frapid = 110 Hz Trapid = 9.1 ms
fbeat = 20 Hz Tbeat = 0.05 s (loud pulsation every 0.05 s)
CP 527
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DOPPLER EFFECT - motion related frequency changes
Doppler 1842, Buys Ballot 1845 - trumpeters on railway
carriage
v  vo
f o  fs
v  vs
formula different to textbook
Source (s)
Observer (o)
Applications: police microwave speed units, speed of a tennis ball,
speed of blood flowing through an artery, heart beat of a developing
fetous, burglar alarms, sonar – ships & submarines to detect
submerged objects, detecting distance planets, observing the motion
of oscillating stars, weather (Doppler radar), plaet detection
note: formula is very different to textbook
CP 495
DOPPLER RADAR
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DOPPLER EFFECT
• Consider source of sound at frequency fs, moving
v=f
speed vs, observer at rest (vo = 0)
v  vo
• Speed of sound v
f o  fs
• What is frequency fo heard by observer?
v  vs
On right - source approaching
• source catching up on waves
• wavelength reduced
• frequency increased
On left - source receding
• source moving away from waves
• wavelength increased
• frequency reduced
CP 495
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v  vo
f o  fs
v  vs
Source frequency
fs = 1000 Hz
v (source) = 0
v (source) = v (wave)
fo > 1000 Hz
fo < 1000 Hz
click for sounds
v (source) = v (wave) / 2
v (source) = 1.25 v (wave)
CP 495
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source vs
v  vo
f o  fs
v  vs
observer vo
observed
frequency
fo
= fs
< fs
stationary
stationary
stationary
receding
stationary
receding
approaching
receding
approaching
stationary
stationary
receding
> fs
< fs
> fs
< fs
approaching
approaching
approaching
receding
> fs
?
receding
approaching
?
CP 595
SHOCK Waves –
supersonic waves
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bullet travelling at Mach 2.45
fast moving power boat
boat moving through water: speed of boat > speed of
water wave created
sailing boat rocket “violently” wave crests add to give “large wave”
shock wave - bunching of wavefronts ---> abruptive rise and fall of air pressure
plane travelling at supersonic speeds
Mach cone
Mach Number = v / vs vs speed of sound in air
eg Mach Number = 2.3 speed is 2.3 times the speed of sound
sonic boooom
CP 506
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Shock Waves – supersonic waves
CP 506
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DOPPLER EFFECT
Stationary Sound Source
Source moving with
vsource < vsound ( Mach 0.7 )
Source moving with
vsource = vsound ( Mach 1 breaking the sound barrier )
Source moving with
vsource > vsound (Mach 1.4 supersonic)
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Problem 4
A train whistle is blown by the driver who hears the sound at
650 Hz. If the train is heading towards a station at 20.0 m.s-1,
what will the whistle sound like to a waiting commuter? Take
the speed of sound to be 340 m.s-1.
Solution
fs = 650 Hz vs = 20 m.s-1 vo = 0 m.s-1 v = 340 m.s-1
fo = ? Hz (must be higher since train approaching observer).
v  vo
v
 340 
f o  fs
 fs
  650 
 Hz  691 Hz
v  vs
v  vs
 340  20 
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Problem 5
The speed of blood in the aorta is normally about 0.3000 m.s-1.
What beat frequency would you expect if 4.000 MHz ultrasound
waves were directed along the blood flow and reflected from the
end of red blood cells?
Assume that the sound waves travel through the blood with a
velocity of 1540 m.s-1.
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Solution 5
fs1 = 4.0 MHz = 4.0x106 Hz
v01= 0.30 m.s-1
v = 1.54x103 m.s-1
fo1 = ? Hz
fo2 = ? Hz
fs2 = fo1
vs2 = 0.30 m.s-1
reflected wave
Doppler Effect
v  vo
f o  fs
v  vs
Beats
fbeat  f2  f1
Blood is moving away from source  observer moving away from source  fo < fs
v  vo1
f o1  fs1
 4 106
v  vs1


 1.54 103  0.30 
6

3.999221

10
Hz


3

 1.54 10

Wave reflected off red blood cells  source moving away from observer  fo < fs
f o2
v  vo2
 fs2
 3.999221106
v  vs2


 1.54 103

6

3.998442

10
Hz


3
 1.54 10  0.30 
Beat frequency = | 4.00 – 3.998442| 106 Hz = 1558 Hz
In this type of calculation you must keep extra significant figures.
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Problem 6
An ambulance travels down a highway at a speed of 33.5 m.s-1, its
siren emitting sound at a frequency of 4.00x102 Hz. What frequency
is heard by a passenger in a car traveling at 24.6 m.s-1 in the opposite
direction as the car and ambulance: (a) approach each other and
(b) pass and move away from each others?
Speed of sound in air is 345 m.s-1.
Solution 6
(a)
(b)
 v  vO 
2  345  24.6 
fO  f S 

(4.00

10
)

 Hz  475 Hz
 345  33.5 
 v  vS 
 v  vO 
 345  24.6) 
2
fO  f S 

(4.00

10
)
Hz


  339 Hz
 345  33.5 
 v  vS 
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Problem 7
An ultrasonic wave at 8.000104 Hz is emitted into a vein where
the speed of sound is about 1.5 km.s-1. The wave reflects off the red
blood cells moving towards the stationary receiver. If the frequency
of the returning signal is 8.002104 Hz, what is the speed of the
blood flow?
What would be the beat frequency detected and the beat period?
Draw a diagram showing the beat pattern and indicate the beat
period.
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Solution
fs = 8.000×104 Hz fo = 8.002×104 Hz v = 1.5×103 m.s-1
vb = ? m.s-1
Need to consider two Doppler shifts in frequency – blood cells act as observer
and than as source.
v  vo
v  vb
'
f

f

f
Red blood cells (observer)
o
s
s
v  vs
v
moving toward source
 v  vb 
 v  vb   v 
f o   fs

  fs 

Red blood cells (source)

v
v

v
v

v


b
b

moving toward observer
 fo  1
 8.002  1 
 fs 
3 
8  m.s-1
vb  v 

1.5

10

fo
 8.002  1 
 f  1
8 

s




60
vb  0.19 m.s-1
50
40
fbeat = |f2-f1| =
(8.002-8.000)×104
Hz = 20 Hz
f =100
f = 11
beats
30
Tbeat = 1/fbeat = 0.05 s
20
10
tbeat
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STANDING WAVES IN AIR COLUMNS (PIPES)
• If we try to produce a traveling harmonic wave in a pipe,
repeated reflections from an end produces a wave traveling
in the opposite direction - with subsequent reflections we
have waves travelling in both directions
• The result is the superposition (sum) of two waves
traveling in opposite directions
• The superposition of two waves of the same amplitude
travelling in opposite directions is called a standing wave
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STANDING WAVES IN AIR COLUMNS (PIPES)
Flute & clarinet same length, why can a much lower note be played on a
clarinet?
L
Closed at both ends
Closed at one end
open at the other
Open at both ends
Closed end: displacement zero (node), pressure max (antinode)
Open end: displacement max (antinode), pressure zero (node)
CP 516
Organ pipes
are open at
both ends
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STANDING WAVES IN AIR COLUMNS (PIPES)
Sound wave in a pipe with one closed and one open end (stopped pipe)
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STANDING WAVES IN AIR COLUMNS (PIPES)
equilibrium position of particles
14
13
instantaneous position of
particles
12
11
10
sine curve showing
instantaneous displacement of
particles from equilibrium
9
8
7
6
instantaneous pressure
distribution
5
4
3
2
time averaged pressure
fluctuations
1
0
Enter t/T
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
CP 516
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Search google or YouTube for
Rubens or Rubins tube
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STANDING WAVES IN AIR COLUMNS (PIPES)
L
N
L
2
N  1, 2,3,
L
Nv
fN 
2L


L  (2 N  1)  
4
(2 N  1)v
fN 
4L
N
2
Nv
fN 
2L
L
CP 516
42
STANDING WAVES IN AIR COLUMNS (PIPES)
Normal modes in a pipe with an open and a closed end (stopped pipe)
Fundamental
L

4
  4L
n
f1 
3rd harmonic or 2nd overtone
v
4L
4L
L  n or n 
( n  1,3,5,...)
4
n
L3

4

4L
3
f3  3
v
 3 f1
4L
v
fn  n
( n  1,3,5,...)
4L
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Pipe closed at one end and open at
the other
closed end
particle displacement zero
 node
open end
max particle displacement
 antinode
120
100
80
60
40
20
0
0
0.1
0.2
0.3
0.4
0.5
0.6
position along column
0.7
0.8
0.9
1
CP 523
44
Musical instruments – wind
An air stream produced by mouth by blowing the instruments
interacts with the air in the pipe to maintain a steady oscillation.
All brass instruments are closed at one end by the mouth of the
player.
Flute and piccolo – open at atmosphere and mouth piece
(embouchure) – covering holes L      f 
Trumpet – open at atmosphere and closed at mouth – covering holes
adds loops of tubing into air stream L      f 
Woodwinds – vibrating reed used to produce oscillation of the air
molecules in the pipe.
CP 516
45
Woodwind instruments are not necessarily made of wood eg saxophone, but
they do require wind to make a sound. They basically consist of a tube with a
series of holes. Air is blow into the top of the tube, either across a hole or past a
flexible reed. This makes the air inside the tube vibrate and give out a note. The
pitch of the note depends upon the length of the tube. A shorter tube produces a
higher note, and so holes are covered. Blowing harder makes a louder sound. To
produce deep notes woodwind instruments have to be quite long and therefore the
tube is curved.
Brass instruments (usually made of brass) consist of a long pipe that is usually
coiled and has no holes. The player blows into a mouthpiece at one end of the
pipe, the vibration of the lips setting the air column vibrating throughout the pipe.
The trombone has a section of pipe called a slide that can be moved in and out.
To produce a lower note the slide is moved out. The trumpet has three pistons that
are pushed down to open extra sections of tubing. Up to six different notes are
obtained by using combinations of the three pistons.
CP 516
46
Natural frequencies of vibration (open – closed air column)
Speed of sound in air (at room temperature v ~ 344 m.s-1) v = f 
Boundary conditions
Reflection of sound wave at ends of air column: Open end – a
compression is reflected as a rarefaction and a rarefaction as a
compression ( phase shift). Zero phase change at closed end.
2 N 1 
4L
2N 1
f 2 N 1 
v
2 N 1


L   2 N  1  2 N 1 
 4 
N  1, 2,3,...
 2N  1 

 v   2 N  1 f1
 4L 
odd harmonics exit: f1, f3, f5, f7 , …
CP 516
47
Problem 8
A narrow glass tube 0.50 m long and sealed at its bottom end
is held vertically just below a loudspeaker that is connected to
an audio oscillator and amplifier. A tone with a gradually
increasing frequency is fed into the tube, and a loud resonance
is first observed at 170 Hz. What is the speed of sound in the
room?
f = 170 Hz
Solution 8
pressure node
Pressure distribution in
tube for fundamental mode
L

v = ? m.s-1
L = 0.50 m
  4L
4
pressure antinode
speed of wave
N
v  f   170 4 0.5 m.s-1  340 m.s-1
A
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49
Problem 9
What are the natural frequencies of vibration for a human ear?
Why do sounds ~ (3000 – 4000) Hz appear loudest?
Solution
Assume the ear acts as pipe open at the atmosphere and closed at the eardrum.
The length of the auditory canal is about 25 mm.
Take the speed of sound in air as 340 m.s-1.
L = 25 mm = 0.025 m
v = 340 m.s-1
For air column closed at one end and open at the other
L = 1 / 4  1 = 4 L  f1 = v / 1 = (340)/{(4)(0.025)} = 3400 Hz
When the ear is excited at a natural frequency of vibration  large amplitude
oscillations (resonance)  sounds will appear loudest ~ (3000 – 4000) Hz.
50
RESONANCE
• When we apply a periodically varying force to a system that can
oscillate, the system is forced to oscillate with a frequency equal
to the frequency of the applied force (driving frequency): forced
oscillation. When the applied frequency is close to a characteristic
frequency of the system, a phenomenon called resonance occurs.
• Resonance also occurs when a
periodically varying force is applied
to a system with normal modes.
When the frequency of the applied
force is close to one of normal
modes of the system, resonance
occurs.
51
Problem 10
Why does a clarinet play a lower note than a flute
when both instruments are about the same length ?
A flute is an open-open tube.
open
open
A clarinet is open at one end and closed at the other end by the player’s lips and reed.
closed
open
A
A
Solution 10
A
FLUTE is open at both
ends  particle
displacement antinodes
at the open ends


2
Fundamental (1st harmonic)
L

2
  2L
v f
Particle
displacement
variations
f1 
v


v
2L
2nd harmonic (1st overone)
L
v f
f2 
v


All harmonics can be excited
v
 2 f1
L
Lm

2

2L
m
f m  m f1
m  1, 2,3,
52
Solution 10
N
A
N
CLARINET is open at one
and closed at the other 
pressure node at open end &
antinode at closed end

4
A
N
Fundamental (1st harmonic)
L

4
  4L
v f
f1 
v



pressure
variations
v
4L
3nd harmonic (1st overone)
L
3

4

4L
v f
3
f3 
3v
 3 f1
4L
All odd harmonics can be excited

4L
L   2m  1

4
 2m 1
f 2m1   2m  1 f1
m  1, 2,3,
53
54
Problem 11
Resonance
The sound waves generated by the
fork are reinforced when the length
of the air column corresponds to one
of the resonant frequencies of the
tube. Suppose the smallest value of
L for which a peak occurs in the
sound intensity is 90.0 mm.
(a) Find the frequency of the
tuning fork.
n 1
f1 
v  345 m.s-1
Lsmalles t= 9.00 cm
L1  90.0 103 m
v
345

Hz  958 Hz
2
4 L1 4(9.00 10 )
(b) Find the wavelength and the next two water levels giving resonance.
  4L1  4(9.00 102 ) m  0.360 m
L2  L1   / 2  0.270m, L3  L2   / 2  0.450m.
Solution 11 (alternative)
Tube closed at one end and open at the other  odd harmonics can be excited
L   2m  1

4

4L
 2m 1
f 2m1   2m  1 f1
m  1, 2,3,
The frequency of the tuning fork is fixed and hence the wavelength also has a
fixed value. The length of the tube is variable.
A
pressure variation
A
N
N
L1= /4
N
N
L3= 3/4
1st resonance
L1= 90.0 mm = 90.0×10-3 m v = 343 m.s-1 f1 = ? Hz
 = 4 L = 0.360 m v = f  f1 = v /  = (343)/(0.36) Hz = 958 Hz
2nd and 3rd resonances
L3 = 3/4 = (3)(0.36)/(4) m = 0.270 m
L5 = 5/4 = (5)(0.36)/(4) m = 0.450 m
L5= 5/4
55
56
Why does a chimney moan ?
Problem 12
Chimney acts like an organ pipe open at both ends
Pressure node
N
Speed of sound in air v = 340 m.s-1
Length of chimney L = 3.00 m
A
L=/2 =2L v=f
f = v /  = 340 / {(2)(3)} Hz
Pressure node
N
Fundamental mode of vibration
f = 56 Hz
low moan
-0.8
0
0.5
1
1.5
2
1.5
2
1.5
2
Natural frequencies of a-0.6trumpet closed at mouth
and open at the flared
Shape of pipe
0.6
-0.4
end
0.4
-0.2
0.2
0
0
0.2
-0.2
0.4
-0.4
0.6
-0.6
Fundamental f1 = 60 Hz
-0.8
Shape of pipe
0
0.5
1
end
end pressure
pressure at
at (+)
(+) == -0.997
0.260
0.253
0.195
0.130
0.001
1st
overtone
163 Hz
pressure P (a.u)
1
0.5
0
-0.5
-1
2nd overtone 247 Hz
NOT a harmonic
sequence for the natural
frequencies of vibration
0
0.5
1
position x (m)
Simulation of the human voice tract – natural frequencies
58

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