1. waves_04
SOUND WAVES
flute
clarinet
click for sounds

2. 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

3. 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
Infrasound: f < 20 Hz
Atoms/molecules are displaced in the direction of propagation about there equilibrium positions
For medical imaging f = 1-10 MHz
Why so large?

4. SOUND WAVES IN AIR 0 max 0 max 0 displacement
min max min pressure
Displacement of molecules
Pressure variation

5. 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.

6. What is the physics of this image?
CP 495

7. ULTRASOUND
Flow of blood through the placenta

8. Speed of sound wave in a fluid
The speed of a sound wave in a fluid depends
on the fluid’s compressibility and inertia.
B : bulk modulus of the fluid
r : equilibrium density of the fluid
Speed of sound wave in a solid rod
Y : Young’s modulus of the rod
r : density of the fluid
Speed of sound wave in air
v = 343 m.s-1 at T = 20oC
Above formulae not examinable

9. 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:
I0 = 1.0x10-12 W.m-2 : the reference intensity
the sound intensity at the threshold of hearing
Threshold of hearing
Threshold of pain
Not examinable

10. 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.
(b) If a second machine is added, then:
(c) Find the intensity corresponding to an
intensity level of 77.0 dB.
Not examinable

11. 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 W = I A = 4.0×10-6 J

12. (a) Find the intensity 3.00 m from the source.
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.
(b) At what distance would the intensity be one-fourth as much as it
is at r = 3.00 m?
(c) Find the distance at which the sound level is 40.0 dB?

13. Used to tune musical instruments to same pitch
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
CP 52

14. 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.

15. frequency of pulses is | f1-f2 |
BEATS
Superimpose oscillations of equal amplitude, but different frequencies
Modulation of amplitude
Oscillation at the average frequency
frequency of pulses is | f1-f2 |
CP 527

16. 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 =
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
CP 527

17. f1 = 100 Hz f2 = 104 Hz frapid = 102 Hz Trapid = 9.8 ms
click for sound
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

18. f1 = 100 Hz f2 = 110 Hz frapid = 105 Hz Trapid = 9.5 ms
click for sound
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

19. f1 = 100 Hz f2 = 120 Hz frapid = 110 Hz Trapid = 9.1 ms
24 25
click for sound
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

20. DOPPLER EFFECT - motion related frequency changes
Doppler 1842, Buys Ballot trumpeters on railway carriage
Source (s) Observer (o)
formula different to textbook
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

21. DOPPLER RADAR

22. What is frequency fo heard by observer? v =  f
DOPPLER EFFECT
Consider source of sound at frequency fs, moving speed vs, observer at rest (vo = 0)
Speed of sound v
What is frequency fo heard by observer?
v =  f
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

23. fs = 1000 Hz fo > 1000 Hz fo < 1000 Hz Source frequency
click for sounds
CP 495

24. source vs observer vo observed frequency fo
stationary
= fs
receding
< fs
approaching
> fs ?
CP 595

25. SHOCK Waves – supersonic waves
bullet travelling at Mach 2.45
CP 506

26. Shock Waves – supersonic waves
CP 506

27. DOPPLER EFFECT Stationary Sound Source Source moving with
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 supersonic)

28. fs = 650 Hz vs = 20 m.s-1 vo = 0 m.s-1 v = 340 m.s-1
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).

29. The speed of blood in the aorta is normally about 0.3000 m.s-1.
Problem 5
The speed of blood in the aorta is normally about m.s-1.
What beat frequency would you expect if 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.

30. Solution 5
Doppler Effect Beats

31. Blood is moving away from source  observer moving away from source  fo < fs
Wave reflected off red blood cells  source moving away from observer  fo < fs
Beat frequency = | 4.00 – | 106 Hz = 1558 Hz
In this type of calculation you must keep extra significant figures.

32. An ambulance travels down a highway at a speed of 33.5 m.s-1, its
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)

33. 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. 30

34. fs = 8.000×104 Hz fo = 8.002×104 Hz v = 1.5×103 m.s-1 vb = ? m.s-1
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.
Red blood cells (observer) moving toward source
Red blood cells (source) moving toward observer
fbeat = |f2-f1| = ( )×104 Hz = 20 Hz
Tbeat = 1/fbeat = 0.05 s
tbeat

35. 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

36. 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

37. Organ pipes
are open at
both ends

38. STANDING WAVES IN AIR COLUMNS (PIPES)
Sound wave in a pipe with one closed and one open end (stopped pipe)

39. STANDING WAVES IN AIR COLUMNS (PIPES)
CP 516

40. Search google or YouTube for
Rubens or Rubins tube

41. STANDING WAVES IN AIR COLUMNS (PIPES)
CP 516

42. STANDING WAVES IN AIR COLUMNS (PIPES)
Normal modes in a pipe with an open and a closed end (stopped pipe)
Fundamental
3rd harmonic or 2nd overtone

43. 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.
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?

48. f = 170 Hz
Solution 8
pressure node
Pressure distribution in tube for fundamental mode
L = 0.50 m
v = ? m.s-1
pressure antinode
speed of wave N A

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 = 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.
A clarinet is open at one end and closed at the other end by the player’s lips and reed.
open
open
open
closed

52. A A A
Solution 10
FLUTE is open at both ends  particle displacement antinodes at the open ends
Particle displacement variations
Fundamental (1st harmonic)
2nd harmonic (1st overone)
All harmonics can be excited

53. Solution 10 A N N
CLARINET is open at one and closed at the other  pressure node at open end & antinode at closed end
pressure variations A N
Fundamental (1st harmonic)
3nd harmonic (1st overone)
All odd harmonics can be excited

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.
Find the frequency of the
tuning fork.
Lsmalles t= 9.00 cm
(b) Find the wavelength and the next two water levels giving resonance.

55. = 4 L = 0.360 m v = f  f1 = v /  = (343)/(0.36) Hz = 958 Hz
Solution 11 (alternative)
Tube closed at one end and open at the other  odd harmonics can be excited
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 N N N A N
L1= /4
L3= 3/4
1st resonance
L1= 90.0 mm = 90.0×10-3 m v = 343 m.s-1 f1 = ? Hz
= 4 L = m v = f  f1 = v /  = (343)/(0.36) Hz = 958 Hz
2nd and 3rd resonances
L3 = 3/4 = (3)(0.36)/(4) m = m
L5 = 5/4 = (5)(0.36)/(4) m = m
L5= 5/4

56. Chimney acts like an organ pipe open at both ends
Problem 12
Why does a chimney moan ?
Chimney acts like an organ pipe open at both ends
Pressure node
Speed of sound in air v = 340 m.s-1
Length of chimney L = 3.00 m
L =  / 2  = 2 L v = f 
f = v /  = 340 / {(2)(3)} Hz
f = 56 Hz low moan N A
Pressure node N
Fundamental mode of vibration

57. Natural frequencies of a trumpet closed at mouth and open at the flared end
Fundamental f1 = 60 Hz
1st overtone Hz
2nd overtone Hz
NOT a harmonic sequence for the natural frequencies of vibration

58. Simulation of the human voice tract – natural frequencies