Chapter 2

Sound Waves

Dr Mohamed Saudy

1

Introduction to Sound Waves

Sound waves are the most important

example of longitudinal waves.

They can travel through any material, except

vacuum (no one can hear you scream in

outer space).

Speed of sound depends on material (and

temperature)

2

Sound

- is a wave (sound wave)

- Rarefied and compressed regions

- Longitudinal wave

- air molecules move back and forth

3

Categories of Sound Waves

The categories cover different frequency ranges

Audible waves are within the sensitivity of the human ear Range is approximately 20 Hz to 20 kHz

Infrasonic waves have frequencies below the audible range.<20 Hz

Ex: African elephant, volcanoes, earthquakes

Ultrasonic waves have frequencies above the audible range. (20-100) KHz

Ex: bats, dolphins

4

How we hear?

When you speak or shout, your vocal chords vibrate .

These vibrations travel in all directions through the

air as waves. When the waves reach our ears, they

make our eardrums vibrate too, so we can hear the

words.

5

• Outer ear collects sound

• Middle ear amplifies sound

• Inner ear converts sound

Sound waves travel on a MEDIUM:

Any SOLID, LIQUID OR GAS

Sound travels by pushing the particles of a

substance. The particles push into the particles

next to them, and then return to their original

position. And the sound continues to travel in

this form until it reaches your ear!

6

Speed of Sound Waves,

General

The speed of sound waves in a medium

depends on the compressibility and the

density of the medium

The compressibility can sometimes be

expressed in terms of the elastic modulus of

the material

The speed of all mechanical waves follows a

general form:

elastic property

inertial propertyv

7

Which state of substance would

sound travel through faster?

Why? 8

Mediums: Sound travels through a solid faster, than

through a liquid, which is faster, than through

a gas.

Our ears are custom to hear sound through a

gas…

Solid : Fast speed

Liquid : Medium speed

Gas : Slow Speed

Vacuum : No Sound

9

Speed of Sound in Liquid or

Gas

The bulk modulus of the material is B

The density of the material is r

The speed of sound in that medium is

Bv

r

10

Speed of Sound in a Solid Rod

The Young’s modulus of the material is Y

The density of the material is r

The speed of sound in the rod is

Yv

r

11

Speed of Sound Waves

r

Bv

Y… Young’s modulus (

B… Bulk modulus of medium

r…density of material

Bulk modules

determines the volume

change of an object due

to an applied pressure P.

ii VV

P

VV

AFB

//

/

strain volume

stress volume

r

Yv

In gas and liquids: In solids:

iLL

AFY

/

/

strain tensile

stress tensile

Young’s modules

determines the length

change of an object

due to an applied force

F.

12

Speed of Sound in Gases,

Example Values

Note temperatures, speeds are in m/s 13

Speed of Sound in Liquids,

Example Values

Speeds are in m/s

14

Speed of Sound in Solids,

Example Values

Speeds are in m/s; values are for bulk solids

15

Speed of Sound in Air

The speed of sound also depends on the temperature of the medium

This is particularly important with gases

For air, the relationship between the speed and temperature is

The 331 m/s is the speed at 0o C

TK is the air temperature in Kelvin

K(331 m/s)273

Tv

16

Example 1: Find the speed of sound in a steel rod?

r = 7800 kg/m3 , Y = 2.07 x 1011 Pa

11

3

2.07 105150 /

7800 /

Y Pav m S

Kg mr

Example 2: What is the speed of sound in air when

the temperature is 270C?

(27 273)331 347 /

273v m S

17

Sound Level

The range of intensities detectible by the

human ear is very large

It is convenient to use a logarithmic scale to

determine the intensity level, b

10log

o

I

Ib

18

Intensity, cont

In the case of our example wave in air,

I = ½ rv(wsmax)2

Therefore, the intensity of a periodic sound

wave is proportional to the

Square of the displacement amplitude

Square of the angular frequency

In terms of the pressure amplitude,

2

max

2

PI

vr

19

Intensity of a Point Source

A point source will emit sound waves

equally in all directions

This results in a spherical wave

Identify an imaginary sphere of radius r

centered on the source

The power will be distributed equally through

the area of the sphere

20

Intensity of a Point Source,

cont

This is an inverse-

square law

The intensity decreases

in proportion to the

square of the distance

from the source

24av avIA r

21

Sound Level, cont

I0 is called the reference intensity

It is taken to be the threshold of hearing

I0 = 1.00 x 10-12 W/ m2

I is the intensity of the sound whose level is to be determined

b is in decibels (dB)

Threshold of pain (ألم): I = 1 W/m2; b = 120 dB

Threshold of hearing: I0 = 10-12 W/ m2 corresponds to b = 0 dB

22

10logo

I

Ib

23

Sound Intensity

Sound is characterized in decibels (dB), according

to:

sound level b= Decibel level=10log(I/I0) =

20log(P/P0) dB

I0 = 1012 W/m2 is the threshold power

intensity(0 dB)

P0 = 2105 N/m2 is the threshold pressure (0

dB)

atmospheric pressure is about 105 N/m2

Example 3:

60 dB means log(I/I0) = 6, so I = 106 W/m2

and log(P/P0) = 3, so P = 2102 N/m2

0 0

6 6 12 6 2

0

6060 10log log 6

10

10 10 10 10 /

I IdB

I I

II W m

I

0 0

3 3 5 2 2

0

6060 20log log 3

20

10 10 2 10 2 10 /

P PdB

P P

PP N m

P

24

• 120 dB (pain threshold) means log (I/I0) = 12,

so I = 1 W/m2

and log(P/P0) = 6, so P = 20 N/m2

• 10 dB (barely detectable) means log(I/I0) = 1, so I =

1011 W/m2

and log(P/P0) = 0.5, so P 6105 N/m2

• What is the sound level that corresponds to an intensity of

2 × 10-7 W/m2 ?

b = 10 log (2 × 10-7 W/m2 / 10-12 W/m2) = 10 log 2 ×105 = 53 dB 25

Example 4: Sound system A produces an intensity

level of 107dB. Sound system B produces an

intensity level of 110dB. Compute the ratio of

intensity for the two sound systems.

1 22 1

0 0

2

1

0.32 2

1 1

10log( ) 10log( )

3 10log( )

3log( ) 10 2

10

I I

I I

I

I

I I

I I

b b

26

Example 5: The maximum amplitude pressure P of a

sound wave that is tolerable to a human ear is about 28

Pa. (a) What fraction is P of normal atmospheric

pressure? (b) What intensity of sound does P

correspond to in air at room temperature?

r=1.2 Kg/m3, v=344m/s

4

5

22

3

282.77 10

1.013 10

( ) (28 )20.95 /

2 2(1.2 / )(344 / )

P Pa

P Pa

P PaI W m

v Kg m m sr

27

Sound Levels

28

Loudness and Intensity

Sound level in decibels relates to a physical measurement of the strength of a sound

We can also describe a psychological “measurement” of the strength of a sound

Our bodies “calibrate” a sound by comparing it to a reference sound

This would be the threshold of hearing

Actually, the threshold of hearing is this value for 1000 Hz

29

Loudness and Frequency, cont

There is a complex relationship between loudness and frequency

The white area shows average human response to sound

The lower curve of the white area shows the threshold of hearing

The upper curve shows the threshold of pain

30

The Doppler Effect

The Doppler effect is the apparent change in frequency (or wavelength) that occurs because of motion of the source or observer of a wave

When the relative speed of the source and observer is higher than the speed of the wave, the frequency appears to increase

When the relative speed of the source and observer is lower than the speed of the wave, the frequency appears to decrease

31

The Doppler Effect

Doppler Effect – 4 cases

Source moving toward receiver

Source moving away from receiver

Receiver (observer) moving towards source

Receiver (observer) moving away from source. 32

When the observer and source are

moving:

f=the frequency of the source

f’=the frequency of the apparent listener (observer).

C=the speed of sound

V0=the velocity of the observer (+ observer moving toward

the source, - observer moving away from a source)

Vs=the velocity of the source (- source moving toward the

observer, + source moving away from the observer) 33

• The word toward is associated with an increase in the

observer frequency.

• The word away from is associated with decrease in

the observer frequency.

• Strationary source means Vs=0, and stationary

observer means V0=0.

• The observed frequency of wave is increased when

the source and observer are approaching each other

and is decreased when they are receding from each

other. 34

s

c Cf f

C V

s

c Cf f

C V

1. Source moving case, V0=0 (stationary observer)

When the source is moving toward from the listener

When the source is moving away from the listener

35

2. Observer moving case, Vs=0 (stationary source)

0c C Vf f

C

0c C Vf f

C

•When the observer is moving away the source

• When the observer is moving toward the source

36

Example 6: A train moving at a speed of 40 m/s

sounds it whistle, which has a frequency of 500 Hz.

Determine the frequencies heard by a stationary

observer as the train approaches and then receds

from the observer.

566 ,s

Cf f Hz f f

C V

448 ,s

Cf f Hz f f

C V

Solution: Vs=40 m/s, f=500Hz, Vo=0, C=343 m/s

For approaching:

For receding:

37

475 ,o

s

C Vf f Hz f f

C V

0 338 ,s

C Vf f Hz f f

C V

Example 7: an ambulance travels down a highway at a speed of

33.5 m/s. Its siren emits sound at a frequency of 400 Hz. What is

the frequency heard by a passanger in a car traveling at 24.6 m/s

in the opposite directions as the car approaches the ambulance

and as the car moves away from the ambulance?

Solution: Vs=33.5 m/s, f=400Hz, Vo=24.6 m/s, C=343 m/s

For car approaching:

For car receding:

38

956 ,oC Vf f Hz f f

C

0 1044 ,C V

f f Hz f fC

Example 8: A stationary civil defense siren has a frequency of

1000 Hz. What frequency will be heard by drivers of cars

moving at 15 m/s (a) away from the siren; (b) toward the

siren? The velocity of sound in air is 344 m/s.

Solution: Vs=0, f=1000 Hz, Vo=15 m/s, C=344 m/s

For moving away:

For moving towards:

39

958 ,s

Cf f Hz f f

C V

1046 ,s

Cf f Hz f f

C V

Example 9: A police car with a 1000 Hz siren is moving at 15

m/s. What frequency is heard by a stationary listener when the

police car ia (a) receding from and (b) approaching the

listener?.

Solution: Vs=15 m/s, f=1000 Hz, Vo=0, C=344 m/s

For receding:

For approaching:

40

Shock Wave

The speed of the source can exceed the speed of the wave

The envelope of these wave fronts is a cone whose apex half-angle is given by sin q v/vS

This is called the Mach angle

41

Mach Number

The ratio vs / v is referred to as the Mach

number

The relationship between the Mach angle and

the Mach number is

sin

s s

vt v

v t vq

42

Shock Wave, final

The conical wave front

produced when vs > v is known

as a shock wave

This is supersonic

The shock wave carries a

great deal of energy

concentrated on the surface of

the cone

There are correspondingly

great pressure variations

43

Application of Doppler Effect

Nexrad: Next Generation Weather

Radar

44

Applications of Sound in Medicine

1.Ultrasonic Scanner

2.The cavitron ultrasonic surgical aspirator

(CUSA)

3.The Doppler flow meter

45

Ultrasonic Scanner

46

The cavitron ultrasonic

surgical aspirator (CUSA)

Neurosurgeons use a cavitron ultrasonic surgical aspirator (CUSA) to “cut

out” brain tumors without adversely affecting the surrounding healthy tissue. 47

Doppler Flow Meter

A Doppler flow meter measures the speed of red

blood cells. 48

The Reflection of Sound

Echoes are sounds that reflect off surfaces.

Repeated echoes are called reverberation.

The reflection of sound can be used to

locate or identify objects.

Echolocation is the process of locating

objects by bounding sounds off them.

Some animals emit short, high frequency

sound waves toward a certain area.

By interpreting the reflected waves, the

animals can locate and determine properties of

other animals.

49

SONAR

SONAR stands for Sound Navigation ;and Ranging (المالحة)

It relies on the reflection of ultrasonic pulses bouncing off an object. By timing how long it takes for the signal to return, the depth of the object can be calculated. It has been used to map the ocean floor, as well as finding shoals of fish by fishermen

50

51

SONAR Applications

What is Ultrasound? Ultrasound is defined as any sound wave above 20000Hz. Sound waves of this frequency are above the human audible range and therefore cannot be heard by humans. All sound waves, including ultrasound are longitudinal waves. Medical ultrasounds are usually of the order of MEGAHERTZ (1-15MHz). Ultrasound as all sound waves are caused by vibrations and therefore cause no ionisation and are safe to use on pregnant women. Ultrasound is also able to distinguish between muscle and blood and therefore show blood movement.

When an ultrasound wave meets a boundary between two different materials some of it is refracted and some is reflected. The reflected wave is detected by the ultrasound scanner and forms the image.

52

Uses of Ultrasound

Cardiology

Seeing the inside of the heart to identify abnormal structures or functions and measuring blood flow through the heart and major blood vessels Urology

•measuring blood flow through the kidney •seeing kidney stones •detecting prostate cancer early

Obstetrics and Gynecology

The development and monitoring of a

developing foetus

53

The Piezoelectric Effect piezoelectric means pressure electricity

Discovered by Pierre and Jacques

Curie in 1880.

Ultrasound waves are produced using the piezoelectric effect.

When a potential difference is applied across certain crystals (piezoelectric) the crystals themselves deform and contract a little. If the p.d. applied is alternating then the crystal vibrates at the same frequency and sends out ultrasonic waves. For ultrasound – quartz crystals are used. This process also works in reverse. The piezoelectric crystal acts a receiver of ultrasound by converting sound waves to alternating voltages and as a transmitter by converting alternating voltages to sound waves

54

The Transducer The transducer probe is the main part of the ultrasound machine. The

transducer probe transmits and receives the ultrasound. The curved faceplate shapes the ultrasound waves into a narrow beam.

Transducer probes come in many shapes and sizes. The shape of the probe determines its field of view, and the frequency of emitted sound waves

(controlled by the tuning device) determines how deep the sound waves penetrate and the resolution of the image. The ultrasound is pulsed. There must be a pause

to allow the reflected wave to be detected. 55

Why Ultrasound? ADVANTAGES

•No known hazards – non ionizing for patient and sonographer.

•Good for imaging soft tissue.

•Relatively cheap and portable.

DISADVANTAGES

•Cannot pass through bone

•Cannot pass through air spaces.

•Poor resolution.

56

Exam Style Question

a) Explain what an Ultrasound wave is.

(2 marks)

b) Describe how ultrasound images are carried out

(4 marks)

c) Discuss the uses, advantages and disadvantages of Ultrasound in medicine.

(4 marks)

57

Acoustic Impedance, Z As stated earlier, when an ultrasound wave meets a boundary between two different materials some of it is refracted and some is reflected. The reflected wave is detected by the ultrasound scanner and forms the image. The proportion of the incident wave that is reflected depends on

the change in the acoustic impedance, Z.

Acoustic Impedance, Z of a medium is defined as:

Z = rc

Where r = the density of the material, kgm-3

c = speed of sound in that material, ms-1

TASK: What are the units of Z? 58

Intensity reflection coefficient,

At a boundary between mediums, the ratio of the intensity reflected, Ir to

the intensity incident, I0 is known as the intensity reflection coefficient, .

The intensity of both the reflected and incident ultrasound waves depend on the acoustic impedance, Z of the two mediums. Therefore the fraction of the wave intensity reflected can be calculated for an ultrasound wave travelling from medium 1, (acoustic impedance Z1) to medium 2 (acoustic impedance Z2).

If 2 mediums have a large difference in impedance, then most of the wave is reflected. If they have a similar impedance then none is reflected.

r

o

I

I

2

2 1

2 1

r

o

I Z Z

I Z Z

59

Impedance Matching / Gel When ultrasound passes through two very different materials the majority of it is reflected. This happens between air and the body, meaning that most ultrasound waves never enter the body. To prevent this large difference in impedance a coupling medium (gel) is used between the air and the skin. The need to match up similar impedances to ensure the waves pass through the body is known as impedance matching.

60

A-Scan

A-Scan (Amplitude scan) • Gives no photo image

•Pulses of ultrasound sent into the body, reflected ultrasound is detected and appear as vertical spikes on a CRO screen.

•The horizontal positions of the ‘spikes’ indicate the time it took for the wave to be reflected.

•Commonly used to measure size of foetal head.

61

B-Scan

B-Scan (Brightness scan) • An array of transducers are used and the ultrasound beam is spread out across the body.

•Returning waves are detected and appear as spots of varying brightness.

•These spots of brightness are used to build up a picture.

62

Applications of Ultrasonic

Waves in Engineering

(1) Detection of flaws in metals (Non Destructive Testing –NDT)

Principle:

Ultrasonic waves are used to detect the presence of flaws or defects

in the form of cracks, blowholes porosity etc., in the internal structure

of a material

By sending out ultrasonic beam and by measuring the time interval of

the reflected beam, flaws in the metal block can be determined.

It consists of an ultrasonic frequency generator and a cathode ray

oscilloscope (CRO),transmitting transducer(A), receiving

transducer(B) and an amplifier. 63

64

In flaws, there is a change of medium and this produces

reflection of ultrasonic at the cavities or cracks.

The reflected beam (echoes) is recorded by using cathode ray

oscilloscope.

The time interval between initial and flaw echoes depends on

the range of flaw.

By examining echoes on CRO, flaws can be detected and

their sizes can be estimated.

65

(2) Ultrasonic Drilling

•Ultrasonics are used for making holes in very hard

materials like glass, diamond etc.

•For this purpose, a suitable drilling tool bit is fixed at the

end of a powerful ultrasonic generator.

•Some slurry (a thin paste of carborundum powder and

water) is made to flow between the bit and the plate in

which the hole is to be made

•Ultrasonic generator causes the tool bit to move up and

down very quickly and the slurry particles below the bit

just remove some material from the plate.

• This process continues and a hole is drilled in the plate.

66

67