Sound waves chapter 2 by yaman mohamed

Chapter 2 Sound Waves

Dr Mohamed Saudy

Introduction to Sound Waves 

Sound waves are the most important example of longitudinal waves.

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They can travel through any material, except vacuum (no one can hear you scream in outer space).

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Speed of sound depends on material (and temperature)

Sound - is a wave (sound wave) - Rarefied and compressed regions

- Longitudinal wave - air molecules move back and forth

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.

• Outer ear collects sound • Middle ear amplifies sound • Inner ear converts sound 5

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!

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 v inertial property

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

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 v

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

v

Speed of Sound Waves In gas and liquids:

v

In solids:

v

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.

volume stress F/A P B   volume strain V / Vi V / Vi

Young’s modules determines the length change of an object due to an applied force F.

tensile stress F / A Y  tensile strain L / Li12

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

Speed of Sound in Solids, Example Values Speeds are in m/s; values are for bulk solids

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 TK v  (331 m/s) 273  

The 331 m/s is the speed at 0o C TK is the air temperature in Kelvin 16

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

r = 7800 kg/m3 ,

v

Y = 2.07 x 1011 Pa

2.07  10 Pa   5150 m / S 3 r 7800 Kg / m 11

Example 2: What is the speed of sound in air when the temperature is 270C?

(27  273) v  331  347 m / S 273 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 I  b  10log    Io 

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, Pmax   I 2 rv

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

Intensity of a Point Source, cont 

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av av I  A 4 r 2 This is an inversesquare law 

The intensity decreases in proportion to the square of the distance from the source

Sound Level, cont  I0   



 

I  b  10log    Io 

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

Sound Intensity 

Sound is characterized in decibels (dB), according to:  sound level b= Decibel level=10log(I/I0) = 20log(P/P0) dB  I0 = 1012 W/m2 is the threshold power intensity(0 dB)  P0 = 2105 N/m2 is the threshold pressure (0 dB)  atmospheric pressure is about 105 N/m2 23

 

Example 3: 60 dB means log(I/I0) = 6, so I = 106 W/m2 I I 60 60dB  10 log  log  6 I0 I0 10 I  106  I  106  10 12  10 6W / m 2 I0



and log(P/P0) = 3, so P = 2102 N/m2 P P 60 60dB  20 log  log   3 P0 P0 20 P  103  P  103  2  10 5  2  10 2 N / m 2 P0

• 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 = 1011 W/m2

and log(P/P0) = 0.5, so P  6105 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. I1 I2 b 2  b1  10log( )  10log( ) I0 I0 I2 3  10log( ) I1 I2 3 I2  log( )    100.3  2 I1 10 I1

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

P 28 Pa 4   2.77  10 P 1.013  105 Pa ( P )2 (28Pa )2 2 I   0.95 W / m 2r v 2(1.2 Kg / m3 )(344 m / s ) 27

Sound Levels

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 â&#x20AC;&#x201C; 4 cases Source moving toward receiver Source moving away from receiver

Receiver (observer) moving towards source Receiver (observer) moving away from source. 32

ď Ź

When the observer and source are moving:

f=the frequency of the source fâ&#x20AC;&#x2122;=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)

• 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

1. Source moving case, V0=0 (stationary observer) When the source is moving toward from the listener

 C  c f   f     C  Vs  When the source is moving away from the listener

 C  c f   f     C  Vs  35

2. Observer moving case, Vs=0 (stationary source) • When the observer is moving toward the source

c C  V0   f   f     C 

•When the observer is moving away the source

c C  V0   f   f     C 

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. Solution: Vs=40 m/s, f=500Hz, Vo=0, For approaching:

For receding:

C=343 m/s

 C  f f  f f   566 Hz,  C  Vs   C  f f  f   448 Hz,  C  Vs 

f 37

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, For car approaching:

For car receding:

C=343 m/s

 C  Vo  f f    475 Hz,  C  Vs 

 C  V0  f f    338 Hz,  C  Vs 

f

f 38

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, For moving away:

For moving towards:

C=344 m/s

 C  Vo  f f    956 Hz,  C  C  V0   f f    1044 Hz,  C 

f

f 39

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, For receding:

C=344 m/s

 C  f f    958 Hz,  C  Vs 

 C  For approaching: f   f    1046 Hz,  C  Vs 

f

f 40

Shock Wave 

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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

Mach Number 



The ratio vs / v is referred to as the Mach number The relationship between the Mach angle and the Mach number is

vt v sinq   vst v s

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

Application of Doppler Effect Nexrad: Next Generation Weather Radar

Applications of Sound in Medicine 1. Ultrasonic Scanner 2. The cavitron ultrasonic surgical aspirator (CUSA)

3. The Doppler flow meter

Ultrasonic Scanner

The cavitron ultrasonic surgical aspirator (CUSA)

Neurosurgeons use a cavitron ultrasonic surgical aspirator (CUSA) to â&#x20AC;&#x153;cut outâ&#x20AC;? brain tumors without adversely affecting the surrounding healthy tissue.47

Doppler Flow Meter

A Doppler flow meter measures the speed of red blood cells.

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 (‫ )المالحة‬and Ranging; 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

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 Obstetrics and Gynecology The development and monitoring of a developing foetus

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

The Piezoelectric Effect piezoelectric means pressure electricity 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 â&#x20AC;&#x201C; 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

Discovered by Pierre and Jacques Curie in 1880.

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 55 to allow the reflected wave to be detected.

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)

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, .

Ir  Io 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).

I r  Z 2  Z1      I o  Z 2  Z1 

If 2 mediums have a large difference in impedance, then most of the wave is reflected. If they have a similar impedance 59 then none is reflected.

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.

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.

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.

Applications of Ultrasonic Waves in Engineering (1) Detection of flaws in metals (Non Destructive Testing â&#x20AC;&#x201C;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(B) and an amplifier.

transducer(A),

receiving 63

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