SSBasic ScienceBasic SciencePhysics of SoundPhysics of Sound

Dr. Omed Ghareb AbdullahUniversity of Sulaimani

Faculty of Science and Science EducationSchool of Science / Physics Department

omed.abdullah@univsul.net 1

The Nature of WavesThe Nature of Waves• A wave is a traveling disturbance

– two broad classifications of waves given how the– two broad classifications of waves, given how the medium (water, ground, air,…) is disturbed

• Longitudinal: the disturbance is parallel to the direction of motion of the wave (e g sound waves)of motion of the wave (e.g. sound waves).

• Transverse: the disturbance is perpendicular to the direction of motion of the wave (e.g. Electromagnetic waves).waves).

TransverseLongitudinal

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Sound WavesSound WavesSound is produced due to the vibration of objects.

Sound waves are longitudinal. They can be represented by either variations in pressure (gauge

) b di l t f i l tpressure) or by displacements of an air element.

Infrasonic – 20 Hz – 20000 Hz - Ultrasonic

Audible Range

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Propagation of SoundThe sound produced by a vibrating object travels through a

medium to a listener. The medium can be solid, liquid or gas.

When an object vibrates the particles around the mediumWhen an object vibrates, the particles around the medium vibrates. The particle in contact with the vibrating object is first displaced from its equilibrium position. It then exerts a force on the adjacent particle and the adjacent particle is displaced from itsthe adjacent particle and the adjacent particle is displaced from its position of rest. After displacing the adjacent particle the first particle comes back to its original position. This process repeats in the medium till the sound reaches the ear.in the medium till the sound reaches the ear.

The disturbance produced by the vibrating body travels through the medium but the particles do not move forward themselves.

A wave is a disturbance which moves through a medium by theA wave is a disturbance which moves through a medium by the vibration of the particles of the medium. So sound is considered as a wave. Since sound waves are produced due to the vibration of particles of the medium sound waves are called mechanicalof particles of the medium sound waves are called mechanical waves.

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The middle of a compressioncompression (rarefaction) corresponds to a pressure maximumpressure maximum (minimum).

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The Speed of Sound WavesThe Speed of Sound WavesThe speed of sound in different materials can be determined as follows:as follows:

In fluids B

v B is the bulk modulus of the fluid and its density.

Y Y is the Young’s modulus ofIn thin solid rods

Y

v Y is the Young s modulus of the solid and its density.

m/s6060331 Tv

For air, a useful approximation to the above expression is

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m/s606.0331 CTv where Tc is the air temperature in C.

Materials that have a high restoring force (stiffer) will have a higher sound speed.g p

Materials that are denser (more inertia) will have a lower sound speed.

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Example

A copper alloy has a Young’s Modulus of 1.11011 Paand a density of 8 92103 kg/m3 What is the speed

p

and a density of 8.92103 kg/m3. What is the speed of sound in a thin rod made of this alloy?

m/s 3500k /1098

Pa 101.133

11

Y

vkg/m109.8 33

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Example

Bats emit ultrasonic sound waves with a frequency as high as 1 0105 Hz What is the wavelength of

p

as high as 1.010 Hz. What is the wavelength of such a wave in air of temperature 15.0 C?

m/s 606.0331 CTv The speed of sound in air of this temperature is 340 m/s.

m/s340vm 104.3

Hz 101.0

m/s 340 35

f

v

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ExampleA lightning flash is seen in the sky and 8.2 secondslater the boom of thunder is heard. The temperature

p

of the air is 12.0 C.(a) What is the speed of sound in air at that temperature?

(b) How far away is the lightning strike?

The speed of sound in m/s 606.0331 CTv

The speed of sound in air of this temperature is 338 m/s.

(a)

k2 8280028/338td10

km 2.8m 2800s2.8m/s 338 vtd(b)

Amplitude & Intensity of Sound Waves

Intensity: the amount of energy transported through aIntensity: the amount of energy transported through a given area per unit of time. Amplitude and intensity are directly proportional. The louder the sound, the larger the amplitude. SI unit: decibels (dB)

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20pI p0 is the pressure amplitude and

20sI s0 is the displacement amplitude.

The intensity of sound waves also

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The intensity of sound waves also follow an inverse square law.

Sound Loudness

Loudness of a sound is measured by the logarithm of the intensityof the intensity.

The threshold of hearing is at an intensity of:The threshold of hearing is at an intensity of: 1012 W/m2.

S d i t it l l i d fi d bSound intensity level is defined by

I 0

logdB10I

I dB are decibels

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Example

The sound level 25 m from a loudspeaker is 71 dB. What is the rate at which sound energy is being

p

What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source, and Io=10-12 W/m2?p o

dB 71logdB100

I

IGiven:

1.7log I

ISolve for I, the intensity of a sound wave:

1.7

0

10

g

I

I

I

13 251.72121.70

0

W/m103.110W/m1010 II

I

The intensity of an isotropic source is defined by:The intensity of an isotropic source is defined by:

2

PI

4

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2

rIP

r

Watts100

m 254)W/m103.1( 225

Watts10.0

P: rate of sound energy (energy per unit of time)

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ExampleTwo sounds have levels of 80 dB and 90 dB. What is the difference in the sound intensities?

p

dB 90logdB100

2 I

I dB 80logdB100

1 I

I

0

1

0

212 loglogdB 10dB 10

I

I

I

ISubtracting:

1

2logdB 10dB 10

I

I

1

2 1log

I

I

15

1

1

2 10I

Iand I2 = 10I1

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Standing WavesStanding WavesStanding Wave – When two identical waves move

through each other in opposite directions, creating a repeating pattern of waves that seem not to travel.

Which standing wave is the most basic (fundamental)? How much of a full wavelength is it? 17

Standing Sound WavesStanding Sound WavesConsider a pipe open at both ends:

The ends of the pipe are open to the atmosphere. The

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open ends must be pressure nodes (and displacement antinodes).

The distance between two adjacent antinodes is ½.

Each pair of antinodes must have a node in between.

The fundamental mode (it has the fewest number of antinodes) will have a wavelength of 2L.

The next standing wave pattern to satisfy the conditions at th d f th i ill h d dthe ends of the pipe will have one more node and one more antinode than the previous standing wave. Its wavelength will be L.

The general result for standing waves in a tube open at both ends is:

n

Ln

2 where n = 1, 2, 3,…

nvv

1912

nfL

nvvf

nn

f1 is the fundamental frequency.

Now consider a pipe open at one end and closed at the other

As before, the end of the pipe open to the atmosphere must be a pressure node (and a displacement antinode).p ( p )

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The closed end of the pipe must be a displacement node (and a pressure antinode).

One end of the pipe is a pressure node, the other a pressure antinode. The distance between a consecutive node and antinode is one-quarter of a wavelength.

Here, the fundamental mode will have a wavelength of 4L.

The next standing wave to satisfy the conditions at the ends of the pipe will have one more node and one more antinode than the previous standing wave Its wavelength will bethan the previous standing wave. Its wavelength will be (4/3)L.

The general result for standing waves in a tube open at g g pone end and closed at the other is

L4 where n = 1 3 5 (odd values only!!)

nn where n = 1, 3, 5,… (odd values only!!)

nfnvv

f f1 is the fundamental frequency

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14nf

Lf

nn

f1 is the fundamental frequency.

Example

An organ pipe that is open at both ends has a fundamental frequency of 382 Hz at 0 0 °C

p

fundamental frequency of 382 Hz at 0.0 C.What is the fundamental frequency for this pipe at 20.0 °C?

At Tc = 0.0 °C, the speed of sound is 331 m/s.c p

At Tc = 20.0 °C, the speed of sound is 343 m/s.

The fundamental frequency is L

vvf

211

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The ratio of the fundamental frequencies at the two t t itemperatures is:

220 vL

vf

04.1

2

2

0

20

00,1

20,1 v

v

Lv

Lf

f

Hz 39604.1 0,120,1 ff

How long is this organ pipe?

v

430

21

vL

L

vfUsing either

set of v and f1.

23

m43.02 1

f

L

How we hear?

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How we hear?How we hear?

Ear consists of 3 main partsEar consists of 3 main parts

• Outer Ear – gathers and focuses sound

• Middle Ear – receives and amplifies vibrations

• Inner Ear – uses nerve endings to sense vibrations and send signals to the braing

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a) Threshold of hearing (0 db)

b) Th h ld f i (120 db)b) Threshold of pain (120 db)27

BeatsBeats

When two waves with nearly the same frequency are y ysuperimposed, the result is a pulsation called beats.

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Two waves of different frequency

Superposition of the above waveswaves

The beat frequency is 21 fff

If th b t f d b t 15 H th ill29

If the beat frequency exceeds about 15 Hz, the ear will perceive two different tones instead of beats.

The Doppler EffectThe Doppler EffectWhen a moving object emits a sound, the wave crests appear bunched up in front of the object and appear to beappear bunched up in front of the object and appear to be more spread out behind the object. This change in wave crest spacing is heard as a change in frequency.

The results will be similar when the observer is in motion and the sound source is stationary and also when both the sound source and observer are in motionsound source and observer are in motion.

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The Doppler Effect formula

v

fo is the observed frequency.

s

o

o fvvv

f

1 fs is the frequency emitted by the source.

vo is the observer’s velocity.

s

vv

1 vs is the source’s velocity.

v is the speed of sound.

Note: take vs and vo to be positive when they move in the direction of wave propagation andmove in the direction of wave propagation and negative when they are opposite to the direction of wave propagation.

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p p g

Example

A source of sound waves of frequency 1.0 kHz is stationary An observer is traveling at 0 5 times the

p

stationary. An observer is traveling at 0.5 times the speed of sound.

(a) What is the observed frequency if the observer(a) What is the observed frequency if the observer moves toward the source?

fo is unknown; fs= 1.0 kHz; vo = 0.5v; vs = 0; and v is the speed of sound.

kHz 5.15.10

5.011

ffvv

fvv

f ss

o

o

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011

ff

v

f

vv

f sss

o

(b) Repeat, but with the observer moving in the other direction.

f i k f 1 0 kH 0 5 0fo is unknown; fs = 1.0 kHz; vo = +0.5v; vs = 0; and v is the speed of sound.

5.011

vvo

kHz 5.05.00

11

ff

v

vf

vvvf sss

o

vv

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Echo locationEcho locationSound waves can be sent out from a transmitter of

t th ill fl t ff bj t thsome sort; they will reflect off any objects they encounter and can be received back at their source. The time interval between emission and reception canThe time interval between emission and reception can be used to build up a picture of the scene.

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ExampleA boat is using sonar to detect the bottom of a freshwater lake. If the echo from a sonar signal is

p

gheard 0.540 s after it is emitted, how deep is the lake? Assume the lake’s temperature is uniform and at 25 C ,(the speed of sound in freshwater is 1493 m/s).

The signal travels two times the depth of the lake so the one-way travel time is 0.270 s.

m403s2700m/s1493

depth

tv

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m 403s 270.0m/s 1493

Example

A bat emits chirping sounds of frequency 82.0 kHz while h ti f th t t If th b t i fl i t d

p

hunting for moths to eat. If the bat is flying toward a moth at a speed of 4.40 m/s and the moth is flying away from the bat at 1.20 m/s, what is the frequency of thefrom the bat at 1.20 m/s, what is the frequency of the wave reflected from the moth as observed by the bat? Assume T = 10.0 C.

The speed of sound in air of this temperature is 337 m/s.

m/s 606.0331 CTv

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The flying bat emits sound of f = 82.0 kHz that is received by a moving moth. The frequency observed by the moth is:by a moving moth. The frequency observed by the moth is:

/337m/s 2.1

11

ov

kHz 8.82kHz 0.82

m/s337m/s 4.4

1

m/s337

1

ss

o f

vvvf

m/s337 v

Some of the sound received by the moth will be reflected back toward the bat The moth becomes the sound sourceback toward the bat. The moth becomes the sound source (f = 82.8 kHz) and the bat is now the observer.

m/s44

kHz 6.83kHz 8.82m/s2.1m/s 337m/s 4.4

11

s

o

o fvv

v

f

37m/s 337m/s 2.1

11

s

vv

UltrasoundUltrasound

• Most people hear sounds between 20 andMost people hear sounds between 20 and 20,000 Hz.

Infrasound sound at frequencies lower than– Infrasound – sound at frequencies lower than people usually hear (less than 20 Hz).

– Ultrasound – sound at frequencies higherUltrasound sound at frequencies higher than people usually hear.

• Used in technologies such as sonar and• Used in technologies such as sonar and ultrasound imaging

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UltrasoundUltrasound• Sonar: a technique used to determine the

distance to an object under waterdistance to an object under water.

• Ultrasound: medical technique used to take pictures of different organs (or a fetus!)

SOund NAvigation and Ranging39

Fourier TransformFourier Transform

• It is possible to take any periodic function of time x(t)p y p ( )and resolve it into an equivalent infinite summation of sine waves and cosine waves with frequencies that start at 0 and increase in integer multiples of a base g pfrequency = 1/T, where T is the period of x(t).

• Mathematically, we can say the same thing with this equation:equation:

))2sin()2cos()( 001

0tkfbtkfaatx k

kk

• This equation does NOT tell how to compute the Fourier transform, that is, how we get the coefficients a1…a and

1k

g 1 b1…b.

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Fourier AnalysisFourier Analysis

Most Complex Sounds Are Composites ofMost Complex Sounds Are Composites of Multiple ‘Simple’ Frequencies

Fundamentals and Harmonics

Sound Complexity:  Perceived ‘psychologically’ as  TIMBRE (sound quality).  Timbre is a function of the number, intensity, and duration of harmonicfrequencies associated with a complex sound

‘complex’ sound wave‘simple’ sound wave

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aaav

eav

e

Fundamental (pitch)

sis

of

sis

of

d W

ad

Wa

Ana

lys

Ana

lys

Sou

nS

oun

rier A

rier A

plex

Spl

ex S Harmonics (timbre)

Fou

rF

our

Com

pC

omp

Is there an easier way CC

to see this?

+ + + +=42

Sharp bends imply high frequenciesSharp bends imply high frequencies

Leaving out the high frequency

componentscomponents smoothes the

curves Low pass filterremoves high frequencies –frequencies

Makes the sound less shrill or bright

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ExamplesExamples

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