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Unit 12 Physical Properties of Waves I

12.1 Types of waves

12.2 Useful quantities in describing waves

12.3 Waves on a string

12.4 Sound waves

12.5 The frequency of a sound wave

12.6 Sound intensity

12.7 Human perception of sound

12.8 The Doppler effect

12.9 Superposition and interference12.10 Standing waves12.11 Beats

12.1 Types of waves

A disturbance that propagates from one place to another is referred to as a wave.

Waves propagate with well-defined speeds determined by the properties of the

material through which they travel. For example, sound waves have different speeds

in different materials. The following table lists a sampling of sound speed in various

materials.

Material Speed (m/s)

Aluminum 6420

Steel 5960

Copper 5010

Plastic 2680

Fresh water (20oC) 1480

Air (20oC) 343

Waves carry energy and propagate it when the waves travel. There are two typical

waves, namely, the transverse waves and the longitudinal waves.

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Transverse waves:

The displacement of individual particles is perpendicular to the direction of

propagation of the wave, e.g. holding one end of a string with another end fixed on

the wall. When you swing your hands vertically, the waves propagate horizontally

along the string and the particles of string moves up and down.

Longitudinal waves:

The displacement of individual particles is in the same direction as the direction of

propagation of the waves, e.g. sound waves. The particles of air move back and forth

such that a series of compression and rarefaction are observed. Note that the particle

does not travel with the wave, but vibrating about its equilibrium position.

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12.2 Useful quantities in describing wavesWavelength: The distance over which a wave repeats,

e.g. the distance between successive crests and the

distance between successive troughs. Wavelength is

quite often labeled as . The SI unit is, of course,

meter, m.

Angular wave number: The angular wave number is

defined as

2=k . The SI unit is radian per meter.

Angular frequency: It is the measure of how many

radians the waves change in one second. It is labeled as

.

Frequency: The number of oscillation per unit time, f,

where f 2= .

Period: The time for one oscillation, it is labeled as T,

where

21==

fT .

Velocity: The distance that the wave travels per unit time is referred to as the velocity.

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The wave equation is: )sin(),( tkxytxy m = , in general we have

)sin(),( += tkxytxy m ,

where is the phase angle.

A useful relation in waves: kfv

== .

12.3 Waves on a stringThe speed of a wave is determined by the properties of the medium through which it

propagates. For a string of length L, there are two factors that vary the speed of a

wave: (i) the tension in the string F, and (ii) the mass of the string. For the second

factor, we should say it more precisely that the speed of a wave varies with the

density of the string (mass per length) . The definition of is m/L. The unit is kg/m.

We can obtain the velocity v by dimensional analysis. Let the velocity relates the

tension of string Fand the mass per unit length by

Fv =

The proof is simple. LetbaFv = and consider the dimensions of the following

quantities.

[v] = [L][T]1

(Unit: ms1

)

[F] = [M][L][T]2 (Unit: kg ms2)

[] = [M][L]1

(Unit: kg m1

)

Comparing the dimension on both sides ofbaFv = , we have three equations

[L]: 1 = a b

[T]: 1 = 2a

[M]: 0 = a + b

After solving, we find that

2

1== ba . Hence, we have

Fv = .

Example

A rope of length L and mass Mhangs from a ceiling. If the bottom of the

rope is given a gentle wiggle, a wave will travel to the top of the rope. As

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the wave travels upward does its speed (a) increase, (b) decrease, or (c) stay the same?

Answer:

Since the tension increases with the height, the speed of the wave increases when it

climbs up the rope. Note also that the tension of the rope increases from almost zero

at the bottom toMg at the top of rope.

12.4 Sound wavesA mechanical model of a sound wave is provided by a slinky. Consider if a slinky is

oscillated at one end back and forth horizontally. Longitudinal wave travels in

horizontal direction with some regions are compressed and some regions are more

widely spaced, but these regions are distributed alternatively. If we plot the density

variation against the displacement x, we observe classical wave shape in the graph.

The rarefactions and compressions oscillate in a wave-like fashion. In the

compressions regions, the pressure is high, and in the rarefaction regions, the pressure

is low. The speed of sound is determined by the properties of the medium through

which it propagates. In air, under normal atmospheric pressure and temperature, the

speed of sound is approximately 343 m/s 770 mi/h. As the air is heated up to a

higher temperature, the air molecules moves faster and the speed of sound increases

as expected.

In a solid, the speed of sound is determined in part by the stiffness of the material.

The stiffer the material, the faster the sound wave, just as having more tension in a

string causes a faster wave. The speed of sound in steel is greater than that in plastic.

And both speeds are much higher than that in air.

ExampleYou drop a stone into a well that is 7.35 m deep. How long does it take

before you hear the splash?

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Answer:

The time until the splash is heard is the sum of two time intervals.

t1: the time for the stone to drop a distance dand

t2: the time for the sound to travel a distance d.

Since2

12

1gtd= , we obtain s

g

dt 22.1

81.9

)35.7(221 === .

To calculate t2, we have 2vtd= , and sv

dt 0214.0

343

35.72 === .

Hence, the sum of the two time intervals is (1.22 +0.0214) s = 1.24 s.

12.5 The frequency of a sound waveHuman can hear sounds between 20 Hz on the low frequency and 20,000 Hz on the

high frequency end. Sounds with frequencies above this range are referred as

ultrasonic, while those with frequencies lower than 20 Hz are classified as infrasonic.

12.6 Sound intensityIntensity is a quantitative scale by which loudness may be measured. The intensity is

defined as the amount of energy that passes through a given area in a given time. This

is illustrated in the figure. If the energy Epasses through the areaA in the time tthe

intensity, I, of the wave carrying the energy isAt

EI = , where E/t is the power.

Rewrite the expression again, we haveA

PI= .

The SI unit is W/m2. An example of intensity of light on the Earths upper atmosphere

coming from the Sun is about 1380 W/m2. A rock concert has an intensity of 0.1

W/m2, while the intensity of a classroom is 0.0000001 W/m2. The threshold of

hearing is 1012

W/m2.

When we listen to a source of sound, such as a person speaking or a radio playing a

song, the loudness of the sound decreases as we move away from the source. The

surface area of a sphere from a distance ris2

4 r . The intensity of such sound is

24 r

PI

= .

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12.7 Human perception of soundWe can detect sounds that are about a million times fainter than a typical

conversation, and listen to sounds that are a million times louder before experiencing

pain. We are able to hear sounds over a wide range of frequencies, from 20 Hz to

20,000 Hz. Our perception of sound, for example the loudness seems to be twice as

loud if the intensity of the sound is about 10 times the original one. In the study of

sound, the loudness is measured by a convenient scale, which depends on the

logarithm of intensity.

Mathematically, the intensity level is expressed in the form )log(100I

I= . The

intensity level is dimensionless and the unit is given as decibel (dB), where I0is the

intensity of the faintest sounds that can be heard. Experiments show that the lowest

detectable intensity is212

0 /10 mWI= . The smallest increase in intensity level that

can be detected by the human ear is about 1 dB. And, the loudness of a sound doubles

with each increase in intensity level of 10 dB.

Example

If a sound has an intensityI = I0, the corresponding intensity level is

dBI

I01log10)log(10

0

0 === .

Increasing the intensity by a factor of 10 makes the sound seem twice as loud. In

terms of decibels, we have

Sound Decibels

Ear drum ruptures 160

Jet taking off 140

Loud rock band 120

Heavy traffic 90

Classroom 50

Whisper 20

Threshold of hearing 0

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dBI

I1010log10)

10log(10

0

0 === .

A further increase in intensity by a factor of 10 double the loudness again.

dB

I

I20100log10)

100log(10

0

0 === .

Thus, the loudness of a sound doubles with each increase in intensity level of 10 dB.

The smallest increase in intensity level that can be detected by the human ear is about

1 dB.

Example

A crying child emits sound with an intensity of26

/100.8 mW . Find

(a) the intensity level in decibels forthe childs sounds, and

(b) the intensity level for this child andits twin, both crying with identical

intensities.

Answer:

(a) As the intensity level is given by )log(100I

I= , we substitute

26 /100.8 mWI = and the lowest detectable intensity 2120 /10 mWI= ,

hence [ ] dB69)10log()100.8log(10)10

100.8log(10

126

12

6

==

=

.

(b) When the twins cry, the intensity will be doubled,2526

/106.1)/100.8(2 mWmWI == .

The intensity level is dB72)10

106.1log(10

12

5

=

=

.

Or, we can write

[ ] dB72)10(log)100.8log()2(log10)10

100.82log(10 126

12

6

=+=

=

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N.B. We should note that double the intensity increases the intensity level by 3 dB,

since 32log10 . Halved the intensity leads to a decrease of intensity level by 3 dB.

Obviously, ten times the intensity of sound gives an increase of 10 dB.

Example

Many animal species use sound waves that are too high or too low for human ears to

detect, e.g. bats and blue whales.

12.8 The Doppler effect

The relative motion between a source of sound and the receiver gives a change in

pitch. This is the Doppler effect. There are two cases for Doppler effect: Moving

observer and moving the source. For example, there is a change in pitch of a train

whistle or a car horn as the vehicle moves past us. Doppler effect applies to all wave

phenomena, not just to sound.

Example

For light, we observe a change in color, e.g. red-shifted in the color of their light when

the galaxies are moving away from the Earth. However, some galaxies are movingtoward us, and their light shows a blue shift.

12.81 Moving observerA sound wave is emitted from a stationary source. The wave travels in the air with

velocity v, having frequency f and wavelength , where v = f. For an observer

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moving toward the source with a speed u, the sound seems to have a higher speed, e.g.

v + u. As a result, more wavefronts move past the observer in a given time than if the

observer had been at rest. To the observer, the sound has a frequency,f, that is higher

than the frequency of the source.

ffv

u

f

v

u

v

v

u

uvvf >+=

+

=

+

=

+

== )1(1

11'

'

If the observer moves away from the source, the sound seems to have a lower speed,

e.g. v u. As a result, less wavefronts move past the observer in a given time than if

the observer had been at rest. To the observer, the sound has a frequency, f, that is

lower than the frequency of the source.

ffv

u

f

v

u

v

v

u

uvvf

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the source is at rest. The new frequency of the sound waves is obtained by v = f,

that is

ffvu

Tv

uTuv

vf >

=

=

=/1

1

)1(

1

)(' .

When the source reverses its direction, the new wavelength of the sound waves,

TuvuTvT )( +=+ , is longer than that when the source is at rest.

ffvu

Tv

uTuv

vf

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Example

A car moving at 18 m/s sounds its 550 Hz horn. A bicyclist on the sidewalk, moving

with a speed of 7.2 m/s, approaches the car. What frequency is heard by the bicyclist?

Answer:

As the car (source) and the bicyclist (observer) approach each other, we apply the

formula HzHzfvu

vuf

s

o 7.592)550(343/181

343/2.71

/1

/1' =

+=

+= .

The right figure shows the Doppler shifted

frequency versus speed for a 400-Hz sound

source. The upper curve corresponds to a

moving source, the lower curve to a moving

observer. Notice that while the two cases

give similar results for low speed, the high-

speed behavior is quite different. In fact, the

Doppler frequency for the moving source

grows without limit for speeds near the speed

of sound, while the Doppler frequency for

the moving observer is relatively small.

12.84 Supersonic speed and shock wavesWhat happen when the speed of the source exceeds the speed of sound? The equations

derived above are no longer valid. For supersonic speeds, a V-shaped envelope is

observed, all wavefronts bunch are along this

envelop, which is in three dimensions. This cone is

called the Mach cone. A shock wave is said to exist

along the surface of this cone, because the bunching

of wavefronts causes an abrupt rose and fall of air

pressure as the surface passes through any point. The

Mach cone angle is given by

ssv

v

tv

vt==sin .

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The ratio vs/v is the Mach number. The shock wave generated by a supersonic aircraft

or projectile produces a burst of sound, called a sonic boom.

12.9 Superposition and interference

The combination of two or more waves to form a resultant

wave is referred to as superposition. When waves are of

small amplitude, they superpose in the simplest of ways

they just add.

For example, consider two waves on a string, as shown in

figure.

Example

Since two waves add, does the resultant wavey always have a greater amplitude than

the individual wavesy1andy2?

Answer:

The wave y is the sum ofy1 and y2, but remember that y1 and y2 are sometimes

positive and sometimes negative. Thus, ify1 is positive at a given time, for example,

andy2 is negative, the sumy1 +y2 can be zero or even negative.

As simple as the principle of superposition is, it still leads to interesting

consequences. For example, consider the wave pulse on a string shown in the above

figure (a). When they combine, the resulting pulse has an amplitude equal to the sum

of the amplitudes of the individual pulses. This is referred to as constructive

interference. When two pulses like those in figure (b) may combine and gives a net

displacement of zero. That is the pulses momentarily cancel one another. This is

destructive interference.

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Answer:

The wavelength of sound: mHzsmfv 55.1221//343/ === .

To determine the path difference, d = d2 d1, we need to find d2 first, and

2 2 2 2

2 1 (4.30 ) (2.80 ) 5.13d D d m m m= + = + = .

Now d= 5.13 m 2.80 m = 2.33 m. The number of wavelength that fit into the path

difference: 50.155.1

33.2==

m

md

. Since the path difference is 3/2 we expect destructive

interference. In the ideal case, the person would hear no sound. As a practical matter,

some sound will be reflected from objects in the vicinity, resulting in a finite sound

intensity.

12.10 Standing wavesIf you plucked a guitar string, or blown across the

mouth of a pop bottle to create a tone, you have

generated standing waves. In general, a standing

wave is one that oscillates with time, but remains

in its location. It is in this sense that the wave is

said to be standing. In some respects, a standing

wave can be considered as resulting from

constructive interference of a wave with itself.

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12.10.1 Waves on a string

A string is tied down at both ends. If the string is plucked

in the middle a standing wave results. This is the

fundamental mode of oscillation of the string. The

fundamental consists of one-half a wavelength between

the two ends of the string. Hence, its wavelength is 2L,

or we write L2= .

If the speed of waves on the string is v, it follows that the

frequency of the fundamental, f1, is determined by

11 )2( fLfv == . Therefore,

L

vvf

21 ==

.

Note that the fundamental frequency increases with the speed of the waves, and

decreases as the string is lengthened. Other than the fundamental frequency, there are

an infinite number of standing wave modes or harmonics for any given string.

Remarks:

Points on a standing wave that stay fixed arereferred to as nodes,N.

Halfway between any two nodes is a point on thewave that has a maximum displacement, is called

an anti-node,A.

The second harmonics can be constructed by including

one more half wavelength in the standing wave. This

mode has one complete wavelength between the walls.

12 2fL

vvf ===

.

Similarly, the third harmonic has one-and-a-half

wavelength in the lengthL, therefore, its frequency

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13 3

3

2f

L

vvf ===

.

Remark:

In general, we have L

vf 2

1 = , 1nffn = , and nLn /2= where n = 1, 2, 3, . That is,

all harmonics are present.

12.10.2 Vibrating columns of air

If you blow across the open end of a pop bottle, you hear a

tone of a certain frequency. If you pour some water into the

bottle and repeat the experiment, the sound you hear has a

higher frequency. The standing wave will have an antinode,

A, at the top (where the air is moving) and a node, N, at the

bottom (where the air cannot move.) The lowest frequency

standing wave has one-quarter of a wavelength fits into the

column of air in the bottom. Thus, we have the wave form N-

A in the pipe

L

L

4

4

1

=

=

The fundamental frequency,f1, is given by 11 )4( fLfv == . Or we can write

L

vf

41 = .

The second harmonic is produced by adding

half a wavelength, i.e. N-A-N-A, therefore,

L=4/3 , and hence L3

4= . The frequency

is1

3)4

(3

34

fL

v

L

vv===

.

Similarly, the next-higher harmonic is

represented by N-A-N-A-N-A. Inside the pipe,

we have standing waves L=4/5 , the frequency is 15)4

(5

5

4f

L

v

L

vv===

.

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Remark:

In general, we haveL

vf

41 = , 1nffn = and nLn /4= , where n = 1, 3, 5, . That is

odd harmonics are present.

Standing waves in a pipe that is open at both ends

have the following modes, as shown in the figure.

Remark:

In general, we haveL

vf

21 = , 1nffn = and

nLn /2= , where n = 1, 2, 3, . That is all

harmonics are present.

Example

An empty pop bottle is to be used as a musical instrument in a

band. In order to be tuned properly the fundamental frequency

of the bottle must be 440.0 Hz. If the bottle is 26.0 cm tall,

how high should it be filled with water to produce the desired

frequency?

Answer:

Since Lvf 4/1 = , we have

mHz

smfvL 195.0

)0.440(4

/3434/ 1 === .

The depth of water to be filled: cmmmLHh 5.6195.0260.0 === .

Example

If you fill your lungs with helium and speak you sound something like Donald Duck.

From this observation, we can conclude that the speed of sound in helium must be (a)

less than (b) the same as, or (c) greater than the speed of sound in air.

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Answer:

When we speak with helium our words are higher pitched. Looking at the relation,

e.g.L

vf

21 = , the velocity of sound is increased if the length of vocal chords is fixed

while the frequency is increased.

12.11 Beats

Beats can be considered as the interference pattern in time. To be specific, imagine

plucking two guitar strings that have slightly different frequencies. If you listen

carefully, you notice that the sound produced by the strings is not constant is time. In

fact, the intensity increases and decreases with a definite period. These fluctuations in

intensity are the beats, and the frequency of successive maximum intensities is the

beat frequency.

Consider two waves, with frequencies f1 = 1/T1 and f2 = 1/T2, interfere at a given,

fixed location. At this location, each wave moves up and down to the vertical

position,y, of each wave yields the following:

)2cos(2

cos

)2cos(2

cos

2

2

2

1

1

1

tfAtT

Ay

tfAtT

Ay

=

=

=

=

IfA = 1,we have the following plots, where 21 yyytotal += . Mathematically, we have

1 2 1 2

1 2 1 2

cos(2 ) cos(2 )

2 cos 2 cos 22 2

total

total

y y y A f t A f t

f f f f y A t t

= + = +

+ =

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The first part of the ytotal is

t

ffA

22cos2 21 which gives the slowly-varying

amplitude of the beats. Since a loud sound is heard whenever this term is 2A or 2A,

the beat frequency is 21 fffbeat = . The rapid oscillations within each beat are due

to the second part ofytotal,

+t

ff

22cos 21 . Now, beats can be understood as

oscillations at the average frequency, modulated by a slowly varying amplitude.

Example

Suppose two guitar strings have frequencies 438 Hz and 442 Hz. If you sound them

simultaneously you will hear the average frequency, 440 Hz, increasing and

decreasing in loudness with a beat frequency of 4 Hz. Beats can be used to tune amusical instrument to a desired frequency. To tune a guitar string to 440 Hz, for

example, the string can be played simultaneously with a 440 Hz fork. Listening to

the beats, the tension in the string can be increased or decreased until the beat

frequency becomes vanishingly small.

Example (Challenging)

An experimental way to tune the pop bottle is to

compare its frequency with that of a 440-Hz tuning fork.

Initially, a beat frequency of 4 Hz is heard. As a small

amount of water is added to that already present, the

beat frequency increases steadily to 5 Hz. What were the

initial and final frequencies of the bottle?

Answer:

Before extra water is added, possible frequency of the bottle is either 436 Hz or 444

Hz. After water is added, possible frequency of the bottle is either 435 Hz or 445 Hz.

But the frequency of the bottle should be increased as water is added. Hence, the

frequency of the bottle before adding extra water should be 444 Hz.