HKPhO_13vbnvbnvbnv bnv bnvbn

8/9/2019 HKPhO_13vbnvbnvbnv bnv bnvbn

1/28

1

Hong Kong Physics Olympiad Lesson 13

13.1 Types of waves

13.2 Useful quantities in describing waves

13.3 Waves on a string

13.4 Sound waves

13.5 The frequency of a sound wave

13.6 Sound intensity

13.7 Human perception of sound

13.8 The Doppler effect

13.9 Superposition and interference

13.10 Standing waves

13.11 Beats

13.1 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.


2/28

2

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.


3/28

3

13.2 Useful quantities in describing waves

Wavelength: 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 λ . The SI unit is, of course, meter, m.

Angular wave number: The angular wave number is defined asλ

π 2=k . The SI unit

is radian per meter.


4/28


5/28

5

density of the string (mass per length) µ . The definition of µ 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 F and the mass per unit length µ by

µ

F v =

The proof is simple. Let baF v µ = and consider the dimensions of the following

quantities.

[v] = [L][T]−1

(Unit: ms−1

)

[F ] = [M][L][T]−2

(Unit: kg ms−2

)

[ µ ] = [M][L]−1 (Unit: kg m−1)

Comparing the dimension on both sides of baF v µ = , we have three equations

[L]: 1 = a – b

[T]: −1 = − 2a

[M]: 0 = a + b

After solving, we find that2

1=−= ba . Hence, we have

µ

F v = .

Example

A 12-m rope is pulled under tension of 92 N. When one end of the string is swung, it

takes 0.45 s for the disturbance to travel to the other end. What is the mass of rope?

Answer:

The velocity of disturbance is given by 12 m/0.45 s = 27 m/s.

Rewrite the relation µ

F v = as

2v

F = µ which gives the mass density of string as

µ = 92 N/(27 m/s)2 = 0.13 kg/m.


6/28

6

Since, the mass of string relates to its density by L

m= µ , we obtain the mass of string,

m = (0.13 kg/m)(12 m) = 1.6 kg.

ExampleA rope of length L and mass M hangs 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 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 to Mg at the top of rope.

13.4 Sound waves

A 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


7/28

7

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.

Example

You drop a stone into a well that is 7.35 m deep. How long does

it take before you hear the splash?

Answer:

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

intervals.

t 1: the time for the stone to drop a distance d and

t 2: the time for the sound to travel a distance d .

Since2

12

1gt d = , we obtain s

g

d t 22.1

81.9

)35.7(221 === .

To calculate t 2, we have 2vt d = , and sv

d t 0214.0

343

35.72 === .

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

13.5 The frequency of a sound wave

Human 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.

13.6 Sound intensity

Intensity 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. Thisis illustrated in the figure. If the energy E passes through the area A in the time t the

intensity, I , of the wave carrying the energy is At

E I = , where E/t is the power.

Rewrite the expression again, we have

A

P I = .


8/28

8

The SI unit is W/m2. An example of intensity of light on the Earth’s 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/m

2. The threshold of

hearing is 10−12

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 r is2

4 r π . The intensity of such sound is

24 r

P I

π = .

13.7 Human perception of sound

We 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 β is expressed in the form )log(100 I

I

= β . The

intensity level β is dimensionless and the unit is given as decibel (dB), where I 0 is the

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

detectable intensity is212

0 / 10 mW I −= . 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 intensity I = I 0, the corresponding intensity level is

dB I

I 01log10)log(10

0

0 === β .

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

terms of decibels, we have


9/28

9

dB I

I 1010log10)

10log(10

0

0 === β .

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

dB

I

I 20100log10)

100log(10

0

0 === β .

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

Find the intensity that produces an intensity level of 60.0 dB.

Answer:

Since )log(100 I

I = β , we have )

10log(100.60

12−=

I according to the information

provided.

Now, )10

log(0.612−

= I

gives12

6

1010

−=

I and 26 / 10 mW I

−= .

Example

A crying child emits sound with an intensity of26

/ 100.8 mW −× . Find

(a) the intensity level in decibels for the child’s sounds, and

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

intensities.

Sound Decibels

Ear drum ruptures 160

Jet taking off 140

Loud rock band 120Heavy traffic 90

Classroom 50

Whisper 20

Threshold of hearing 0


10/28

10

Answer:

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

I = β , we substitute

26 / 100.8 mW I

−×= and the lowest detectable intensity 2120 / 10 mW I −= ,

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

100.8log(10 126

12

6

=−×=×

= −−−

−

β .

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

2526 / 106.1) / 100.8(2 mW mW I −− ×=××= .

The intensity level is dB72)10

106.1log(10

12

5

=×

=−

−

β .

Or, we can write

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

100.82log(10 126

12

6

=−×+=××

= −−−

−

β

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.


11/28

11

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.

13.8 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 moving

toward us, and their light shows a blue shift.

13.81 Moving observer

A sound wave is emitted from a stationary source. The wave travels in the air with

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

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.


12/28

12

f f v

u

f

v

u

v

v

u

uvv f >+=

+

=

+

=+

== )1(1

11'

'λ λ λ

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.

f f v

u

f

v

u

v

v

u

uvv f


13/28


14/28

14

f f vu

T v

uT uv

v f >

−=

−

=−

= / 1

1

)1(

1

)(' .

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

T uvuT vT )( +=+ , is longer than that when the source is at rest.

f f vu

T v

uT uvv f


15/28

15

Answer:

As the source moves away from the observer, we use the formula

Hz f vu

f 1.97)100(0.340 / 0.101

1

/ 1

1' =

+=

+= .

For the reflected waves, as the source moves toward the vertical wall, the waves

incident on the wall have a shorter wavelength than they would if the source were at

rest. The reflected waves appear to come from a source moving towards him.

Hz f vu

f 103)100(0.340 / 0.101

1

/ 1

1' =

−=

−=

The number of beats per second = (103 – 97.1) Hz = 5.9 Hz.

13.83 General case

Doppler effect for both moving source and observer is concluded in a simple formula:

f vu

vu f

s

o

±=

/ 1

/ 1'

m.

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 Hz Hz f vu

vu f

s

o 7.592)550(343 / 181

343 / 2.71

/ 1

/ 1' =

−

+=

−

+= .

Example

The following 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.


16/28

16

13.84 Supersonic speed and shock waves

What 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

ss v

v

t v

vt ==θ sin .

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.


17/28

17

13.9 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 wave y always have a greater amplitude than

the individual waves y1 and y2?

Answer:

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

positive and sometimes negative. Thus, if y1 is positive at a given time, for example,

and y2 is negative, the sum y1 + 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.

It should also be noted that interference is not limited to waves on a string; all waves

exhibit interference effects. In fact, interference is one of the key characteristics that

define waves.


18/28

18

Suppose the two sources emit waves in phase. At point

A the distance to each source is the same, hence crest

meets crest and constructive interference results. At B

the distance from source 1 is greater than that from

source 2 by half a wavelength. The result is crest

meeting trough and destructive interference. Finally, at

C the distance from source 1 is one wavelength greater

than the distance from source 2. Hence, we find constructive interference at C , and the

waves are in phase again at C. If the sources had been opposite in phase, then A and C

would be points of destructive interference, and B would be a point of constructive

interference.

Remarks:

• The system that when one source emits a crest, the other emits a crest as well

is referred to as synchronized system. The sources are said to be in phase.

• In general, we can say that constructive and destructive interference occur

under the following conditions for two sources that are in phase:

i) Constructive interference occurs when the path length from

the two sources differs by 0, λ, 2λ, 3λ, ….

ii) Destructive interference occurs when the path length from

the two sources differs by λ /2, 3λ /2, 5λ /2, ….


19/28

19

Example

Two speakers separated by a distance of 4.30 m

emit sound of frequency 221 Hz. The speakers

are in phase with one another. A person listens

from a location 2.8 m directly in front of one of

the speakers. Does the person hear constructive

or destructive interference?

Answer:

The wavelength of sound: m Hzsm f v 55.1221 / / 343 / ===λ .

To determine the path difference, d = d 2 – d 1, we need to find d 2 first, and

mmmd Dd 13.5)80.2()30.4( 22212

2 =+=−= .

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

λ . Since the path difference is 3λ /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.

ExampleThe speakers shown below have opposite phase. They are separated by a distance of

5.20 m and emit sound with a frequency of 104 Hz. A person stands 3.00 m in front of

the speakers and 1.30 m to one side of the centerline between them. What type of

interference occurs at the person’s location?


20/28


21/28

21

13.10.1 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 2 L,

or we write L2=λ .

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

frequency of the fundamental, f 1, is determined by

11 )2( f L f v == λ . Therefore,


22/28


23/28

23

13.10.2 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

=

=

λ

λ

The fundamental frequency, f 1, is given by 11 )4( f L f v == λ . Or we can write

L

v f

41 = .

The second harmonic is produced by adding

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

L=4 / 3λ , and hence L3

4=λ . The frequency

is 13)4

(3

3

4 f Lv

Lvv ===λ

.

Similarly, the next-higher harmonic is

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

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

(5

5

4 f

L

v

L

vv===

λ .

Remark:

In general, we have L

v f 4

1 = , 1nf f n = and n Ln / 4=λ , 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:


24/28

24

Remark:

In general, we have L

v f

21 = , 1nf f n = and n Ln / 2=λ , 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 Lv f 4 / 1 = , we have

m Hz

sm f v L 195.0

)0.440(4

/ 3434 / 1 === .

The depth of water to be filled: cmmm L H h 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.

Answer:


25/28

25

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

L

v f

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

while the frequency is increased.

13.11 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 f 1 = 1 /T 1 and f 2 = 1 /T 2, 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

t f At T

A y

t f At T

A y

π π

π π

=

=

=

=

If A = 1,we have the following plots, where 21 y y ytotal += . Mathematically, we have


26/28

26

+

−=

+=

+=

t f f

t f f

A

t f At f A

y y ytotal

22cos

22cos2

)2cos()2cos(

2121

21

21

π π

π π

The first part of the ytotal is

−t

f f A

22cos2 21π 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 f f f beat −= . The rapid oscillations within each beat are due

to the second part of ytotal,

+

t

f f

22cos21

π . 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 a

musical 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.


27/28

27

Example

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 336 Hz or 444

Hz. After water is added, possible frequency of the bottle is either 335 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.

Example

A sonometer wire of length 0.50 m and mass per unit length 1.0 ×10−3 kg m−1 is

stretched by a load of 4.0 kg. If it is plucked at its mid-point, what will be (a) the

wavelength and (b) the frequency, of the note emitted? Take g = 10 N kg−1

.

Answer:

(a) The wire vibrates and emits it fundamental frequency f of wavelength λ . If l is

the length of the wire then2

λ =l gives mml 0.150.022 =×==λ .


28/28

(b) The fundamental frequency f emitted by a wire of length l, mass per unit

length µ and tension T is given by µ

T

l f

2

1= as ( )l

T v f 2 /

µ λ == .

Now T = 4.0 ×10 N kg−1 = 40 N, therefore

Hzmkg

N

m f 200

/ 100.1

40

50.02

13

=××

=−

.

Documents

HKPhO_13vbnvbnvbnv bnv bnvbn