ch16 sound and waves

Chapter 16

Waves and Sound

SEMESTER 1: PHYSICS

Date/weekLearning unit

Theme/Topic Teaching Method

SEMESTER 1

1 -2

Learning unit 1

Dynamics of uniform circular motion ( Chapter 5)

Lecture/ Question and answer/ Whole Class Discussions

3-4Learning unit 2 Waves and sound (Chapter 16)

Lecture/ WCD/ CS/ Practical

5-6Learning unit 3 Electromagnetic waves (Chapter 24)


7Learning unit 4 Particles and waves (Chapter 29)


8-9Learning unit 5 DC circuits (Chapter 20)


10-11 Learning unit 6 Magnetic Forces and Magnetic Fields (Chapter 21)


12-13Learning unit 7 Electromagnetic Induction (Chapter 22)


MAY- JUNE EXAMINATIONS : PHYSICS

Sections to be covered

• Section 16.1: The nature of waves• Section 16.2: Periodic waves• Section 16.3: The speed of a wave on a string• Section 16.4: Mathematical description of a wave• Section 16.5: The nature of sound• Section 16.6: The speed of sound• Section 16.9: The Doppler Effect

16.1 The Nature of Waves

1. A wave is a traveling disturbance.

2. A wave carries energy from place to place.


longitudinal


Longitudinal Wave


Water waves are partially transverse and partially longitudinal.

16.2 Periodic Waves

Periodic waves consist of cycles or patterns that are produced over and over again by the source.

In the figures, every segment of the slinky vibrates in simple harmonicmotion, provided the end of the slinky is moved in simple harmonicmotion.

16.2 Periodic Waves

In the drawing, one cycle is shaded in color.

The amplitude A is the maximum excursion of a particle of the medium fromthe particles undisturbed position.

The wavelength is the horizontal length of one cycle of the wave.

The period is the time required for one complete cycle.

The frequency is related to the period and has units of Hz, or s-1.

Tf

1

16.2 Periodic Waves

f

Tv

16.2 Periodic Waves

Example 1 The Wavelengths of Radio Waves

AM and FM radio waves are transverse waves consisting of electric andmagnetic field disturbances traveling at a speed of 3.00x108m/s. A stationbroadcasts AM radio waves whose frequency is 1230x103Hz and an FM radio wave whose frequency is 91.9x106Hz. Find the distance between adjacent crests in each wave.

f

Tv

f

v

16.2 Periodic Waves

AM m 244Hz101230

sm1000.33

8

f

v

FM m 26.3Hz1091.9

sm1000.36

8

f

v

16.3 The Speed of a Wave on a String

The speed at which the wave moves to the right depends on how quicklyone particle of the string is accelerated upward in response to the net pulling force.

Lm

Fv

tension

linear density


Example 2 Waves Traveling on Guitar Strings

Transverse waves travel on each string of an electric guitar after thestring is plucked. The length of each string between its two fixed endsis 0.628 m, and the mass is 0.208 g for the highest pitched E string and3.32 g for the lowest pitched E string. Each string is under a tension of 226 N. Find the speeds of the waves on the two strings.

sm826m 0.628kg100.208

N 2263-

Lm

Fv

sm207m 0.628kg103.32

N 2263-

Lm

Fv

High E

Low E


To think about

Is the speed of a transverse wave on a string the same as the speed at which a particle on the string moves?

16.4 The Mathematical Description of a Wave

What is the displacement y at time t of a particle located at x?

The quantity in brackets is called the phase angle measured in radians

A particle located at a distance x also exhibits the SHM but with phase angle given by

Your Turn

SolutionSOLUTION The dimensionless term 2 f t in Equation 16.4 corresponds to the term 8.2t in the given wave equation. The time t is measured in seconds (s), so in order for the quantity 8.2t to be dimensionless, the units of the numerical factor 8.2 must be s−1 = Hz, and we have that

Similarly, the dimensionless term 2x/ in Equation 16.4 corresponds to the term 0.54x in the mathematical description of this wave. Because x is measured in meters (m), the term 0.54x is dimensionless if the numerical factor 0.54 has units of m−1. Thus,

Your Turn

Answer: 161 m/s

Tutorial exercises 26 02 2013

Tutorial Exercises 26 February 2013

16.5 The Nature of Sound Waves

LONGITUDINAL SOUND WAVES


The distance between adjacent condensations is equal to the wavelength of the sound wave.


THE FREQUENCY OF A SOUND WAVE

The frequency is the number of cyclesper second.

A sound with a single frequency is calleda pure tone.

The brain interprets the frequency in termsof the subjective quality called pitch.


THE PRESSURE AMPLITUDE OF A SOUND WAVE

Loudness is an attribute ofa sound that depends primarily on the pressure amplitudeof the wave.

16.6 The Speed of Sound

Sound travels through gases, liquids, and solids at considerablydifferent speeds.


In a gas, it is only when molecules collide that the condensations andrarefactions of a sound wave can move from place to place.

Ideal Gasm

kTv

m

kTvrms

3

KJ1038.1 23k

gasesdiatomicidealfor5

7 and

gases ideal monoatomic idealfor 3

5

Do Example 4: The physics of ultrasonic Ruler


Conceptual Example 5 Lightning, Thunder, and a Rule of Thumb

There is a rule of thumb for estimating how far away a thunderstorm is.After you see a flash of lighting, count off the seconds until the thunder is heard. Divide the number of seconds by five. The result gives theapproximate distance (in miles) to the thunderstorm. Why does thisrule work?


LIQUIDS SOLID BARS

adB

v Y

v

Where B and Y are Bulk and Young’s modulus defined in section 10.7

16.9 The Doppler Effect

The Doppler effect is the change in frequency or pitchof the sound detected byan observer because the soundsource and the observer havedifferent velocities with respectto the medium of sound propagation.

The Doppler effect is the apparent change in frequency and wavelength of a wave when the observer and the source of the wave move relative to each other.We experience the Doppler effect quite often in our lives, without realizing that it is science taking place. For example, the changing sound of a taxi hooter or ambulance as it drives past are the most common examples.


MOVING SOURCE

Tvs

sssso fvfv

v

Tv

vvf

vv

ffs

so 1

1


vv

ffs

so 1

1source movingtoward a stationaryobserver

source movingaway from a stationaryobserver

vv

ffs

so 1

1


Example 10 The Sound of a Passing Train

A high-speed train is traveling at a speed of 44.7 m/s when the engineersounds the 415-Hz warning horn. The speed of sound is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a personstanding at the crossing, when the train is (a) approaching and (b) leavingthe crossing?

vv

ffs

so 1

1

vv

ffs

so 1

1


Hz 4771

1Hz 415

sm343sm7.44

of

approaching

leaving

Hz 3671

1Hz 415

sm343sm7.44

of


MOVING OBSERVER

v

vf

f

vf

vff

os

s

os

oso

1

1


v

vff o

so 1

v

vff o

so 1

Observer movingtowards stationarysource

Observer movingaway from stationary source


v

vvv

ffs

o

so

1

1

GENERAL CASE

Numerator: plus sign applies when observer moves towards the source

Denominator: minus sign applies when source moves towards the observer

16.10 Applications of Sound in Medicine

By scanning ultrasonic waves across the body and detecting the echoesfrom various locations, it is possible to obtain an image.


Ultrasonic sound waves causethe tip of the probe to vibrate at23 kHz and shatter sections ofthe tumor that it touches.


When the sound is reflected from the red blood cells, itsfrequency is changed in a kind of Doppler effect becausethe cells are moving.

Class Exercises 28 02 201376. A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1500 Hz. The bird-watcher, however, hears a frequency of 1560 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

REASONING:• The observer of the sound (the bird-watcher) is stationary, while the source (the

bird) is moving toward the observer. • Therefore, the Doppler-shifted observed frequency is given by Equation 16.11. • This expression can be solved to give the ratio of the bird’s speed to the speed of

sound, from which the desired percentage follows directly. • The observed frequency fo is related to the frequency fs of the source, and the

ratio of the speed of the source vs to the speed of sound v by:

Solving for gives

Hence ,the ratio corresponds to 3.8 %

77. From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.75 times as small as the frequency emitted by the car when it is stationary. The speed of sound is 343 m/s. What is the speed of the car?

.REASONING :Since you detect a frequency that is smaller than that emitted by the car when the car is stationary, the car must be moving away from you. Therefore, according to Equation 16.12, the frequency fo heard by a stationary observer from a source moving away from the observer is given by:

where fs is the frequency emitted from the source when it is stationary with respect to the observer, v is the speed of sound, and vs is the speed of the moving source. This expression can be solved for vs .

Solving for vs and noting that

We get

80. The security alarm on a parked car goes off and produces a frequency of 1000 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 100 Hz. At what speed are you driving?

REASONING The observed frequency changes because of the Doppler effect. As you drive toward the parked car (a stationary source of sound), the Doppler effect is that given by Equation 16.13. As you drive away from the parked car, Equation 16.14 applies.

o, toward s o o, away s o

Driving toward parked car Driving away from parked car

1 / and 1– /f f v v f f v v

Subtracting the equation on the right from the one on the left gives the change in the observed frequency

o, toward o, away s o– 2 /f f f v v

Solving for the observer’s speed (which is your speed), we obtain:

o, toward o, awayo

s

– 343 m/s 100 Hz17 m/s

2 2 1000 Hz

v f fv

f

78. Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 10 m/s, the marine biologist measures the number of clicks occurring per second to be at a frequency of 2300 Hz. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?REASONING

• The dolphin is the source of the clicks, and emits them at a frequency fs. The marine biologist measures a lower, Doppler-shifted click frequency fo, because the dolphin is swimming directly away.

• The difference between the frequencies is the source frequency minus the observed frequency: fs − fo. • We will use the equation :

o ss

1

1f f

v

v

where vs is the speed of the dolphin and v is the speed of sound in seawater, to determine the difference between the frequencies.

ss o 1

vf f

v

Solving for fs we get:

Therefore, the difference between the source and observed frequencies is:

s ss o o o o

so

1 1 1

10.0 m/s2300 Hz 15 Hz

1522 m/s

v vf f f f f

v v

vf

v

Tutorial Exercises Due on Monday: the 4th of March 2013Chapter 16: Problem number, 81, 82, 86 and 87 :

Chapter 24

Electromagnetic Waves

What are electromagnetic waves

• Electromagnetic waves consist of a combination of oscillating electrical and magnetic fields, perpendicular to each other.

This is difficult to visualize, however the waveform has similar characteristics of other types of waves.

What are electromagnetic waves

• Although they seem different, radio waves, microwaves, x-rays, and even visible light are all electromagnetic waves. They are part of the electromagnetic spectrum, and each has a different range of wavelengths, which cause the waves to affect matter differently.

• The creation and detection of the wave depend much on the range of wavelengths.

Questions you may have include:

• What is the electromagnetic spectrum?• What are the characteristics of

electromagnetic waves?• How are these waves created and

detected?

Electromagnetic spectrum• The range of wavelengths for electromagnetic waves--from the very long to

the very short--is called the Electromagnetic Spectrum:

Electromagnetic Spectrum• Radio and TV waves are the longest usable waves, having a wavelength of

1 mile (1.5 kilometer) or more.• Microwaves are used in telecommunication as well as for cooking food.• Infrared waves are barely visible. They are the deep red rays you get from

a heat lamp.• Visible light waves are the radiation you can see with your eyes. Their

wavelengths are in the range of 1/1000 centimeter.• Ultraviolet rays are what give you sunburn and are used in "black lights"

that make object glow.• X-rays go through the body and are used for medical purposes.• Gamma rays are dangerous rays coming from nuclear reactors and atomic

bombs. They have the shortest wavelength in the electromagnetic spectrum of about 1/10,000,000 centimeter.

http://www.school-for-champions.com/science/electromagnetic_waves.htm


Properties

• They do not need a medium for transmission. Other waves, such as sound waves, can not travel through a vacuum. An electromagnetic wave is perfectly happy to do that.

• Electromagnetic waves are transverse waves, similar to water waves in the ocean or the waves seen on a guitar

string.• The velocity of electromagnetic waves in a vacuum is

approximately 186,000 miles per second or 300,000 kilometers per second, the same as the speed of light. When these waves pass through matter, they slow down slightly, according to their wavelength.



The speed of light

• In 1865, Maxwell determined theoretically that electromagnetic waves propagate through a vacuum at a speed given by:

They all obey

• Electromagnetic waves are split into different categories based on their frequency (or, equivalently, on their wavelength).

• In other words, we split up the electromagnetic spectrum based on frequency.

• Visible light, for example, ranges from violet to red. • Violet light has a wavelength of 400 nm, and a frequency of 7.5 x

1014 Hz.• Red light has a wavelength of 700 nm, and a frequency of 4.3 x 1014

Hz.• Any electromagnetic wave with a frequency (or wavelength)

between those extremes can be seen by humans.

Properties continued…

• An electromagnetic wave, although it carries no mass, does carry energy. It also has momentum, and can exert pressure (known as radiation pressure)

• The energy carried by an electromagnetic wave is proportional to the frequency of the wave.

24.2 The Electromagnetic Spectrum

The Wavelength of Visible Light

Find the range in wavelengths for visible light in the frequency rangebetween 4.0x1014Hz and 7.9x1014Hz.

nm 750m105.7Hz104.0

sm1000.3 714

8

f

c

nm 380m108.3Hz107.9

sm1000.3 714

8

f

c

Creating an electromagnetic wave

• From high school we already learned how moving charges (currents) produce magnetic fields.

• A constant current produces a constant magnetic field, while a changing current produces a changing field.

• We can go the other way, and use a magnetic field to produce a current, as long as the magnetic field is changing.

• This is what induced emf is all about. A steadily-changing magnetic field can induce a constant voltage, while an oscillating magnetic field can induce an oscillating voltage.

To Note:• an oscillating electric field generates an oscillating magnetic field • an oscillating magnetic field generates an oscillating electric field • What this means in practice is that the source has created oscillating electric and

magnetic fields, perpendicular to each other, that travel away from the source. • The E and B fields, along with being perpendicular to each other, are perpendicular

to the direction the wave travels, meaning that an electromagnetic wave The energy of the wave is stored in the electric and magnetic fields

Energy in an electromagnetic wave

• The energy in an electromagnetic wave is tied up in the electric and magnetic fields.

• In general, the energy per unit volume in an electric field is given by:

And the magnetic energy density:

In an electromagnetic wave propagating through a vacuum or air, the electric field and the magnetic field carry equal amounts of energy per unit volume of space.

24.5 The Doppler Effect and Electromagnetic Waves

Electromagnetic waves also can exhibit a Doppler effect, but itdiffers for two reasons:

a) Sound waves require a medium, whereas electromagneticwaves do not.

b) For sound, it is the motion relative to the medium that is important. For electromagnetic waves, only the relative motion of the sourceand observer is important.

cvc

vff so

rel

rel if 1

Sign Conventions for Relative Motion

In this expression, is the observed frequency, and is the frequency emitted by the source. The symbol stands for the speed of the source and the observer relative to one another, and c is the speed of light in a vacuum. Equation 24.6 applies only if is very small compared to —that is, if .

It is essential to realize that is the relative speed of the source and the observer. Thus, if the source is moving due east at a speed of 28 m/s with respect to the earth, while the observer is moving due east at a speed of 22 m/s, the value for is /

Plus Sign (source and observer come together)

Minus sign (source and observer move apart)

1. The source is catching up with the observer

2. The observer is catching up with the source

3. The source and the observer move toward one another

1. The source is pulling away from the observer

2. The observer is pulling away from the source

3. The source and the observer both move away from one another

Focus on Concepts Problem 6The drawing shows four situations—A, B, C, and D—in which an observer and a source of electromagnetic waves can move along the same line. In each case the source emits a wave of the same frequency, and in each case only the source or the observer is moving. The arrow in each situation denotes the velocity vector, which has the same magnitude in each situation. When there is no arrow, the observer or the

source is stationary. Rank the frequencies of the observed electromagnetic waves in descending order (largest first) according to magnitude.

(a) A and B (a tie), C and D (a tie)

(b) C and D (a tie), A and B (a tie)

(c) A and D (a tie), B and C (a tie)

(d) B and D (a tie), A and C (a tie)

(e) B and C (a tie), A and D (a tie)

Problem 1

The team monitoring a space probe exploring the outer solar system finds that radio transmissions from the probe take 3.82 hours to reach earth. How distant (in meters) is the probe?Solution:

The reasoning part: The distance d between earth and the probe is determined by (Equation 2.1), where is the speed of light in a vacuum and t is the time for the radio signal to reach earth.

The calculation part:

The elapsed time t is given in hours, so it must be converted to seconds:

The distance, then, between earth and the probe is

Problem 5

In a traveling electromagnetic wave, the electric field is represented mathematically as

where is the maximum field strength.

(a)What is the frequency of the wave?

(b)This wave and the wave that results from its reflection can form a standing wave, in a way similar to that in which standing waves can arise on a string. What is the separation between adjacent nodes in the standing wave?

Solution:

REASONING According to Equation 16.3, the displacement y of a wave that travels in the +x direction and has amplitude A, frequency f, and wavelength is given by:

For this equation, with y = E, applies to the traveling electromagnetic wave in the problem, which is represented mathematically as

As is the maximum field strength, it represents the amplitude A of the wave. We can find the frequency and wavelength of this electromagnetic wave by comparing the mathematical form of the electric field with Equation 16.3.

Problem 9In a certain UHF radio wave, the shortest distance between positions at which the electric and magnetic fields are zero is 0.34 m. Calculate the frequency of this UHF radio wave.

The reasoning part:The frequency of the UHF wave is related to its wavelength by

The electric and magnetic fields are both zero at the same positions, which are separated by a distance d equal to half a wavelength.

We can express the wavelength in terms of the distance between adjacent positions of zero field as:

The calculation part:

Substituting Equation (1) into Equation (2), we obtain:

2d (1)

2d (1)

Solving c f (Equation 16.1) for f yields

cf

(2)

Problem 38

Reasoning part:The observed frequency is given by:

The frequency fs emitted by the source is the same in each case, so that only the direction of the relative motion and the relative speed determine the observed frequency.

In situations A and B the observer and the source move away from each other, and the minus sign in Equation 24.6 applies. In situation C the observer and the source move toward each other, and the plus sign applies. Thus, the observed frequency is largest in C.

To distinguish between A and B, we note that the relative speed in A is , whereas in B the relative speed is . The greater relative speed means that the term is greater in B than in A, and since the minus sign applies, the observed frequency is more reduced in B than in A.

[Situation A, minus sign in Equation 24.6]

Calculation part[Situation A, minus sign in Equation 24.6]

rel

614 14rel

o s s 8

2

1.50 10 m/s1 1 4.57 10 Hz 1 4.55 10 Hz

3.00 10 m/s

v v v v

v vf f f

c c

[Situation B, minus sign in Equation 24.6]

rel

614 14rel

o s s 8

2 3

3 1.50 10 m/s31 1 4.57 10 Hz 1 4.50 10 Hz

3.00 10 m/s

v v v v

v vf f f

c c

[Situation C, plus sign in Equation 24.6]

rel

614 14rel

o s s 8

2

2 1.50 10 m/s21 1 4.57 10 Hz 1 4.62 10 Hz

3.00 10 m/s

v v v v

v vf f f

c c


Your Chance: Study Example: Radar Guns and Speed Traps

Police use radar guns and the Doppler effect to catch speeders. A moving car approaches a stationary police car. A radar gun emits an electromagnetic waves that reflects form the oncoming car. The reflected Wave returns to the police car with a frequency measured by on-board equipment that is different from the emitted frequency. One such radar emits a wave whose frequency is 8.0x109Hz. When the speed is 39m/s and the approach is essentially head on, what is the difference between the frequency of the wave returning to the police car and that emitted by the radar gun.


1 rel

c

vff so

1 rel

c

vff oo

frequency “observed”by speeding car

frequency observedby police car

soso fc

vfff

rel1

c

vfff sso

rel2

ss fc

v

c

vf

relrel 11

ss fc

vf

rel21

c

v

c

vf s

relrel 2

The value v rel/c is which is very small compared to 2 so that approaches 2

c

vrel2

)(Hz

C Y U 7 page 24.5• An astronomer measures a Doppler change in frequency for light reaching

the earth from a distant star, from this measurement, can the astronomer tell whether the star is moving away from the earth or the earth moving away from the star?

• Answer: No The same Doppler change results when the star moves away from the earth and when the earth moves away from the star

ch16 sound and waves

Documents