What Yoursquoll Learnbull You will describe sound in

terms of wave properties and behavior

bull You will examine some ofthe sources of sound

bull You will explain propertiesthat differentiate betweenmusic and noise

Why Itrsquos ImportantSound is an importantmeans of communicationand in the form of musiccultural expression

Musical Groups A smallmusical group might containtwo or three instrumentswhile a marching band cancontain 100 or more Theinstruments in these groupsform sounds in differentways but they can createexciting music when theyare played together

Think About This How do the instruments in a musical group createthe sounds that you hearWhy do various instrumentssound different even whenthey play the same note

402

physicsppcom

Tim Fuller

402-410 CH15-S1-845813 33004 653 AM Page 402

How can glasses produce musical notesQuestionHow can you use glasses to produce different musical notes and how do glasses with stems compare to those without stems

Procedure

1 Select a stemmed glass with a thin rim2 Prepare Carefully inspect the top edge of

the glass for sharp edges Notify your teacherif you observe any sharp edges Be sure torepeat this inspection every time you select a different glass

3 Place the glass on the table in front of youFirmly hold the base with one hand Wet yourfinger and slowly rub it around the top edgeof the glass CAUTION Glass is fragileHandle carefully

4 Record your observations Increase ordecrease the speed a little What happens

5 Select a stemmed glass that is larger orsmaller than the first glass Repeat steps 2-4

6 Select a glass without a stem and repeatsteps 2ndash4

Analysis

Summarize your observations Which glassesmdashstemmed not stemmed or bothmdashwere able toproduce ringing tones What factors affected the tones producedCritical ThinkingPropose a methodfor producingdifferent notes from the sameglass Test yourproposed methodSuggest a test tofurther investigatethe properties of glasses that can produceringing tones

151 Properties and Detection of Sound

Objectivesbull Demonstrate the

properties that soundshares with other waves

bull Relate the physicalproperties of sound wavesto our perception of sound

bull Identify some applicationsof the Doppler effect

Vocabulary

sound wavepitchloudnesssound leveldecibelDoppler effect

Sound is an important part of existence for many living thingsAnimals can use sound to hunt attract mates and warn of the

approach of predators In humans the sound of a siren can heighten ourawareness of our surroundings while the sound of music can soothe andrelax us From your everyday experiences you already are familiar with several of the characteristics of sound including volume tone and pitchWithout thinking about it you can use these and other characteristics tocategorize many of the sounds that you hear for example some sound pat-terns are characteristic of speech while others are characteristic of a musicalgroup In this chapter you will study the physical principles of soundwhich is a type of wave

In Chapter 14 you learned how to describe waves in terms of speed fre-quency wavelength and amplitude You also discovered how waves interactwith each other and with matter Knowing that sound is a type of wave allowsyou to describe some of its properties and interactions First however thereis a question that you need to answer exactly what type of wave is sound

Section 151 Properties and Detection of Sound 403Horizons Companies

402-410 CH15-S1-845813 33004 656 AM Page 403

PPP-403

null

7418787

Figure 15-1 Before the bell isstruck the air around it is a regionof average pressure (a) Once the bell is struck however thevibrating edge creates regions ofhigh and low pressure The darkareas represent regions of higherpressure the light areas representregions of lower pressure (b) Forsimplicity the diagram shows theregions moving in one direction in reality the waves move outfrom the bell in all directions

Sound WavesPut your fingers against your throat as you hum or speak Can you feel

the vibrations Have you ever put your hand on the loudspeaker of a boombox Figure 15-1 shows a vibrating bell that also can represent your vocalcords a loudspeaker or any other sound source As it moves back andforth the edge of the bell strikes the particles in the air When the edgemoves forward air particles are driven forward that is the air particlesbounce off the bell with a greater velocity When the edge moves backwardair particles bounce off the bell with a lower velocity

The result of these velocity changes is that the forward motion of thebell produces a region where the air pressure is slightly higher than averageThe backward motion produces slightly below-average pressure Collisionsamong the air particles cause the pressure variations to move away fromthe bell in all directions If you were to focus at one spot you would seethe value of the air pressure rise and fall not unlike the behavior of a pen-dulum In this way the pressure variations are transmitted through matter

Describing sound A pressure variation that is transmitted through matteris a sound wave Sound waves move through air because a vibratingsource produces regular variations or oscillations in air pressure The air

particles collide transmitting the pressure variations away from the source of the soundThe pressure of the air oscillates about the meanair pressure as shown in Figure 15-2 The fre-quency of the wave is the number of oscillationsin pressure each second The wavelength is thedistance between successive regions of high orlow pressure Because the motion of the parti-cles in air is parallel to the direction of thewaversquos motion sound is a longitudinal wave

The speed of sound in air depends on thetemperature with the speed increasing by about06 ms for each 1degC increase in air temperature At room temperature (20degC) sound movesthrough air at sea level at a speed of 343 msSound also travels through solids and liquidsIn general the speed of sound is greater in

Figure 15-2 A coiled springmodels the compressions andrarefactions of a sound wave (a)The pressure of the air rises andfalls as the sound wave propagatesthrough the atmosphere (b)You can use a sine curve alone to model changes in pressureNote that the positions of x y andz show that the wave not mattermoves forward (c)

a

b

a b c

404 Chapter 15 Sound

402-410 CH15-S1-845813 61004 515 PM Page 404

PPP-404

null

1460231

>

solids and liquids than in gases Table 15-1 lists the speeds of sound wavesin various media Sound cannot travel in a vacuum because there are noparticles to collide

Sound waves share the general properties of other waves For examplethey reflect off hard objects such as the walls of a room Reflected soundwaves are called echoes The time required for an echo to return to thesource of the sound can be used to find the distance between the sourceand the reflective object This principle is used by bats by some camerasand by ships that employ sonar Two sound waves can interfere causingdead spots at nodes where little sound can be heard As you learned inChapter 14 the frequency and wavelength of a wave are related to thespeed of the wave by the equation vf

Table 15-1Speed of Soundin Various Media

Medium ms

Air (0deg)Air (20deg)Helium (0deg)Water (25deg)Seawater (25deg)Copper (25deg)Iron (25deg)

331

343

972

1493

1533

3560

5130

1 Find the wavelength in air at 20degC of an 18-Hz sound wave which isone of the lowest frequencies that is detectable by the human ear

2 What is the wavelength of an 18-Hz sound wave in seawater at 25degC

3 Find the frequency of a sound wave moving through iron at 25degCwith a wavelength of 125 m

4 If you shout across a canyon and hear the echo 080 s later howwide is the canyon

5 A 2280-Hz sound wave has a wavelength of 0655 m in an unknownmedium Identify the medium

Detection of Pressure WavesSound detectors convert sound energymdashthe kinetic energy of the vibrat-

ing air particlesmdashinto another form of energy A common detector is amicrophone which converts sound waves into electrical energy A micro-phone consists of a thin disk that vibrates in response to sound waves andproduces an electrical signal You will learn about this transformationprocess in Chapter 25 during your study of electricity and magnetism

The human ear As shown in Figure 15-3 the human ear is a detectorthat receives pressure waves and converts them to electrical impulsesSound waves entering the auditory canal cause vibrations of the tympanicmembrane Three tiny bones then transfer these vibrations to fluid in the cochlea Tiny hairs lining the spiral-shaped cochlea detectcertain frequencies in the vibrating fluid These hairs stimulatenerve cells which send impulses to the brain and produce thesensation of sound

The ear detects sound waves over a wide range of frequenciesand is sensitive to an enormous range of amplitudes In addi-tion human hearing can distinguish many different qualities of sound Knowledge of both physics and biology is required to understand the complexities of the ear The interpretation of sounds by the brain is even more complex and it is nottotally understood

Auditorycanal

Auditorynerve

Cochlea

MalleusIncus

Stapes

Tympanic membrane(eardrum)

Figure 15-3 The human ear is a complex sense organ thattranslates sound vibrations intonerve impulses that are sent tothe brain for interpretation Themalleus incus and stapes are the three bones of the middle earthat sometimes are referred to asthe hammer anvil and stirrup

Section 151 Properties and Detection of Sound 405

402-410 CH15-S1-845813 6404 944 AM Page 405

PPP-405

null

14633661

406 Chapter 15 Sound

10 dBBarely

audible

30 dBWhisper1 m away

50 dBCasual

conversation

80 dBAlarmclock

110 dBRock

concert

140 dBJet engine

70 dBHeavytraffic

100 dBSiren

Figure 15-4 This decibel scaleshows the sound levels of somefamiliar sounds

Perceiving SoundPitch Marin Mersenne and Galileo first determined that the pitch we heardepends on the frequency of vibration Pitch can be given a name on themusical scale For instance the middle C note has a frequency of 262 Hz Theear is not equally sensitive to all frequencies Most people cannot hear soundswith frequencies below 20 Hz or above 16000 Hz Older people are lesssensitive to frequencies above 10000 Hz than are young people By age 70most people cannot hear sounds with frequencies above 8000 Hz Thisloss affects the ability to understand speech

Loudness Frequency and wavelength are two physical characteristics ofsound waves Another physical characteristic of sound waves is amplitudeAmplitude is the measure of the variation in pressure along a wave Inhumans sound is detected by the ear and interpreted by the brain Theloudness of a sound as perceived by our sense of hearing depends pri-marily on the amplitude of the pressure wave

The human ear is extremely sensitive to pressure variations in soundwaves which is the amplitude of the wave Recall from Chapter 13 that 1 atm of pressure equals 101105 Pa The ear can detect pressure-waveamplitudes of less than one-billionth of an atmosphere or 2105 Pa Atthe other end of the audible range pressure variations of approximately 20 Pa or greater cause pain It is important to remember that the ear detectsonly pressure variations at certain frequencies Driving over a mountainpass changes the pressure on your ears by thousands of pascals but thischange does not take place at audible frequencies

Because humans can detect a wide range in pressure variations theseamplitudes are measured on a logarithmic scale called the sound levelThe unit of measurement for sound level is the decibel (dB) The soundlevel depends on the ratio of the pressure variation of a given sound waveto the pressure variation in the most faintly heard sound 2105 Pa Suchan amplitude has a sound level of 0 dB A sound with a pressure amplitudeten times larger (2104 Pa) is 20 dB A pressure amplitude ten timeslarger than this is 40 dB Most people perceive a 10-dB increase in soundlevel as about twice as loud as the original level Figure 15-4 shows thesound level for a variety of sounds In addition to pressure variationspower and intensity of sound waves can be described by decibel scales

Exposure to loud sounds in the form of noise or music has been shownto cause the ear to lose its sensitivity especially to high frequencies Thelonger a person is exposed to loud sounds the greater the effect A personcan recover from short-term exposure in a period of hours but the effects

402-410 CH15-S1-845813 33004 658 AM Page 406

PPP-406

null

1996316

ObserverB

ObserverA

S

vs

λB

λA

of long-term exposure can last for days or weeks Long exposure to 100-dBor greater sound levels can produce permanent damage Many rock musi-cians have suffered serious hearing loss some as much as 40 percentHearing loss also can result from loud music being transmitted to stereoheadphones from personal radios and CD players In some cases the lis-teners are unaware of just how high the sound levels really are Cottonearplugs reduce the sound level only by about 10 dB Special ear inserts canprovide a 25-dB reduction Specifically designed earmuffs and inserts asshown in Figure 15-5 can reduce the sound level by up to 45 dB

Loudness as perceived by the human ear is not directly proportional tothe pressure variations in a sound wave The earrsquos sensitivity depends onboth pitch and amplitude Also perception of pure tones is different fromperception of a mixture of tones

The Doppler EffectHave you ever noticed that the pitch of an ambulance fire or police siren

changed as the vehicle sped past you The pitch was higher when the vehiclewas moving toward you then it dropped to a lower pitch as the source movedaway This frequency shift is called the Doppler effect and is shown inFigure 15-6 The sound source S is moving to the right with a speed of vsThe waves that it emits spread in circles centered on the source at the time itproduced the waves As the source moves toward the sound detectorObserver A in Figure 15-6a more waves are crowded into the space betweenthem The wavelength is shortened to A Because the speed of sound is notchanged more crests reach the ear per second which means that the fre-quency of the detected sound increases When the source is moving awayfrom the detector Observer B in Figure 15-6a the wavelength is lengthenedto B and the detected frequency is lower Figure 15-6b illustrates theDoppler effect for a moving source of sound on water waves in a ripple tank

A Doppler shift also occurs if the detector is moving and the source isstationary In this case the Doppler shift results from the relative velocityof the sound waves and the detector As the detector approaches the sta-tionary source the relative velocity is larger resulting in an increase in thewave crests reaching the detector each second As the detector recedes fromthe source the relative velocity is smaller resulting in a decrease in thewave crests reaching the detector each second

Figure 15-5 Continuousexposure to loud sounds cancause serious hearing loss Inmany occupations workers such as this flight controller must wear ear protection

Figure 15-6 As a sound-producing source moves towardan observer the wavelength isshortened to A the wavelength is B for waves produced by asource moving away from anobserver (a) A moving wave-producing source illustrates theDoppler effect in a ripple tank (b)

a b

Section 151 Properties and Detection of Sound 407(t)Willie MaldonadGetty Images (b)Max KurtzZtek

402-410 CH15-S1-845813 33004 659 AM Page 407

PPP-407

null

17627586

408 Chapter 15 Sound

Reducing Equations When an element in a complex equation is equal tozero the equation might reduce to a form that is easier to use

fd fs(vv

vvd

s) fd fs(vv

vvd

s)

fs(v v

vs) fs(v

vvd)

fs( ) fs( ) fs( ) fs( )

fs(1 vvd)

1 vvd

1

11

vvs

vv

vvd

vv

vv

vv

vvs

Stationary detector source inmotion vd 0

Stationary source detector inmotion vs 0

For both a moving source and a moving observer the frequency that theobserver hears can be calculated using the equation below

In the Doppler effect equation v is the velocity of the sound wave vd isthe velocity of the detector vs is the velocity of the soundrsquos source fs is thefrequency of the wave emitted by the source and fd is the frequencyreceived by the detector This equation applies when the source is movingwhen the observer is moving and when both are moving

As you solve problems using the above equation be sure to define thecoordinate system so that the positive direction is from the source to thedetector The sound waves will be approaching the detector from the sourceso the velocity of sound is always positive Try drawing diagrams to confirmthat the term (v vd)(v vs) behaves as you would predict based on whatyou have learned about the Doppler effect Notice that for a source movingtoward the detector (positive direction which results in a smaller denom-inator compared to a stationary source) and for a detector moving towardthe source (negative direction and increased numerator compared to a stationary detector) the detected frequency fd increases Similarily if thesource moves away from the detector or if the detector moves away fromthe source then fd decreases Read the Connecting Math to Physics featurebelow to see how the Doppler effect equation reduces when the source orobserver is stationary

Doppler Effect fd fsvv

vvd

s

The frequency perceived by a detector is equal to the velocity of the detectorrelative to the velocity of the wave divided by the velocity of the sourcerelative to the velocity of the wave multiplied by the waversquos frequency

402-410 CH15-S1-845813 33004 659 AM Page 408

PPP-408

null

12439261

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

Figure 15-1 Before the bell isstruck the air around it is a regionof average pressure (a) Once the bell is struck however thevibrating edge creates regions ofhigh and low pressure The darkareas represent regions of higherpressure the light areas representregions of lower pressure (b) Forsimplicity the diagram shows theregions moving in one direction in reality the waves move outfrom the bell in all directions

Sound WavesPut your fingers against your throat as you hum or speak Can you feel

the vibrations Have you ever put your hand on the loudspeaker of a boombox Figure 15-1 shows a vibrating bell that also can represent your vocalcords a loudspeaker or any other sound source As it moves back andforth the edge of the bell strikes the particles in the air When the edgemoves forward air particles are driven forward that is the air particlesbounce off the bell with a greater velocity When the edge moves backwardair particles bounce off the bell with a lower velocity

The result of these velocity changes is that the forward motion of thebell produces a region where the air pressure is slightly higher than averageThe backward motion produces slightly below-average pressure Collisionsamong the air particles cause the pressure variations to move away fromthe bell in all directions If you were to focus at one spot you would seethe value of the air pressure rise and fall not unlike the behavior of a pen-dulum In this way the pressure variations are transmitted through matter

Describing sound A pressure variation that is transmitted through matteris a sound wave Sound waves move through air because a vibratingsource produces regular variations or oscillations in air pressure The air

particles collide transmitting the pressure variations away from the source of the soundThe pressure of the air oscillates about the meanair pressure as shown in Figure 15-2 The fre-quency of the wave is the number of oscillationsin pressure each second The wavelength is thedistance between successive regions of high orlow pressure Because the motion of the parti-cles in air is parallel to the direction of thewaversquos motion sound is a longitudinal wave

The speed of sound in air depends on thetemperature with the speed increasing by about06 ms for each 1degC increase in air temperature At room temperature (20degC) sound movesthrough air at sea level at a speed of 343 msSound also travels through solids and liquidsIn general the speed of sound is greater in

Figure 15-2 A coiled springmodels the compressions andrarefactions of a sound wave (a)The pressure of the air rises andfalls as the sound wave propagatesthrough the atmosphere (b)You can use a sine curve alone to model changes in pressureNote that the positions of x y andz show that the wave not mattermoves forward (c)

a

b

a b c

404 Chapter 15 Sound

402-410 CH15-S1-845813 61004 515 PM Page 404

PPP-404

null

1460231

>

solids and liquids than in gases Table 15-1 lists the speeds of sound wavesin various media Sound cannot travel in a vacuum because there are noparticles to collide

Sound waves share the general properties of other waves For examplethey reflect off hard objects such as the walls of a room Reflected soundwaves are called echoes The time required for an echo to return to thesource of the sound can be used to find the distance between the sourceand the reflective object This principle is used by bats by some camerasand by ships that employ sonar Two sound waves can interfere causingdead spots at nodes where little sound can be heard As you learned inChapter 14 the frequency and wavelength of a wave are related to thespeed of the wave by the equation vf

Table 15-1Speed of Soundin Various Media

Medium ms

Air (0deg)Air (20deg)Helium (0deg)Water (25deg)Seawater (25deg)Copper (25deg)Iron (25deg)

331

343

972

1493

1533

3560

5130

1 Find the wavelength in air at 20degC of an 18-Hz sound wave which isone of the lowest frequencies that is detectable by the human ear

2 What is the wavelength of an 18-Hz sound wave in seawater at 25degC

3 Find the frequency of a sound wave moving through iron at 25degCwith a wavelength of 125 m

4 If you shout across a canyon and hear the echo 080 s later howwide is the canyon

5 A 2280-Hz sound wave has a wavelength of 0655 m in an unknownmedium Identify the medium

Detection of Pressure WavesSound detectors convert sound energymdashthe kinetic energy of the vibrat-

ing air particlesmdashinto another form of energy A common detector is amicrophone which converts sound waves into electrical energy A micro-phone consists of a thin disk that vibrates in response to sound waves andproduces an electrical signal You will learn about this transformationprocess in Chapter 25 during your study of electricity and magnetism

The human ear As shown in Figure 15-3 the human ear is a detectorthat receives pressure waves and converts them to electrical impulsesSound waves entering the auditory canal cause vibrations of the tympanicmembrane Three tiny bones then transfer these vibrations to fluid in the cochlea Tiny hairs lining the spiral-shaped cochlea detectcertain frequencies in the vibrating fluid These hairs stimulatenerve cells which send impulses to the brain and produce thesensation of sound

The ear detects sound waves over a wide range of frequenciesand is sensitive to an enormous range of amplitudes In addi-tion human hearing can distinguish many different qualities of sound Knowledge of both physics and biology is required to understand the complexities of the ear The interpretation of sounds by the brain is even more complex and it is nottotally understood

Auditorycanal

Auditorynerve

Cochlea

MalleusIncus

Stapes

Tympanic membrane(eardrum)

Figure 15-3 The human ear is a complex sense organ thattranslates sound vibrations intonerve impulses that are sent tothe brain for interpretation Themalleus incus and stapes are the three bones of the middle earthat sometimes are referred to asthe hammer anvil and stirrup

Section 151 Properties and Detection of Sound 405

402-410 CH15-S1-845813 6404 944 AM Page 405

PPP-405

null

14633661

406 Chapter 15 Sound

10 dBBarely

audible

30 dBWhisper1 m away

50 dBCasual

conversation

80 dBAlarmclock

110 dBRock

concert

140 dBJet engine

70 dBHeavytraffic

100 dBSiren

Figure 15-4 This decibel scaleshows the sound levels of somefamiliar sounds

Perceiving SoundPitch Marin Mersenne and Galileo first determined that the pitch we heardepends on the frequency of vibration Pitch can be given a name on themusical scale For instance the middle C note has a frequency of 262 Hz Theear is not equally sensitive to all frequencies Most people cannot hear soundswith frequencies below 20 Hz or above 16000 Hz Older people are lesssensitive to frequencies above 10000 Hz than are young people By age 70most people cannot hear sounds with frequencies above 8000 Hz Thisloss affects the ability to understand speech

Loudness Frequency and wavelength are two physical characteristics ofsound waves Another physical characteristic of sound waves is amplitudeAmplitude is the measure of the variation in pressure along a wave Inhumans sound is detected by the ear and interpreted by the brain Theloudness of a sound as perceived by our sense of hearing depends pri-marily on the amplitude of the pressure wave

The human ear is extremely sensitive to pressure variations in soundwaves which is the amplitude of the wave Recall from Chapter 13 that 1 atm of pressure equals 101105 Pa The ear can detect pressure-waveamplitudes of less than one-billionth of an atmosphere or 2105 Pa Atthe other end of the audible range pressure variations of approximately 20 Pa or greater cause pain It is important to remember that the ear detectsonly pressure variations at certain frequencies Driving over a mountainpass changes the pressure on your ears by thousands of pascals but thischange does not take place at audible frequencies

Because humans can detect a wide range in pressure variations theseamplitudes are measured on a logarithmic scale called the sound levelThe unit of measurement for sound level is the decibel (dB) The soundlevel depends on the ratio of the pressure variation of a given sound waveto the pressure variation in the most faintly heard sound 2105 Pa Suchan amplitude has a sound level of 0 dB A sound with a pressure amplitudeten times larger (2104 Pa) is 20 dB A pressure amplitude ten timeslarger than this is 40 dB Most people perceive a 10-dB increase in soundlevel as about twice as loud as the original level Figure 15-4 shows thesound level for a variety of sounds In addition to pressure variationspower and intensity of sound waves can be described by decibel scales

Exposure to loud sounds in the form of noise or music has been shownto cause the ear to lose its sensitivity especially to high frequencies Thelonger a person is exposed to loud sounds the greater the effect A personcan recover from short-term exposure in a period of hours but the effects

402-410 CH15-S1-845813 33004 658 AM Page 406

PPP-406

null

1996316

ObserverB

ObserverA

S

vs

λB

λA

of long-term exposure can last for days or weeks Long exposure to 100-dBor greater sound levels can produce permanent damage Many rock musi-cians have suffered serious hearing loss some as much as 40 percentHearing loss also can result from loud music being transmitted to stereoheadphones from personal radios and CD players In some cases the lis-teners are unaware of just how high the sound levels really are Cottonearplugs reduce the sound level only by about 10 dB Special ear inserts canprovide a 25-dB reduction Specifically designed earmuffs and inserts asshown in Figure 15-5 can reduce the sound level by up to 45 dB

Loudness as perceived by the human ear is not directly proportional tothe pressure variations in a sound wave The earrsquos sensitivity depends onboth pitch and amplitude Also perception of pure tones is different fromperception of a mixture of tones

The Doppler EffectHave you ever noticed that the pitch of an ambulance fire or police siren

changed as the vehicle sped past you The pitch was higher when the vehiclewas moving toward you then it dropped to a lower pitch as the source movedaway This frequency shift is called the Doppler effect and is shown inFigure 15-6 The sound source S is moving to the right with a speed of vsThe waves that it emits spread in circles centered on the source at the time itproduced the waves As the source moves toward the sound detectorObserver A in Figure 15-6a more waves are crowded into the space betweenthem The wavelength is shortened to A Because the speed of sound is notchanged more crests reach the ear per second which means that the fre-quency of the detected sound increases When the source is moving awayfrom the detector Observer B in Figure 15-6a the wavelength is lengthenedto B and the detected frequency is lower Figure 15-6b illustrates theDoppler effect for a moving source of sound on water waves in a ripple tank

A Doppler shift also occurs if the detector is moving and the source isstationary In this case the Doppler shift results from the relative velocityof the sound waves and the detector As the detector approaches the sta-tionary source the relative velocity is larger resulting in an increase in thewave crests reaching the detector each second As the detector recedes fromthe source the relative velocity is smaller resulting in a decrease in thewave crests reaching the detector each second

Figure 15-5 Continuousexposure to loud sounds cancause serious hearing loss Inmany occupations workers such as this flight controller must wear ear protection

Figure 15-6 As a sound-producing source moves towardan observer the wavelength isshortened to A the wavelength is B for waves produced by asource moving away from anobserver (a) A moving wave-producing source illustrates theDoppler effect in a ripple tank (b)

a b

Section 151 Properties and Detection of Sound 407(t)Willie MaldonadGetty Images (b)Max KurtzZtek

402-410 CH15-S1-845813 33004 659 AM Page 407

PPP-407

null

17627586

408 Chapter 15 Sound

Reducing Equations When an element in a complex equation is equal tozero the equation might reduce to a form that is easier to use

fd fs(vv

vvd

s) fd fs(vv

vvd

s)

fs(v v

vs) fs(v

vvd)

fs( ) fs( ) fs( ) fs( )

fs(1 vvd)

1 vvd

1

11

vvs

vv

vvd

vv

vv

vv

vvs

Stationary detector source inmotion vd 0

Stationary source detector inmotion vs 0

For both a moving source and a moving observer the frequency that theobserver hears can be calculated using the equation below

In the Doppler effect equation v is the velocity of the sound wave vd isthe velocity of the detector vs is the velocity of the soundrsquos source fs is thefrequency of the wave emitted by the source and fd is the frequencyreceived by the detector This equation applies when the source is movingwhen the observer is moving and when both are moving

As you solve problems using the above equation be sure to define thecoordinate system so that the positive direction is from the source to thedetector The sound waves will be approaching the detector from the sourceso the velocity of sound is always positive Try drawing diagrams to confirmthat the term (v vd)(v vs) behaves as you would predict based on whatyou have learned about the Doppler effect Notice that for a source movingtoward the detector (positive direction which results in a smaller denom-inator compared to a stationary source) and for a detector moving towardthe source (negative direction and increased numerator compared to a stationary detector) the detected frequency fd increases Similarily if thesource moves away from the detector or if the detector moves away fromthe source then fd decreases Read the Connecting Math to Physics featurebelow to see how the Doppler effect equation reduces when the source orobserver is stationary

Doppler Effect fd fsvv

vvd

s

The frequency perceived by a detector is equal to the velocity of the detectorrelative to the velocity of the wave divided by the velocity of the sourcerelative to the velocity of the wave multiplied by the waversquos frequency

402-410 CH15-S1-845813 33004 659 AM Page 408

PPP-408

null

12439261

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

solids and liquids than in gases Table 15-1 lists the speeds of sound wavesin various media Sound cannot travel in a vacuum because there are noparticles to collide

Sound waves share the general properties of other waves For examplethey reflect off hard objects such as the walls of a room Reflected soundwaves are called echoes The time required for an echo to return to thesource of the sound can be used to find the distance between the sourceand the reflective object This principle is used by bats by some camerasand by ships that employ sonar Two sound waves can interfere causingdead spots at nodes where little sound can be heard As you learned inChapter 14 the frequency and wavelength of a wave are related to thespeed of the wave by the equation vf

Table 15-1Speed of Soundin Various Media

Medium ms

Air (0deg)Air (20deg)Helium (0deg)Water (25deg)Seawater (25deg)Copper (25deg)Iron (25deg)

331

343

972

1493

1533

3560

5130

1 Find the wavelength in air at 20degC of an 18-Hz sound wave which isone of the lowest frequencies that is detectable by the human ear

2 What is the wavelength of an 18-Hz sound wave in seawater at 25degC

3 Find the frequency of a sound wave moving through iron at 25degCwith a wavelength of 125 m

4 If you shout across a canyon and hear the echo 080 s later howwide is the canyon

5 A 2280-Hz sound wave has a wavelength of 0655 m in an unknownmedium Identify the medium

Detection of Pressure WavesSound detectors convert sound energymdashthe kinetic energy of the vibrat-

ing air particlesmdashinto another form of energy A common detector is amicrophone which converts sound waves into electrical energy A micro-phone consists of a thin disk that vibrates in response to sound waves andproduces an electrical signal You will learn about this transformationprocess in Chapter 25 during your study of electricity and magnetism

The human ear As shown in Figure 15-3 the human ear is a detectorthat receives pressure waves and converts them to electrical impulsesSound waves entering the auditory canal cause vibrations of the tympanicmembrane Three tiny bones then transfer these vibrations to fluid in the cochlea Tiny hairs lining the spiral-shaped cochlea detectcertain frequencies in the vibrating fluid These hairs stimulatenerve cells which send impulses to the brain and produce thesensation of sound

The ear detects sound waves over a wide range of frequenciesand is sensitive to an enormous range of amplitudes In addi-tion human hearing can distinguish many different qualities of sound Knowledge of both physics and biology is required to understand the complexities of the ear The interpretation of sounds by the brain is even more complex and it is nottotally understood

Auditorycanal

Auditorynerve

Cochlea

MalleusIncus

Stapes

Tympanic membrane(eardrum)

Figure 15-3 The human ear is a complex sense organ thattranslates sound vibrations intonerve impulses that are sent tothe brain for interpretation Themalleus incus and stapes are the three bones of the middle earthat sometimes are referred to asthe hammer anvil and stirrup

Section 151 Properties and Detection of Sound 405

402-410 CH15-S1-845813 6404 944 AM Page 405

PPP-405

null

14633661

406 Chapter 15 Sound

10 dBBarely

audible

30 dBWhisper1 m away

50 dBCasual

conversation

80 dBAlarmclock

110 dBRock

concert

140 dBJet engine

70 dBHeavytraffic

100 dBSiren

Figure 15-4 This decibel scaleshows the sound levels of somefamiliar sounds

Perceiving SoundPitch Marin Mersenne and Galileo first determined that the pitch we heardepends on the frequency of vibration Pitch can be given a name on themusical scale For instance the middle C note has a frequency of 262 Hz Theear is not equally sensitive to all frequencies Most people cannot hear soundswith frequencies below 20 Hz or above 16000 Hz Older people are lesssensitive to frequencies above 10000 Hz than are young people By age 70most people cannot hear sounds with frequencies above 8000 Hz Thisloss affects the ability to understand speech

Loudness Frequency and wavelength are two physical characteristics ofsound waves Another physical characteristic of sound waves is amplitudeAmplitude is the measure of the variation in pressure along a wave Inhumans sound is detected by the ear and interpreted by the brain Theloudness of a sound as perceived by our sense of hearing depends pri-marily on the amplitude of the pressure wave

The human ear is extremely sensitive to pressure variations in soundwaves which is the amplitude of the wave Recall from Chapter 13 that 1 atm of pressure equals 101105 Pa The ear can detect pressure-waveamplitudes of less than one-billionth of an atmosphere or 2105 Pa Atthe other end of the audible range pressure variations of approximately 20 Pa or greater cause pain It is important to remember that the ear detectsonly pressure variations at certain frequencies Driving over a mountainpass changes the pressure on your ears by thousands of pascals but thischange does not take place at audible frequencies

Because humans can detect a wide range in pressure variations theseamplitudes are measured on a logarithmic scale called the sound levelThe unit of measurement for sound level is the decibel (dB) The soundlevel depends on the ratio of the pressure variation of a given sound waveto the pressure variation in the most faintly heard sound 2105 Pa Suchan amplitude has a sound level of 0 dB A sound with a pressure amplitudeten times larger (2104 Pa) is 20 dB A pressure amplitude ten timeslarger than this is 40 dB Most people perceive a 10-dB increase in soundlevel as about twice as loud as the original level Figure 15-4 shows thesound level for a variety of sounds In addition to pressure variationspower and intensity of sound waves can be described by decibel scales

Exposure to loud sounds in the form of noise or music has been shownto cause the ear to lose its sensitivity especially to high frequencies Thelonger a person is exposed to loud sounds the greater the effect A personcan recover from short-term exposure in a period of hours but the effects

402-410 CH15-S1-845813 33004 658 AM Page 406

PPP-406

null

1996316

ObserverB

ObserverA

S

vs

λB

λA

of long-term exposure can last for days or weeks Long exposure to 100-dBor greater sound levels can produce permanent damage Many rock musi-cians have suffered serious hearing loss some as much as 40 percentHearing loss also can result from loud music being transmitted to stereoheadphones from personal radios and CD players In some cases the lis-teners are unaware of just how high the sound levels really are Cottonearplugs reduce the sound level only by about 10 dB Special ear inserts canprovide a 25-dB reduction Specifically designed earmuffs and inserts asshown in Figure 15-5 can reduce the sound level by up to 45 dB

Loudness as perceived by the human ear is not directly proportional tothe pressure variations in a sound wave The earrsquos sensitivity depends onboth pitch and amplitude Also perception of pure tones is different fromperception of a mixture of tones

The Doppler EffectHave you ever noticed that the pitch of an ambulance fire or police siren

changed as the vehicle sped past you The pitch was higher when the vehiclewas moving toward you then it dropped to a lower pitch as the source movedaway This frequency shift is called the Doppler effect and is shown inFigure 15-6 The sound source S is moving to the right with a speed of vsThe waves that it emits spread in circles centered on the source at the time itproduced the waves As the source moves toward the sound detectorObserver A in Figure 15-6a more waves are crowded into the space betweenthem The wavelength is shortened to A Because the speed of sound is notchanged more crests reach the ear per second which means that the fre-quency of the detected sound increases When the source is moving awayfrom the detector Observer B in Figure 15-6a the wavelength is lengthenedto B and the detected frequency is lower Figure 15-6b illustrates theDoppler effect for a moving source of sound on water waves in a ripple tank

A Doppler shift also occurs if the detector is moving and the source isstationary In this case the Doppler shift results from the relative velocityof the sound waves and the detector As the detector approaches the sta-tionary source the relative velocity is larger resulting in an increase in thewave crests reaching the detector each second As the detector recedes fromthe source the relative velocity is smaller resulting in a decrease in thewave crests reaching the detector each second

Figure 15-5 Continuousexposure to loud sounds cancause serious hearing loss Inmany occupations workers such as this flight controller must wear ear protection

Figure 15-6 As a sound-producing source moves towardan observer the wavelength isshortened to A the wavelength is B for waves produced by asource moving away from anobserver (a) A moving wave-producing source illustrates theDoppler effect in a ripple tank (b)

a b

Section 151 Properties and Detection of Sound 407(t)Willie MaldonadGetty Images (b)Max KurtzZtek

402-410 CH15-S1-845813 33004 659 AM Page 407

PPP-407

null

17627586

408 Chapter 15 Sound

Reducing Equations When an element in a complex equation is equal tozero the equation might reduce to a form that is easier to use

fd fs(vv

vvd

s) fd fs(vv

vvd

s)

fs(v v

vs) fs(v

vvd)

fs( ) fs( ) fs( ) fs( )

fs(1 vvd)

1 vvd

1

11

vvs

vv

vvd

vv

vv

vv

vvs

Stationary detector source inmotion vd 0

Stationary source detector inmotion vs 0

For both a moving source and a moving observer the frequency that theobserver hears can be calculated using the equation below

In the Doppler effect equation v is the velocity of the sound wave vd isthe velocity of the detector vs is the velocity of the soundrsquos source fs is thefrequency of the wave emitted by the source and fd is the frequencyreceived by the detector This equation applies when the source is movingwhen the observer is moving and when both are moving

As you solve problems using the above equation be sure to define thecoordinate system so that the positive direction is from the source to thedetector The sound waves will be approaching the detector from the sourceso the velocity of sound is always positive Try drawing diagrams to confirmthat the term (v vd)(v vs) behaves as you would predict based on whatyou have learned about the Doppler effect Notice that for a source movingtoward the detector (positive direction which results in a smaller denom-inator compared to a stationary source) and for a detector moving towardthe source (negative direction and increased numerator compared to a stationary detector) the detected frequency fd increases Similarily if thesource moves away from the detector or if the detector moves away fromthe source then fd decreases Read the Connecting Math to Physics featurebelow to see how the Doppler effect equation reduces when the source orobserver is stationary

Doppler Effect fd fsvv

vvd

s

The frequency perceived by a detector is equal to the velocity of the detectorrelative to the velocity of the wave divided by the velocity of the sourcerelative to the velocity of the wave multiplied by the waversquos frequency

402-410 CH15-S1-845813 33004 659 AM Page 408

PPP-408

null

12439261

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

406 Chapter 15 Sound

10 dBBarely

audible

30 dBWhisper1 m away

50 dBCasual

conversation

80 dBAlarmclock

110 dBRock

concert

140 dBJet engine

70 dBHeavytraffic

100 dBSiren

Figure 15-4 This decibel scaleshows the sound levels of somefamiliar sounds

Perceiving SoundPitch Marin Mersenne and Galileo first determined that the pitch we heardepends on the frequency of vibration Pitch can be given a name on themusical scale For instance the middle C note has a frequency of 262 Hz Theear is not equally sensitive to all frequencies Most people cannot hear soundswith frequencies below 20 Hz or above 16000 Hz Older people are lesssensitive to frequencies above 10000 Hz than are young people By age 70most people cannot hear sounds with frequencies above 8000 Hz Thisloss affects the ability to understand speech

Loudness Frequency and wavelength are two physical characteristics ofsound waves Another physical characteristic of sound waves is amplitudeAmplitude is the measure of the variation in pressure along a wave Inhumans sound is detected by the ear and interpreted by the brain Theloudness of a sound as perceived by our sense of hearing depends pri-marily on the amplitude of the pressure wave

The human ear is extremely sensitive to pressure variations in soundwaves which is the amplitude of the wave Recall from Chapter 13 that 1 atm of pressure equals 101105 Pa The ear can detect pressure-waveamplitudes of less than one-billionth of an atmosphere or 2105 Pa Atthe other end of the audible range pressure variations of approximately 20 Pa or greater cause pain It is important to remember that the ear detectsonly pressure variations at certain frequencies Driving over a mountainpass changes the pressure on your ears by thousands of pascals but thischange does not take place at audible frequencies

Because humans can detect a wide range in pressure variations theseamplitudes are measured on a logarithmic scale called the sound levelThe unit of measurement for sound level is the decibel (dB) The soundlevel depends on the ratio of the pressure variation of a given sound waveto the pressure variation in the most faintly heard sound 2105 Pa Suchan amplitude has a sound level of 0 dB A sound with a pressure amplitudeten times larger (2104 Pa) is 20 dB A pressure amplitude ten timeslarger than this is 40 dB Most people perceive a 10-dB increase in soundlevel as about twice as loud as the original level Figure 15-4 shows thesound level for a variety of sounds In addition to pressure variationspower and intensity of sound waves can be described by decibel scales

Exposure to loud sounds in the form of noise or music has been shownto cause the ear to lose its sensitivity especially to high frequencies Thelonger a person is exposed to loud sounds the greater the effect A personcan recover from short-term exposure in a period of hours but the effects

402-410 CH15-S1-845813 33004 658 AM Page 406

PPP-406

null

1996316

ObserverB

ObserverA

S

vs

λB

λA

of long-term exposure can last for days or weeks Long exposure to 100-dBor greater sound levels can produce permanent damage Many rock musi-cians have suffered serious hearing loss some as much as 40 percentHearing loss also can result from loud music being transmitted to stereoheadphones from personal radios and CD players In some cases the lis-teners are unaware of just how high the sound levels really are Cottonearplugs reduce the sound level only by about 10 dB Special ear inserts canprovide a 25-dB reduction Specifically designed earmuffs and inserts asshown in Figure 15-5 can reduce the sound level by up to 45 dB

Loudness as perceived by the human ear is not directly proportional tothe pressure variations in a sound wave The earrsquos sensitivity depends onboth pitch and amplitude Also perception of pure tones is different fromperception of a mixture of tones

The Doppler EffectHave you ever noticed that the pitch of an ambulance fire or police siren

changed as the vehicle sped past you The pitch was higher when the vehiclewas moving toward you then it dropped to a lower pitch as the source movedaway This frequency shift is called the Doppler effect and is shown inFigure 15-6 The sound source S is moving to the right with a speed of vsThe waves that it emits spread in circles centered on the source at the time itproduced the waves As the source moves toward the sound detectorObserver A in Figure 15-6a more waves are crowded into the space betweenthem The wavelength is shortened to A Because the speed of sound is notchanged more crests reach the ear per second which means that the fre-quency of the detected sound increases When the source is moving awayfrom the detector Observer B in Figure 15-6a the wavelength is lengthenedto B and the detected frequency is lower Figure 15-6b illustrates theDoppler effect for a moving source of sound on water waves in a ripple tank

A Doppler shift also occurs if the detector is moving and the source isstationary In this case the Doppler shift results from the relative velocityof the sound waves and the detector As the detector approaches the sta-tionary source the relative velocity is larger resulting in an increase in thewave crests reaching the detector each second As the detector recedes fromthe source the relative velocity is smaller resulting in a decrease in thewave crests reaching the detector each second

Figure 15-5 Continuousexposure to loud sounds cancause serious hearing loss Inmany occupations workers such as this flight controller must wear ear protection

Figure 15-6 As a sound-producing source moves towardan observer the wavelength isshortened to A the wavelength is B for waves produced by asource moving away from anobserver (a) A moving wave-producing source illustrates theDoppler effect in a ripple tank (b)

a b

Section 151 Properties and Detection of Sound 407(t)Willie MaldonadGetty Images (b)Max KurtzZtek

402-410 CH15-S1-845813 33004 659 AM Page 407

PPP-407

null

17627586

408 Chapter 15 Sound

Reducing Equations When an element in a complex equation is equal tozero the equation might reduce to a form that is easier to use

fd fs(vv

vvd

s) fd fs(vv

vvd

s)

fs(v v

vs) fs(v

vvd)

fs( ) fs( ) fs( ) fs( )

fs(1 vvd)

1 vvd

1

11

vvs

vv

vvd

vv

vv

vv

vvs

Stationary detector source inmotion vd 0

Stationary source detector inmotion vs 0

For both a moving source and a moving observer the frequency that theobserver hears can be calculated using the equation below

In the Doppler effect equation v is the velocity of the sound wave vd isthe velocity of the detector vs is the velocity of the soundrsquos source fs is thefrequency of the wave emitted by the source and fd is the frequencyreceived by the detector This equation applies when the source is movingwhen the observer is moving and when both are moving

As you solve problems using the above equation be sure to define thecoordinate system so that the positive direction is from the source to thedetector The sound waves will be approaching the detector from the sourceso the velocity of sound is always positive Try drawing diagrams to confirmthat the term (v vd)(v vs) behaves as you would predict based on whatyou have learned about the Doppler effect Notice that for a source movingtoward the detector (positive direction which results in a smaller denom-inator compared to a stationary source) and for a detector moving towardthe source (negative direction and increased numerator compared to a stationary detector) the detected frequency fd increases Similarily if thesource moves away from the detector or if the detector moves away fromthe source then fd decreases Read the Connecting Math to Physics featurebelow to see how the Doppler effect equation reduces when the source orobserver is stationary

Doppler Effect fd fsvv

vvd

s

The frequency perceived by a detector is equal to the velocity of the detectorrelative to the velocity of the wave divided by the velocity of the sourcerelative to the velocity of the wave multiplied by the waversquos frequency

402-410 CH15-S1-845813 33004 659 AM Page 408

PPP-408

null

12439261

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

ObserverB

ObserverA

S

vs

λB

λA

of long-term exposure can last for days or weeks Long exposure to 100-dBor greater sound levels can produce permanent damage Many rock musi-cians have suffered serious hearing loss some as much as 40 percentHearing loss also can result from loud music being transmitted to stereoheadphones from personal radios and CD players In some cases the lis-teners are unaware of just how high the sound levels really are Cottonearplugs reduce the sound level only by about 10 dB Special ear inserts canprovide a 25-dB reduction Specifically designed earmuffs and inserts asshown in Figure 15-5 can reduce the sound level by up to 45 dB

Loudness as perceived by the human ear is not directly proportional tothe pressure variations in a sound wave The earrsquos sensitivity depends onboth pitch and amplitude Also perception of pure tones is different fromperception of a mixture of tones

The Doppler EffectHave you ever noticed that the pitch of an ambulance fire or police siren

changed as the vehicle sped past you The pitch was higher when the vehiclewas moving toward you then it dropped to a lower pitch as the source movedaway This frequency shift is called the Doppler effect and is shown inFigure 15-6 The sound source S is moving to the right with a speed of vsThe waves that it emits spread in circles centered on the source at the time itproduced the waves As the source moves toward the sound detectorObserver A in Figure 15-6a more waves are crowded into the space betweenthem The wavelength is shortened to A Because the speed of sound is notchanged more crests reach the ear per second which means that the fre-quency of the detected sound increases When the source is moving awayfrom the detector Observer B in Figure 15-6a the wavelength is lengthenedto B and the detected frequency is lower Figure 15-6b illustrates theDoppler effect for a moving source of sound on water waves in a ripple tank

A Doppler shift also occurs if the detector is moving and the source isstationary In this case the Doppler shift results from the relative velocityof the sound waves and the detector As the detector approaches the sta-tionary source the relative velocity is larger resulting in an increase in thewave crests reaching the detector each second As the detector recedes fromthe source the relative velocity is smaller resulting in a decrease in thewave crests reaching the detector each second

Figure 15-5 Continuousexposure to loud sounds cancause serious hearing loss Inmany occupations workers such as this flight controller must wear ear protection

Figure 15-6 As a sound-producing source moves towardan observer the wavelength isshortened to A the wavelength is B for waves produced by asource moving away from anobserver (a) A moving wave-producing source illustrates theDoppler effect in a ripple tank (b)

a b

Section 151 Properties and Detection of Sound 407(t)Willie MaldonadGetty Images (b)Max KurtzZtek

402-410 CH15-S1-845813 33004 659 AM Page 407

PPP-407

null

17627586

408 Chapter 15 Sound

Reducing Equations When an element in a complex equation is equal tozero the equation might reduce to a form that is easier to use

fd fs(vv

vvd

s) fd fs(vv

vvd

s)

fs(v v

vs) fs(v

vvd)

fs( ) fs( ) fs( ) fs( )

fs(1 vvd)

1 vvd

1

11

vvs

vv

vvd

vv

vv

vv

vvs

Stationary detector source inmotion vd 0

Stationary source detector inmotion vs 0

For both a moving source and a moving observer the frequency that theobserver hears can be calculated using the equation below

In the Doppler effect equation v is the velocity of the sound wave vd isthe velocity of the detector vs is the velocity of the soundrsquos source fs is thefrequency of the wave emitted by the source and fd is the frequencyreceived by the detector This equation applies when the source is movingwhen the observer is moving and when both are moving

As you solve problems using the above equation be sure to define thecoordinate system so that the positive direction is from the source to thedetector The sound waves will be approaching the detector from the sourceso the velocity of sound is always positive Try drawing diagrams to confirmthat the term (v vd)(v vs) behaves as you would predict based on whatyou have learned about the Doppler effect Notice that for a source movingtoward the detector (positive direction which results in a smaller denom-inator compared to a stationary source) and for a detector moving towardthe source (negative direction and increased numerator compared to a stationary detector) the detected frequency fd increases Similarily if thesource moves away from the detector or if the detector moves away fromthe source then fd decreases Read the Connecting Math to Physics featurebelow to see how the Doppler effect equation reduces when the source orobserver is stationary

Doppler Effect fd fsvv

vvd

s

The frequency perceived by a detector is equal to the velocity of the detectorrelative to the velocity of the wave divided by the velocity of the sourcerelative to the velocity of the wave multiplied by the waversquos frequency

402-410 CH15-S1-845813 33004 659 AM Page 408

PPP-408

null

12439261

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

408 Chapter 15 Sound

Reducing Equations When an element in a complex equation is equal tozero the equation might reduce to a form that is easier to use

fd fs(vv

vvd

s) fd fs(vv

vvd

s)

fs(v v

vs) fs(v

vvd)

fs( ) fs( ) fs( ) fs( )

fs(1 vvd)

1 vvd

1

11

vvs

vv

vvd

vv

vv

vv

vvs

Stationary detector source inmotion vd 0

Stationary source detector inmotion vs 0

For both a moving source and a moving observer the frequency that theobserver hears can be calculated using the equation below

In the Doppler effect equation v is the velocity of the sound wave vd isthe velocity of the detector vs is the velocity of the soundrsquos source fs is thefrequency of the wave emitted by the source and fd is the frequencyreceived by the detector This equation applies when the source is movingwhen the observer is moving and when both are moving

As you solve problems using the above equation be sure to define thecoordinate system so that the positive direction is from the source to thedetector The sound waves will be approaching the detector from the sourceso the velocity of sound is always positive Try drawing diagrams to confirmthat the term (v vd)(v vs) behaves as you would predict based on whatyou have learned about the Doppler effect Notice that for a source movingtoward the detector (positive direction which results in a smaller denom-inator compared to a stationary source) and for a detector moving towardthe source (negative direction and increased numerator compared to a stationary detector) the detected frequency fd increases Similarily if thesource moves away from the detector or if the detector moves away fromthe source then fd decreases Read the Connecting Math to Physics featurebelow to see how the Doppler effect equation reduces when the source orobserver is stationary

Doppler Effect fd fsvv

vvd

s

The frequency perceived by a detector is equal to the velocity of the detectorrelative to the velocity of the wave divided by the velocity of the sourcerelative to the velocity of the wave multiplied by the waversquos frequency

402-410 CH15-S1-845813 33004 659 AM Page 408

PPP-408

null

12439261

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

6 Repeat Example Problem 1 but with the car moving away from youWhat frequency would you hear

7 You are in an auto traveling at 250 ms toward a pole-mountedwarning siren If the sirenrsquos frequency is 365 Hz what frequency do you hear Use 343 ms as the speed of sound

8 You are in an auto traveling at 55 mph (246 ms) A second auto is moving toward you at the same speed Its horn is sounding at475 Hz What frequency do you hear Use 343 ms as the speed of sound

9 A submarine is moving toward another submarine at 920 ms It emits a 350-MHz ultrasound What frequency would the secondsub at rest detect The speed of sound in water is 1482 ms

10 A sound source plays middle C (262 Hz) How fast would the sourcehave to go to raise the pitch to C sharp (271 Hz) Use 343 ms asthe speed of sound

The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 246 ms If the car is coming toward you what frequency would you hearAssume that the temperature is 20degC

Analyze and Sketch the Problembull Sketch the situationbull Establish a coordinate axis Make sure that the

positive direction is from the source to the detectorbull Show the velocities of the source and detector

Known Unknown

v 343 ms fd vs 246 msvd 0 msfs 524 Hz

Solve for the Unknown

Use fd fs(vv

vvd

s) with vd 0 ms

fd fs( ) 524 Hz( ) Substitute v 343 ms vs 246 ms and fs 524 Hz

564 Hz

Evaluate the Answerbull Are the units correct Frequency is measured in hertzbull Is the magnitude realistic The source is moving toward you so the frequency

should be increased

3

1

1 234436

mm

ss

11

vvs

2

1

Math Handbook

Fractions page 837

v

vs

Section 151 Properties and Detection of Sound 409

402-410 CH15-S1-845813 33004 700 AM Page 409

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

410 Chapter 15 Sound

11 Graph The eardrum moves back and forth inresponse to the pressure variations of a soundwave Sketch a graph of the displacement of theeardrum versus time for two cycles of a 10-kHztone and for two cycles of a 20-kHz tone

12 Effect of Medium List two sound characteristicsthat are affected by the medium through which the sound passes and two characteristics that arenot affected

13 Sound Properties What physical characteristicof a sound wave should be changed to change thepitch of the sound To change the loudness

14 Decibel Scale How much greater is the soundpressure level of a typical rock bandrsquos music (110 dB)than a normal conversation (50 dB)

15 Early Detection In the nineteenth century peopleput their ears to a railroad track to get an early warn-ing of an approaching train Why did this work

16 Bats A bat emits short pulses of high-frequencysound and detects the echoes

a In what way would the echoes from large andsmall insects compare if they were the samedistance from the bat

b In what way would the echo from an insect fly-ing toward the bat differ from that of an insectflying away from the bat

17 Critical Thinking Can a trooper using a radardetector at the side of the road determine thespeed of a car at the instant the car passes thetrooper Explain

151 Section Review

physicsppcomself_check_quiz

The Doppler effect occurs in all wave motion both mechanical and elec-tromagnetic It has many applications Radar detectors use the Dopplereffect to measure the speed of baseballs and automobiles Astronomersobserve light from distant galaxies and use the Doppler effect to measuretheir speeds and infer their distances Physicians can detect the speed of themoving heart wall in a fetus by means of the Doppler effect in ultrasoundBats use the Doppler effect to detect and catch flying insects When aninsect is flying faster than a bat the reflected frequency is lower but whenthe bat is catching up to the insect as in Figure 15-7 the reflected fre-quency is higher Not only do bats use sound waves to navigate and locatetheir prey but they often must do so in the presence of other bats Thismeans they must discriminate their own calls and reflections against abackground of many other sounds of many frequencies Scientists con-tinue to study bats and their amazing abilities to use sound waves

Biology ConnectionBiology Connection

Figure 15-7 In a process calledecholocation bats use the Dopplereffect to locate prey

Stephen DaltonAnimals Animals

402-410 CH15-S1-845813 33004 700 AM Page 410

Objectivesbull Describe the origin of sound

bull Demonstrate anunderstanding of resonanceespecially as applied to aircolumns and strings

bull Explain why there arevariations in sound amonginstruments and amongvoices

Vocabulary

closed-pipe resonatoropen-pipe resonatorfundamentalharmonicsdissonanceconsonancebeat

152 The Physics of Music

Brass instrument

Woodwind instrument

Mouthpiece

Mouthpiece

Reed

Figure 15-8 The shapes of themouthpieces of a brass instrument(a) and a reed instrument (b)help determine the characteristicsof the sound each instrumentproduces

a

b

In the middle of the nineteenth century German physicist HermannHelmholtz studied sound production in musical instruments and the

human voice In the twentieth century scientists and engineers developedelectronic equipment that permits not only a detailed study of sound butalso the creation of electronic musical instruments and recording devicesthat allow us to listen to music whenever and wherever we wish

Sources of SoundSound is produced by a vibrating object The vibrations of the object

create particle motions that cause pressure oscillations in the air A loud-speaker has a cone that is made to vibrate by electrical currents The surfaceof the cone creates the sound waves that travel to your ear and allow youto hear music Musical instruments such as gongs cymbals and drums areother examples of vibrating surfaces that are sources of sound

The human voice is produced by vibrations of the vocal cords which aretwo membranes located in the throat Air from the lungs rushing throughthe throat starts the vocal cords vibrating The frequency of vibration iscontrolled by the muscular tension placed on the vocal cords

In brass instruments such as the trumpet and tuba the lips of the per-former vibrate as shown in Figure 15-8a Reed instruments such as theclarinet and saxophone have a thin wooden strip or reed that vibrates asa result of air blown across it as shown in Figure 15-8b In flutes andorgan pipes air is forced across an opening in a pipe Air moving past theopening sets the column of air in the instrument into vibration

In stringed instruments such as the piano guitar and violin wires orstrings are set into vibration In the piano the wires are struck in the gui-tar they are plucked and in the violin the friction of the bow causes thestrings to vibrate Often the strings are attached to a sounding board thatvibrates with the strings The vibrations of the sounding board cause thepressure oscillations in the air that we hear as sound Electric guitars useelectronic devices to detect and amplify the vibrations of the guitar strings

Section 152 The Physics of Music 411

411-419 CH15-S2-845813 33004 706 AM Page 411

PPP-411

null

14325386

Tube

Water

Aircolumn

Tuningfork

Hammer

Resonance in Air ColumnsIf you have ever used just the mouthpiece of a brass or reed instrument

you know that the vibration of your lips or the reed alone does not makea sound with any particular pitch The long tube that makes up the instru-ment must be attached if music is to result When the instrument is playedthe air within this tube vibrates at the same frequency or in resonancewith a particular vibration of the lips or reed Remember that resonanceincreases the amplitude of a vibration by repeatedly applying a small external force at the same natural frequency The length of the air columndetermines the frequencies of the vibrating air that will be set into reso-nance For many instruments such as flutes saxophones and tromboneschanging the length of the column of vibrating air varies the pitch of theinstrument The mouthpiece simply creates a mixture of different frequen-cies and the resonating air column acts on a particular set of frequenciesto amplify a single note turning noise into music

A tuning fork above a hollow tube can provide resonance in an air column as shown in Figure 15-9 The tube is placed in water so that thebottom end of the tube is below the water surface A resonating tube withone end closed to air is called a closed-pipe resonator The length of theair column is changed by adjusting the height of the tube above the waterIf the tuning fork is struck with a rubber hammer and the length of the aircolumn is varied as the tube is lifted up and down in the water the soundalternately becomes louder and softer The sound is loud when the air column is in resonance with the tuning fork A resonating air columnintensifies the sound of the tuning fork

Standing pressure wave How does resonance occur The vibrating tuning fork produces a sound wave This wave of alternate high- and low-pressure variations moves down the air column When the wave hits thewater surface it is reflected back up to the tuning fork as indicated inFigure 15-10a If the reflected high-pressure wave reaches the tuning forkat the same moment that the fork produces another high-pressure wavethen the emitted and returning waves reinforce each other This reinforce-ment of waves produces a standing wave and resonance is achieved

An open-pipe resonator is a resonating tube with both ends open thatalso will resonate with a sound source In this case the sound wave doesnot reflect off a closed end but rather off an open end The pressure of thereflected wave is inverted for example if a high-pressure wave strikes theopen end a low-pressure wave will rebound as shown in Figure 15-10b

Resonance lengths A standing soundwave in a pipe can be represented by asine wave as shown in Figure 15-11Sine waves can represent either the airpressure or the displacement of the airparticles You can see that standingwaves have nodes and antinodes In thepressure graphs the nodes are regionsof mean atmospheric pressure and at theantinodes the pressure oscillates betweenits maximum and minimum values

Figure 15-9 Raising or loweringthe tube changes the length ofthe air column When the columnis in resonance with the tuningfork the sound is loudest

TimeClosed pipes high pressure

reflects as high pressure

TimeOpen pipes high pressure

reflects as low pressure

ba

Figure 15-10 A tube placed inwater is a closed-pipe resonatorIn closed pipes high pressurewaves reflect as high pressure (a)In open pipes the reflected wavesare inverted (b)

412 Chapter 15 Sound

411-419 CH15-S2-845813 6404 947 AM Page 412

PPP-412

null

20323686

>

>

Displacementantinode

Displacementantinode

Displacementnode

Open Pipe

Air pressure Displacement of air

Pressure node(average-pressure region)

Pressure node(average-pressure region)

Pressure antinode(high- or low-pressure region)

Closed Pipe

Air pressure Displacement of air

Pressure node(average-pressure region)

Pressure antinode(high- or low-pressure region)

Displacementantinode

Displacementnode

Pressure node (average-pressure region)

Antinode

Antinode

Node

Node

λ1 4L

f1 mdash mdash vvλ1 4L

f3 mdash 3f13v4L

λ3 mdash L43

f5 mdash 5f15v4L

λ5 mdash L45

L

Pressure antinode (high-or low-pressure region)

Figure 15-11 Sine wavesrepresent standing waves in pipes

Figure 15-12 A closed piperesonates when its length is anodd number of quarterwavelengths

In the case of the displacement graph the antinodes are regions of high displacement and the nodes are regions of low displacement In both casestwo antinodes (or two nodes) are separated by one-half wavelength

Resonance frequencies in a closed pipe The shortest column of air thatcan have an antinode at the closed end and a node at the open end is one-fourth of a wavelength long as shown in Figure 15-12 As the frequencyis increased additional resonance lengths are found at half-wavelengthintervals Thus columns of length 4 34 54 74 and so on will allbe in resonance with a tuning fork

In practice the first resonance length is slightly longer than one-fourthof a wavelength This is because the pressure variations do not drop to zeroexactly at the open end of the pipe Actually the node is approximately 04 pipe diameters beyond the end Additional resonance lengths how-ever are spaced by exactly one-half of a wavelength Measurements of thespacing between resonances can be used to find the velocity of sound inair as shown in the next Example Problem

Hearing and FrequencyThe human auditory canal acts as a closed-pipe resonator thatincreases the earrsquos sensitivity forfrequencies between 2000 and5000 Hz but the full range offrequencies that people hearextends from 20 to 20000 Hz A dogrsquos hearing extends tofrequencies as high as 45000 Hzand a catrsquos extends to frequenciesas high as 100000 Hz

Section 152 The Physics of Music 413

411-419 CH15-S2-845813 6404 948 AM Page 413

PPP-413

null

79673294

Pressure node (average-pressure region)

Pressure node (average-pressure region)

Pressure antinode (high-or low-pressure region)

Antinode

Antinode

Node

Node

Node

λ1 2L

f1 mdash mdash vvλ1 2L

f2 mdash 2f1vL

λ2 L

f3 mdash 3f13v2L

2Lλ3 mdash3

L

Figure 15-13 An open piperesonates when its length is an even number of quarterwavelengths

Resonance frequencies in an open pipe The shortest column of air thatcan have nodes at both ends is one-half of a wavelength long as shown inFigure 15-13 As the frequency is increased additional resonance lengthsare found at half-wavelength intervals Thus columns of length 2 32 2 and so on will be in resonance with a tuning fork

If open and closed pipes of the same length are used as resonators thewavelength of the resonant sound for the open pipe will be half as long asthat for the closed pipe Therefore the frequency will be twice as high forthe open pipe as for the closed pipe For both pipes resonance lengths arespaced by half-wavelength intervals

Hearing resonance Musical instruments use resonance to increase theloudness of particular notes Open-pipe resonators include flutes and saxophones Clarinets and the hanging pipes under marimbas and xylo-phones are examples of closed-pipe resonators If you shout into a longtunnel the booming sound you hear is the tunnel acting as a resonatorThe seashell in Figure 15-14 acts as a closed-pipe resonator

Resonance on StringsAlthough the waveforms on vibrating strings vary in shape depending

upon how they are produced such as by plucking bowing or strikingthey have many characteristics in common with standing waves on springs

and ropes which you studied in Chapter 14 A string on an instrument is clamped at both ends and therefore thestring must have a node at each end when it vibrates InFigure 15-15 you can see that the first mode of vibrationhas an antinode at the center and is one-half of a wavelengthlong The next resonance occurs when one wavelength fits onthe string and additional standing waves arise when thestring length is 32 2 52 and so on As with an openpipe the resonant frequencies are whole-number multiplesof the lowest frequency

The speed of a wave on a string depends on the tension ofthe string as well as its mass per unit length This makes itpossible to tune a stringed instrument by changing the ten-sion of its strings The tighter the string the faster the wavemoves along it and therefore the higher the frequency of itsstanding waves

Figure 15-14 A seashell acts asa closed-pipe resonator to amplifycertain frequencies from thebackground noise

414 Chapter 15 SoundDavid MechlinIndex Stock Images

411-419 CH15-S2-845813 33004 707 AM Page 414

PPP-414

null

15244986

Pres

sure

Time

Pure sound

Pres

sure

Time

Clarinet

Section 152 The Physics of Music 415

1 2Lf1

v

1

2vL

2 Lf2

vL

2f1

3 23L

f3 32vL 3f1

L Figure 15-15 A string resonateswith standing waves when itslength is a whole number of halfwavelengths

Figure 15-16 A graph of puresound versus time (a) and agraph of clarinet sound wavesversus time (b) are shown

a b

Because strings are so small in cross-sectional area they move very littleair when they vibrate This makes it necessary to attach them to a sound-ing board which transfers their vibrations to the air and produces astronger sound wave Unlike the strings themselves the sounding boardshould not resonate at any single frequency Its purpose is to convey thevibrations of all the strings to the air and therefore it should vibrate wellat all frequencies produced by the instrument Because of the complicatedinteractions among the strings the sounding board and the air the designand construction of stringed instruments are complex processes consideredby many to be as much an art as a science

Sound QualityA tuning fork produces a soft and uninteresting sound This is because

its tines vibrate like simple harmonic oscillators and produce the simple sine wave shown in Figure 15-16a Sounds made by the humanvoice and musical instruments are much more complex like the wave inFigure 15-16b Both waves have the same frequency or pitch but theysound very different The complex wave is produced by using the principle of superposition to add waves of many frequencies The shape of the wave depends on the relative amplitudes of these frequencies In musicalterms the difference between the two waves is called timbre tone color ortone quality

411-419 CH15-S2-845813 33004 708 AM Page 415

PPP-415

null

91114334

18 A 440-Hz tuning fork is held above a closed pipe Find the spacingbetween the resonances when the air temperature is 20degC

19 A 440-Hz tuning fork is used with a resonating column to determinethe velocity of sound in helium gas If the spacings betweenresonances are 110 cm what is the velocity of sound in helium gas

20 The frequency of a tuning fork is unknown A student uses an aircolumn at 27degC and finds resonances spaced by 202 cm What isthe frequency of the tuning fork Use the speed calculated inExample Problem 2 for the speed of sound in air at 27degC

21 A bugle can be thought of as an open pipe If a bugle werestraightened out it would be 265-m long

a If the speed of sound is 343 ms find the lowest frequency thatis resonant for a bugle (ignoring end corrections)

b Find the next two resonant frequencies for the bugle

416 Chapter 15 Sound

Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz

is used with a closed-pipe resonator the loudest sound is heard when the column is 210 cm

and 653 cm long What is the speed of sound in this case Is the temperature warmer or

cooler than normal room temperature which is 20degC Explain your answer

Analyze and Sketch the Problembull Sketch the closed-pipe resonator bull Mark the resonance lengths

Known Unknown

f 392 Hz v LA 210 cmLB 653 cm

Solve for the UnknownSolve for the length of the wave using the length-wavelength relationship for a closed pipe

LB LA 12

2(LB LA) Rearrange the equation for

2(0653 m 0210 m) Substitute LB 0653 m LA 0210 m

0886 m

Use vf

v f Rearrange the equation for v

(392 Hz)(0886 m) Substitute f 392 Hz 0886 m

347 msThe speed is slightly greater than the speed of sound at 20degC indicating that thetemperature is slightly higher than normal room temperature

Evaluate the Answerbull Are the units correct (Hz)(m) (1s)(m) ms The answerrsquos units are correctbull Is the magnitude realistic The speed is slightly greater than 343 ms which is the

speed of sound at 20degC

3

2

1

Math Handbook

Order of Operationspage 843

LA

LB

411-419 CH15-S2-845813 33004 708 AM Page 416

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

Tube

Water

Aircolumn

Tuningfork

Hammer

Resonance in Air ColumnsIf you have ever used just the mouthpiece of a brass or reed instrument

you know that the vibration of your lips or the reed alone does not makea sound with any particular pitch The long tube that makes up the instru-ment must be attached if music is to result When the instrument is playedthe air within this tube vibrates at the same frequency or in resonancewith a particular vibration of the lips or reed Remember that resonanceincreases the amplitude of a vibration by repeatedly applying a small external force at the same natural frequency The length of the air columndetermines the frequencies of the vibrating air that will be set into reso-nance For many instruments such as flutes saxophones and tromboneschanging the length of the column of vibrating air varies the pitch of theinstrument The mouthpiece simply creates a mixture of different frequen-cies and the resonating air column acts on a particular set of frequenciesto amplify a single note turning noise into music

A tuning fork above a hollow tube can provide resonance in an air column as shown in Figure 15-9 The tube is placed in water so that thebottom end of the tube is below the water surface A resonating tube withone end closed to air is called a closed-pipe resonator The length of theair column is changed by adjusting the height of the tube above the waterIf the tuning fork is struck with a rubber hammer and the length of the aircolumn is varied as the tube is lifted up and down in the water the soundalternately becomes louder and softer The sound is loud when the air column is in resonance with the tuning fork A resonating air columnintensifies the sound of the tuning fork

Standing pressure wave How does resonance occur The vibrating tuning fork produces a sound wave This wave of alternate high- and low-pressure variations moves down the air column When the wave hits thewater surface it is reflected back up to the tuning fork as indicated inFigure 15-10a If the reflected high-pressure wave reaches the tuning forkat the same moment that the fork produces another high-pressure wavethen the emitted and returning waves reinforce each other This reinforce-ment of waves produces a standing wave and resonance is achieved

An open-pipe resonator is a resonating tube with both ends open thatalso will resonate with a sound source In this case the sound wave doesnot reflect off a closed end but rather off an open end The pressure of thereflected wave is inverted for example if a high-pressure wave strikes theopen end a low-pressure wave will rebound as shown in Figure 15-10b

Resonance lengths A standing soundwave in a pipe can be represented by asine wave as shown in Figure 15-11Sine waves can represent either the airpressure or the displacement of the airparticles You can see that standingwaves have nodes and antinodes In thepressure graphs the nodes are regionsof mean atmospheric pressure and at theantinodes the pressure oscillates betweenits maximum and minimum values

Figure 15-9 Raising or loweringthe tube changes the length ofthe air column When the columnis in resonance with the tuningfork the sound is loudest

TimeClosed pipes high pressure

reflects as high pressure

TimeOpen pipes high pressure

reflects as low pressure

ba

Figure 15-10 A tube placed inwater is a closed-pipe resonatorIn closed pipes high pressurewaves reflect as high pressure (a)In open pipes the reflected wavesare inverted (b)

412 Chapter 15 Sound

411-419 CH15-S2-845813 6404 947 AM Page 412

PPP-412

null

20323686

>

>

Displacementantinode

Displacementantinode

Displacementnode

Open Pipe

Air pressure Displacement of air

Pressure node(average-pressure region)

Pressure node(average-pressure region)

Pressure antinode(high- or low-pressure region)

Closed Pipe

Air pressure Displacement of air

Pressure node(average-pressure region)

Pressure antinode(high- or low-pressure region)

Displacementantinode

Displacementnode

Pressure node (average-pressure region)

Antinode

Antinode

Node

Node

λ1 4L

f1 mdash mdash vvλ1 4L

f3 mdash 3f13v4L

λ3 mdash L43

f5 mdash 5f15v4L

λ5 mdash L45

L

Pressure antinode (high-or low-pressure region)

Figure 15-11 Sine wavesrepresent standing waves in pipes

Figure 15-12 A closed piperesonates when its length is anodd number of quarterwavelengths

In the case of the displacement graph the antinodes are regions of high displacement and the nodes are regions of low displacement In both casestwo antinodes (or two nodes) are separated by one-half wavelength

Resonance frequencies in a closed pipe The shortest column of air thatcan have an antinode at the closed end and a node at the open end is one-fourth of a wavelength long as shown in Figure 15-12 As the frequencyis increased additional resonance lengths are found at half-wavelengthintervals Thus columns of length 4 34 54 74 and so on will allbe in resonance with a tuning fork

In practice the first resonance length is slightly longer than one-fourthof a wavelength This is because the pressure variations do not drop to zeroexactly at the open end of the pipe Actually the node is approximately 04 pipe diameters beyond the end Additional resonance lengths how-ever are spaced by exactly one-half of a wavelength Measurements of thespacing between resonances can be used to find the velocity of sound inair as shown in the next Example Problem

Hearing and FrequencyThe human auditory canal acts as a closed-pipe resonator thatincreases the earrsquos sensitivity forfrequencies between 2000 and5000 Hz but the full range offrequencies that people hearextends from 20 to 20000 Hz A dogrsquos hearing extends tofrequencies as high as 45000 Hzand a catrsquos extends to frequenciesas high as 100000 Hz

Section 152 The Physics of Music 413

411-419 CH15-S2-845813 6404 948 AM Page 413

PPP-413

null

79673294

Pressure node (average-pressure region)

Pressure node (average-pressure region)

Pressure antinode (high-or low-pressure region)

Antinode

Antinode

Node

Node

Node

λ1 2L

f1 mdash mdash vvλ1 2L

f2 mdash 2f1vL

λ2 L

f3 mdash 3f13v2L

2Lλ3 mdash3

L

Figure 15-13 An open piperesonates when its length is an even number of quarterwavelengths

Resonance frequencies in an open pipe The shortest column of air thatcan have nodes at both ends is one-half of a wavelength long as shown inFigure 15-13 As the frequency is increased additional resonance lengthsare found at half-wavelength intervals Thus columns of length 2 32 2 and so on will be in resonance with a tuning fork

If open and closed pipes of the same length are used as resonators thewavelength of the resonant sound for the open pipe will be half as long asthat for the closed pipe Therefore the frequency will be twice as high forthe open pipe as for the closed pipe For both pipes resonance lengths arespaced by half-wavelength intervals

Hearing resonance Musical instruments use resonance to increase theloudness of particular notes Open-pipe resonators include flutes and saxophones Clarinets and the hanging pipes under marimbas and xylo-phones are examples of closed-pipe resonators If you shout into a longtunnel the booming sound you hear is the tunnel acting as a resonatorThe seashell in Figure 15-14 acts as a closed-pipe resonator

Resonance on StringsAlthough the waveforms on vibrating strings vary in shape depending

upon how they are produced such as by plucking bowing or strikingthey have many characteristics in common with standing waves on springs

and ropes which you studied in Chapter 14 A string on an instrument is clamped at both ends and therefore thestring must have a node at each end when it vibrates InFigure 15-15 you can see that the first mode of vibrationhas an antinode at the center and is one-half of a wavelengthlong The next resonance occurs when one wavelength fits onthe string and additional standing waves arise when thestring length is 32 2 52 and so on As with an openpipe the resonant frequencies are whole-number multiplesof the lowest frequency

The speed of a wave on a string depends on the tension ofthe string as well as its mass per unit length This makes itpossible to tune a stringed instrument by changing the ten-sion of its strings The tighter the string the faster the wavemoves along it and therefore the higher the frequency of itsstanding waves

Figure 15-14 A seashell acts asa closed-pipe resonator to amplifycertain frequencies from thebackground noise

414 Chapter 15 SoundDavid MechlinIndex Stock Images

411-419 CH15-S2-845813 33004 707 AM Page 414

PPP-414

null

15244986

Pres

sure

Time

Pure sound

Pres

sure

Time

Clarinet

Section 152 The Physics of Music 415

1 2Lf1

v

1

2vL

2 Lf2

vL

2f1

3 23L

f3 32vL 3f1

L Figure 15-15 A string resonateswith standing waves when itslength is a whole number of halfwavelengths

Figure 15-16 A graph of puresound versus time (a) and agraph of clarinet sound wavesversus time (b) are shown

a b

Because strings are so small in cross-sectional area they move very littleair when they vibrate This makes it necessary to attach them to a sound-ing board which transfers their vibrations to the air and produces astronger sound wave Unlike the strings themselves the sounding boardshould not resonate at any single frequency Its purpose is to convey thevibrations of all the strings to the air and therefore it should vibrate wellat all frequencies produced by the instrument Because of the complicatedinteractions among the strings the sounding board and the air the designand construction of stringed instruments are complex processes consideredby many to be as much an art as a science

Sound QualityA tuning fork produces a soft and uninteresting sound This is because

its tines vibrate like simple harmonic oscillators and produce the simple sine wave shown in Figure 15-16a Sounds made by the humanvoice and musical instruments are much more complex like the wave inFigure 15-16b Both waves have the same frequency or pitch but theysound very different The complex wave is produced by using the principle of superposition to add waves of many frequencies The shape of the wave depends on the relative amplitudes of these frequencies In musicalterms the difference between the two waves is called timbre tone color ortone quality

411-419 CH15-S2-845813 33004 708 AM Page 415

PPP-415

null

91114334

18 A 440-Hz tuning fork is held above a closed pipe Find the spacingbetween the resonances when the air temperature is 20degC

19 A 440-Hz tuning fork is used with a resonating column to determinethe velocity of sound in helium gas If the spacings betweenresonances are 110 cm what is the velocity of sound in helium gas

20 The frequency of a tuning fork is unknown A student uses an aircolumn at 27degC and finds resonances spaced by 202 cm What isthe frequency of the tuning fork Use the speed calculated inExample Problem 2 for the speed of sound in air at 27degC

21 A bugle can be thought of as an open pipe If a bugle werestraightened out it would be 265-m long

a If the speed of sound is 343 ms find the lowest frequency thatis resonant for a bugle (ignoring end corrections)

b Find the next two resonant frequencies for the bugle

416 Chapter 15 Sound

Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz

is used with a closed-pipe resonator the loudest sound is heard when the column is 210 cm

and 653 cm long What is the speed of sound in this case Is the temperature warmer or

cooler than normal room temperature which is 20degC Explain your answer

Analyze and Sketch the Problembull Sketch the closed-pipe resonator bull Mark the resonance lengths

Known Unknown

f 392 Hz v LA 210 cmLB 653 cm

Solve for the UnknownSolve for the length of the wave using the length-wavelength relationship for a closed pipe

LB LA 12

2(LB LA) Rearrange the equation for

2(0653 m 0210 m) Substitute LB 0653 m LA 0210 m

0886 m

Use vf

v f Rearrange the equation for v

(392 Hz)(0886 m) Substitute f 392 Hz 0886 m

347 msThe speed is slightly greater than the speed of sound at 20degC indicating that thetemperature is slightly higher than normal room temperature

Evaluate the Answerbull Are the units correct (Hz)(m) (1s)(m) ms The answerrsquos units are correctbull Is the magnitude realistic The speed is slightly greater than 343 ms which is the

speed of sound at 20degC

3

2

1

Math Handbook

Order of Operationspage 843

LA

LB

411-419 CH15-S2-845813 33004 708 AM Page 416

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

Displacementantinode

Displacementantinode

Displacementnode

Open Pipe

Air pressure Displacement of air

Pressure node(average-pressure region)

Pressure node(average-pressure region)

Pressure antinode(high- or low-pressure region)

Closed Pipe

Air pressure Displacement of air

Pressure node(average-pressure region)

Pressure antinode(high- or low-pressure region)

Displacementantinode

Displacementnode

Pressure node (average-pressure region)

Antinode

Antinode

Node

Node

λ1 4L

f1 mdash mdash vvλ1 4L

f3 mdash 3f13v4L

λ3 mdash L43

f5 mdash 5f15v4L

λ5 mdash L45

L

Pressure antinode (high-or low-pressure region)

Figure 15-11 Sine wavesrepresent standing waves in pipes

Figure 15-12 A closed piperesonates when its length is anodd number of quarterwavelengths

In the case of the displacement graph the antinodes are regions of high displacement and the nodes are regions of low displacement In both casestwo antinodes (or two nodes) are separated by one-half wavelength

Resonance frequencies in a closed pipe The shortest column of air thatcan have an antinode at the closed end and a node at the open end is one-fourth of a wavelength long as shown in Figure 15-12 As the frequencyis increased additional resonance lengths are found at half-wavelengthintervals Thus columns of length 4 34 54 74 and so on will allbe in resonance with a tuning fork

In practice the first resonance length is slightly longer than one-fourthof a wavelength This is because the pressure variations do not drop to zeroexactly at the open end of the pipe Actually the node is approximately 04 pipe diameters beyond the end Additional resonance lengths how-ever are spaced by exactly one-half of a wavelength Measurements of thespacing between resonances can be used to find the velocity of sound inair as shown in the next Example Problem

Hearing and FrequencyThe human auditory canal acts as a closed-pipe resonator thatincreases the earrsquos sensitivity forfrequencies between 2000 and5000 Hz but the full range offrequencies that people hearextends from 20 to 20000 Hz A dogrsquos hearing extends tofrequencies as high as 45000 Hzand a catrsquos extends to frequenciesas high as 100000 Hz

Section 152 The Physics of Music 413

411-419 CH15-S2-845813 6404 948 AM Page 413

PPP-413

null

79673294

Pressure node (average-pressure region)

Pressure node (average-pressure region)

Pressure antinode (high-or low-pressure region)

Antinode

Antinode

Node

Node

Node

λ1 2L

f1 mdash mdash vvλ1 2L

f2 mdash 2f1vL

λ2 L

f3 mdash 3f13v2L

2Lλ3 mdash3

L

Figure 15-13 An open piperesonates when its length is an even number of quarterwavelengths

Resonance frequencies in an open pipe The shortest column of air thatcan have nodes at both ends is one-half of a wavelength long as shown inFigure 15-13 As the frequency is increased additional resonance lengthsare found at half-wavelength intervals Thus columns of length 2 32 2 and so on will be in resonance with a tuning fork

If open and closed pipes of the same length are used as resonators thewavelength of the resonant sound for the open pipe will be half as long asthat for the closed pipe Therefore the frequency will be twice as high forthe open pipe as for the closed pipe For both pipes resonance lengths arespaced by half-wavelength intervals

Hearing resonance Musical instruments use resonance to increase theloudness of particular notes Open-pipe resonators include flutes and saxophones Clarinets and the hanging pipes under marimbas and xylo-phones are examples of closed-pipe resonators If you shout into a longtunnel the booming sound you hear is the tunnel acting as a resonatorThe seashell in Figure 15-14 acts as a closed-pipe resonator

Resonance on StringsAlthough the waveforms on vibrating strings vary in shape depending

upon how they are produced such as by plucking bowing or strikingthey have many characteristics in common with standing waves on springs

and ropes which you studied in Chapter 14 A string on an instrument is clamped at both ends and therefore thestring must have a node at each end when it vibrates InFigure 15-15 you can see that the first mode of vibrationhas an antinode at the center and is one-half of a wavelengthlong The next resonance occurs when one wavelength fits onthe string and additional standing waves arise when thestring length is 32 2 52 and so on As with an openpipe the resonant frequencies are whole-number multiplesof the lowest frequency

The speed of a wave on a string depends on the tension ofthe string as well as its mass per unit length This makes itpossible to tune a stringed instrument by changing the ten-sion of its strings The tighter the string the faster the wavemoves along it and therefore the higher the frequency of itsstanding waves

Figure 15-14 A seashell acts asa closed-pipe resonator to amplifycertain frequencies from thebackground noise

414 Chapter 15 SoundDavid MechlinIndex Stock Images

411-419 CH15-S2-845813 33004 707 AM Page 414

PPP-414

null

15244986

Pres

sure

Time

Pure sound

Pres

sure

Time

Clarinet

Section 152 The Physics of Music 415

1 2Lf1

v

1

2vL

2 Lf2

vL

2f1

3 23L

f3 32vL 3f1

L Figure 15-15 A string resonateswith standing waves when itslength is a whole number of halfwavelengths

Figure 15-16 A graph of puresound versus time (a) and agraph of clarinet sound wavesversus time (b) are shown

a b

Because strings are so small in cross-sectional area they move very littleair when they vibrate This makes it necessary to attach them to a sound-ing board which transfers their vibrations to the air and produces astronger sound wave Unlike the strings themselves the sounding boardshould not resonate at any single frequency Its purpose is to convey thevibrations of all the strings to the air and therefore it should vibrate wellat all frequencies produced by the instrument Because of the complicatedinteractions among the strings the sounding board and the air the designand construction of stringed instruments are complex processes consideredby many to be as much an art as a science

Sound QualityA tuning fork produces a soft and uninteresting sound This is because

its tines vibrate like simple harmonic oscillators and produce the simple sine wave shown in Figure 15-16a Sounds made by the humanvoice and musical instruments are much more complex like the wave inFigure 15-16b Both waves have the same frequency or pitch but theysound very different The complex wave is produced by using the principle of superposition to add waves of many frequencies The shape of the wave depends on the relative amplitudes of these frequencies In musicalterms the difference between the two waves is called timbre tone color ortone quality

411-419 CH15-S2-845813 33004 708 AM Page 415

PPP-415

null

91114334

18 A 440-Hz tuning fork is held above a closed pipe Find the spacingbetween the resonances when the air temperature is 20degC

19 A 440-Hz tuning fork is used with a resonating column to determinethe velocity of sound in helium gas If the spacings betweenresonances are 110 cm what is the velocity of sound in helium gas

20 The frequency of a tuning fork is unknown A student uses an aircolumn at 27degC and finds resonances spaced by 202 cm What isthe frequency of the tuning fork Use the speed calculated inExample Problem 2 for the speed of sound in air at 27degC

21 A bugle can be thought of as an open pipe If a bugle werestraightened out it would be 265-m long

a If the speed of sound is 343 ms find the lowest frequency thatis resonant for a bugle (ignoring end corrections)

b Find the next two resonant frequencies for the bugle

416 Chapter 15 Sound

Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz

is used with a closed-pipe resonator the loudest sound is heard when the column is 210 cm

and 653 cm long What is the speed of sound in this case Is the temperature warmer or

cooler than normal room temperature which is 20degC Explain your answer

Analyze and Sketch the Problembull Sketch the closed-pipe resonator bull Mark the resonance lengths

Known Unknown

f 392 Hz v LA 210 cmLB 653 cm

Solve for the UnknownSolve for the length of the wave using the length-wavelength relationship for a closed pipe

LB LA 12

2(LB LA) Rearrange the equation for

2(0653 m 0210 m) Substitute LB 0653 m LA 0210 m

0886 m

Use vf

v f Rearrange the equation for v

(392 Hz)(0886 m) Substitute f 392 Hz 0886 m

347 msThe speed is slightly greater than the speed of sound at 20degC indicating that thetemperature is slightly higher than normal room temperature

Evaluate the Answerbull Are the units correct (Hz)(m) (1s)(m) ms The answerrsquos units are correctbull Is the magnitude realistic The speed is slightly greater than 343 ms which is the

speed of sound at 20degC

3

2

1

Math Handbook

Order of Operationspage 843

LA

LB

411-419 CH15-S2-845813 33004 708 AM Page 416

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

Pressure node (average-pressure region)

Pressure node (average-pressure region)

Pressure antinode (high-or low-pressure region)

Antinode

Antinode

Node

Node

Node

λ1 2L

f1 mdash mdash vvλ1 2L

f2 mdash 2f1vL

λ2 L

f3 mdash 3f13v2L

2Lλ3 mdash3

L

Figure 15-13 An open piperesonates when its length is an even number of quarterwavelengths

Resonance frequencies in an open pipe The shortest column of air thatcan have nodes at both ends is one-half of a wavelength long as shown inFigure 15-13 As the frequency is increased additional resonance lengthsare found at half-wavelength intervals Thus columns of length 2 32 2 and so on will be in resonance with a tuning fork

If open and closed pipes of the same length are used as resonators thewavelength of the resonant sound for the open pipe will be half as long asthat for the closed pipe Therefore the frequency will be twice as high forthe open pipe as for the closed pipe For both pipes resonance lengths arespaced by half-wavelength intervals

Hearing resonance Musical instruments use resonance to increase theloudness of particular notes Open-pipe resonators include flutes and saxophones Clarinets and the hanging pipes under marimbas and xylo-phones are examples of closed-pipe resonators If you shout into a longtunnel the booming sound you hear is the tunnel acting as a resonatorThe seashell in Figure 15-14 acts as a closed-pipe resonator

Resonance on StringsAlthough the waveforms on vibrating strings vary in shape depending

upon how they are produced such as by plucking bowing or strikingthey have many characteristics in common with standing waves on springs

and ropes which you studied in Chapter 14 A string on an instrument is clamped at both ends and therefore thestring must have a node at each end when it vibrates InFigure 15-15 you can see that the first mode of vibrationhas an antinode at the center and is one-half of a wavelengthlong The next resonance occurs when one wavelength fits onthe string and additional standing waves arise when thestring length is 32 2 52 and so on As with an openpipe the resonant frequencies are whole-number multiplesof the lowest frequency

The speed of a wave on a string depends on the tension ofthe string as well as its mass per unit length This makes itpossible to tune a stringed instrument by changing the ten-sion of its strings The tighter the string the faster the wavemoves along it and therefore the higher the frequency of itsstanding waves

Figure 15-14 A seashell acts asa closed-pipe resonator to amplifycertain frequencies from thebackground noise

414 Chapter 15 SoundDavid MechlinIndex Stock Images

411-419 CH15-S2-845813 33004 707 AM Page 414

PPP-414

null

15244986

Pres

sure

Time

Pure sound

Pres

sure

Time

Clarinet

Section 152 The Physics of Music 415

1 2Lf1

v

1

2vL

2 Lf2

vL

2f1

3 23L

f3 32vL 3f1

L Figure 15-15 A string resonateswith standing waves when itslength is a whole number of halfwavelengths

Figure 15-16 A graph of puresound versus time (a) and agraph of clarinet sound wavesversus time (b) are shown

a b

Because strings are so small in cross-sectional area they move very littleair when they vibrate This makes it necessary to attach them to a sound-ing board which transfers their vibrations to the air and produces astronger sound wave Unlike the strings themselves the sounding boardshould not resonate at any single frequency Its purpose is to convey thevibrations of all the strings to the air and therefore it should vibrate wellat all frequencies produced by the instrument Because of the complicatedinteractions among the strings the sounding board and the air the designand construction of stringed instruments are complex processes consideredby many to be as much an art as a science

Sound QualityA tuning fork produces a soft and uninteresting sound This is because

its tines vibrate like simple harmonic oscillators and produce the simple sine wave shown in Figure 15-16a Sounds made by the humanvoice and musical instruments are much more complex like the wave inFigure 15-16b Both waves have the same frequency or pitch but theysound very different The complex wave is produced by using the principle of superposition to add waves of many frequencies The shape of the wave depends on the relative amplitudes of these frequencies In musicalterms the difference between the two waves is called timbre tone color ortone quality

411-419 CH15-S2-845813 33004 708 AM Page 415

PPP-415

null

91114334

18 A 440-Hz tuning fork is held above a closed pipe Find the spacingbetween the resonances when the air temperature is 20degC

19 A 440-Hz tuning fork is used with a resonating column to determinethe velocity of sound in helium gas If the spacings betweenresonances are 110 cm what is the velocity of sound in helium gas

20 The frequency of a tuning fork is unknown A student uses an aircolumn at 27degC and finds resonances spaced by 202 cm What isthe frequency of the tuning fork Use the speed calculated inExample Problem 2 for the speed of sound in air at 27degC

21 A bugle can be thought of as an open pipe If a bugle werestraightened out it would be 265-m long

a If the speed of sound is 343 ms find the lowest frequency thatis resonant for a bugle (ignoring end corrections)

b Find the next two resonant frequencies for the bugle

416 Chapter 15 Sound

Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz

is used with a closed-pipe resonator the loudest sound is heard when the column is 210 cm

and 653 cm long What is the speed of sound in this case Is the temperature warmer or

cooler than normal room temperature which is 20degC Explain your answer

Analyze and Sketch the Problembull Sketch the closed-pipe resonator bull Mark the resonance lengths

Known Unknown

f 392 Hz v LA 210 cmLB 653 cm

Solve for the UnknownSolve for the length of the wave using the length-wavelength relationship for a closed pipe

LB LA 12

2(LB LA) Rearrange the equation for

2(0653 m 0210 m) Substitute LB 0653 m LA 0210 m

0886 m

Use vf

v f Rearrange the equation for v

(392 Hz)(0886 m) Substitute f 392 Hz 0886 m

347 msThe speed is slightly greater than the speed of sound at 20degC indicating that thetemperature is slightly higher than normal room temperature

Evaluate the Answerbull Are the units correct (Hz)(m) (1s)(m) ms The answerrsquos units are correctbull Is the magnitude realistic The speed is slightly greater than 343 ms which is the

speed of sound at 20degC

3

2

1

Math Handbook

Order of Operationspage 843

LA

LB

411-419 CH15-S2-845813 33004 708 AM Page 416

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

Pres

sure

Time

Pure sound

Pres

sure

Time

Clarinet

Section 152 The Physics of Music 415

1 2Lf1

v

1

2vL

2 Lf2

vL

2f1

3 23L

f3 32vL 3f1

L Figure 15-15 A string resonateswith standing waves when itslength is a whole number of halfwavelengths

Figure 15-16 A graph of puresound versus time (a) and agraph of clarinet sound wavesversus time (b) are shown

a b

Because strings are so small in cross-sectional area they move very littleair when they vibrate This makes it necessary to attach them to a sound-ing board which transfers their vibrations to the air and produces astronger sound wave Unlike the strings themselves the sounding boardshould not resonate at any single frequency Its purpose is to convey thevibrations of all the strings to the air and therefore it should vibrate wellat all frequencies produced by the instrument Because of the complicatedinteractions among the strings the sounding board and the air the designand construction of stringed instruments are complex processes consideredby many to be as much an art as a science

Sound QualityA tuning fork produces a soft and uninteresting sound This is because

its tines vibrate like simple harmonic oscillators and produce the simple sine wave shown in Figure 15-16a Sounds made by the humanvoice and musical instruments are much more complex like the wave inFigure 15-16b Both waves have the same frequency or pitch but theysound very different The complex wave is produced by using the principle of superposition to add waves of many frequencies The shape of the wave depends on the relative amplitudes of these frequencies In musicalterms the difference between the two waves is called timbre tone color ortone quality

411-419 CH15-S2-845813 33004 708 AM Page 415

PPP-415

null

91114334

18 A 440-Hz tuning fork is held above a closed pipe Find the spacingbetween the resonances when the air temperature is 20degC

19 A 440-Hz tuning fork is used with a resonating column to determinethe velocity of sound in helium gas If the spacings betweenresonances are 110 cm what is the velocity of sound in helium gas

20 The frequency of a tuning fork is unknown A student uses an aircolumn at 27degC and finds resonances spaced by 202 cm What isthe frequency of the tuning fork Use the speed calculated inExample Problem 2 for the speed of sound in air at 27degC

21 A bugle can be thought of as an open pipe If a bugle werestraightened out it would be 265-m long

a If the speed of sound is 343 ms find the lowest frequency thatis resonant for a bugle (ignoring end corrections)

b Find the next two resonant frequencies for the bugle

416 Chapter 15 Sound

Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz

is used with a closed-pipe resonator the loudest sound is heard when the column is 210 cm

and 653 cm long What is the speed of sound in this case Is the temperature warmer or

cooler than normal room temperature which is 20degC Explain your answer

Analyze and Sketch the Problembull Sketch the closed-pipe resonator bull Mark the resonance lengths

Known Unknown

f 392 Hz v LA 210 cmLB 653 cm

Solve for the UnknownSolve for the length of the wave using the length-wavelength relationship for a closed pipe

LB LA 12

2(LB LA) Rearrange the equation for

2(0653 m 0210 m) Substitute LB 0653 m LA 0210 m

0886 m

Use vf

v f Rearrange the equation for v

(392 Hz)(0886 m) Substitute f 392 Hz 0886 m

347 msThe speed is slightly greater than the speed of sound at 20degC indicating that thetemperature is slightly higher than normal room temperature

Evaluate the Answerbull Are the units correct (Hz)(m) (1s)(m) ms The answerrsquos units are correctbull Is the magnitude realistic The speed is slightly greater than 343 ms which is the

speed of sound at 20degC

3

2

1

Math Handbook

Order of Operationspage 843

LA

LB

411-419 CH15-S2-845813 33004 708 AM Page 416

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

18 A 440-Hz tuning fork is held above a closed pipe Find the spacingbetween the resonances when the air temperature is 20degC

19 A 440-Hz tuning fork is used with a resonating column to determinethe velocity of sound in helium gas If the spacings betweenresonances are 110 cm what is the velocity of sound in helium gas

20 The frequency of a tuning fork is unknown A student uses an aircolumn at 27degC and finds resonances spaced by 202 cm What isthe frequency of the tuning fork Use the speed calculated inExample Problem 2 for the speed of sound in air at 27degC

21 A bugle can be thought of as an open pipe If a bugle werestraightened out it would be 265-m long

a If the speed of sound is 343 ms find the lowest frequency thatis resonant for a bugle (ignoring end corrections)

b Find the next two resonant frequencies for the bugle

416 Chapter 15 Sound

Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz

is used with a closed-pipe resonator the loudest sound is heard when the column is 210 cm

and 653 cm long What is the speed of sound in this case Is the temperature warmer or

cooler than normal room temperature which is 20degC Explain your answer

Analyze and Sketch the Problembull Sketch the closed-pipe resonator bull Mark the resonance lengths

Known Unknown

f 392 Hz v LA 210 cmLB 653 cm

Solve for the UnknownSolve for the length of the wave using the length-wavelength relationship for a closed pipe

LB LA 12

2(LB LA) Rearrange the equation for

2(0653 m 0210 m) Substitute LB 0653 m LA 0210 m

0886 m

Use vf

v f Rearrange the equation for v

(392 Hz)(0886 m) Substitute f 392 Hz 0886 m

347 msThe speed is slightly greater than the speed of sound at 20degC indicating that thetemperature is slightly higher than normal room temperature

Evaluate the Answerbull Are the units correct (Hz)(m) (1s)(m) ms The answerrsquos units are correctbull Is the magnitude realistic The speed is slightly greater than 343 ms which is the

speed of sound at 20degC

3

2

1

Math Handbook

Order of Operationspage 843

LA

LB

411-419 CH15-S2-845813 33004 708 AM Page 416

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

1000 2000 3000 1000 2000 3000 1000 2000 3000

Inte

nsi

ty

Inte

nsi

ty

Inte

nsi

ty

Frequency (Hz) Frequency (Hz) Frequency (Hz)

Violin Clarinet Piano

Figure 15-17 A violin a clarinetand a piano produce characteristicsound spectra Each spectrum is unique as is the timbre of the instrument

1 Determine the tension FT in a violin string of mass mand length L that will play the fundamental note at thesame frequency as a closed pipe also of length LExpress your answer in terms of m L and the speed of sound in air v The equation for the speed of a wave

on a string is u FT where FT is the tension in the

string and is the mass per unit length of the string

2 What is the tension in a string of mass 10 g and 400cm long that plays the same note as a closed pipe ofthe same length

The sound spectrum fundamental and harmonics The complex soundwave in Figure 15-16b was made by a clarinet Why does the clarinet pro-duce such a sound wave The air column in a clarinet acts as a closed pipeLook back at Figure 15-12 which shows three resonant frequencies for aclosed pipe Because the clarinet acts as a closed pipe for a clarinet oflength L the lowest frequency f1 that will be resonant is v4L This lowestfrequency is called the fundamental A closed pipe also will resonate at 3f1 5f1 and so on These higher frequencies which are odd-numbermultiples of the fundamental frequency are called harmonics It is theaddition of these harmonics that gives a clarinet its distinctive timbre

Some instruments such as an oboe act as open-pipe resonators Theirfundamental frequency which is also the first harmonic is f1 v2L withsubsequent harmonics at 2f1 3f1 4f1 and so on Different combinationsand amplitudes of these harmonics give each instrument its own unique tim-bre A graph of the amplitude of a wave versus its frequency is called a soundspectrum The spectra of three instruments are shown in Figure 15-17

L

L

Section 152 The Physics of Music 417(l)Geoff Butler (c)HIRBIndex Stock Images (r)PhotodiscArtbase

411-419 CH15-S2-845813 6404 951 AM Page 417

PPP-417

null

9100985

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

418 Chapter 15 Sound

Perfect fourth34

Major third45

Octave12

Perfect fifth23

Figure 15-18 These time graphsshow the superposition of twowaves having the ratios of 12 23 34 and 45

Consonance and dissonance When sounds that have two differentpitches are played at the same time the resulting sound can be either pleas-ant or jarring In musical terms several pitches played together are calleda chord An unpleasant set of pitches is called dissonance If the combi-nation is pleasant the sounds are said to be in consonance

What makes a sound pleasant to listen to Different cultures have dif-ferent definitions but most Western cultures accept the definitions ofPythagoras who lived in ancient Greece Pythagoras experimented byplucking two strings at the same time He noted that pleasing soundsresulted when the strings had lengths in small whole-number ratios forexample 12 23 or 34 This means that their pitches (frequencies) willalso have small whole-number ratios

Musical intervals Two notes with frequencies related by the ratio 12 aresaid to differ by an octave For example if a note has a frequency of 440 Hza note that is one octave higher has a frequency of 880 Hz The funda-mental and its harmonics are related by octaves the first harmonic is oneoctave higher than the fundamental the second is two octaves higher andso on The sum of the fundamental and the first harmonic is shown inFigure 15-18a It is the ratio of two frequencies not the size of the inter-val between them that determines the musical interval

In other musical intervals two pitches may be close together For exam-ple the ratio of frequencies for a ldquomajor thirdrdquo is 45 A typical major thirdis made up of the notes C and E The note C has a frequency of 262 Hz so E has a frequency of (54)(262 Hz) = 327 Hz In the same way notes in a ldquofourthrdquo (C and F) have a frequency ratio of 34 and those in a ldquofifthrdquo(C and G) have a ratio of 23 Graphs of these pleasant sounds are shownin Figure 15-18 More than two notes sounded together also can produceconsonance The three notes called do mi and sol make a major chordFor at least 2500 years this has been recognized as the sweetest of the three-note chords it has the frequency ratio of 456

BeatsYou have seen that consonance is defined in terms of the ratio of fre-

quencies When the ratio becomes nearly 11 the frequencies become veryclose Two frequencies that are nearly identical interfere to produce highand low sound levels as illustrated in Figure 15-19 This oscillation ofwave amplitude is called a beat The frequency of a beat is the magnitudeof difference between the frequencies of the two waves fbeat fA fBWhen the difference is less than 7 Hz the ear detects this as a pulsation ofloudness Musical instruments often are tuned by sounding one againstanother and adjusting the frequency of one until the beat disappears

a c

d

Sounds GoodSometimes it can be pretty toughto tell just by looking whether aninstrument will act as an open-pipe resonator or as a closed-piperesonator Ask a musician whoplays a wind instrument to bring it to class 1 Measure the length of theinstrument2 Have the musician play thelowest possible note on theinstrument3 Determine the frequency of thenote using a frequency generator

Analyze and Conclude4 Draw Conclusions Did thetested instrument behave mostlike a closed-pipe resonator ormost like an open-pipe resonatorWas the frequency thefundamental or one of thesubsequent harmonics5 Determine the frequencies of the notes that would form anoctave a perfect fifth a perfectfourth and a major third with your observed note

b

411-419 CH15-S2-845813 33004 911 AM Page 418

PPP-418

null

21201486

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

Section 152 The Physics of Music 419

22 Origins of Sound What is the vibrating objectthat produces sounds in each of the followinga a human voiceb a clarinetc a tubad a violin

23 Resonance in Air Columns Why is the tube fromwhich a tuba is made much longer than that of acornet

24 Resonance in Open Tubes How must the lengthof an open tube compare to the wavelength of thesound to produce the strongest resonance

25 Resonance on Strings A violin sounds a note of F sharp with a pitch of 370 Hz What are the frequencies of the next three harmonics producedwith this note

26 Resonance in Closed Pipes One closed organpipe has a length of 240 m

a What is the frequency of the note played by thispipe

b When a second pipe is played at the same time a 140-Hz beat note is heard By how much is the second pipe too long

27 Timbre Why do various instruments sound differ-ent even when they play the same note

28 Beats A tuning fork produces three beats per second with a second 392-Hz tuning fork What isthe frequency of the first tuning fork

29 Critical Thinking Strike a tuning fork with a rub-ber hammer and hold it at armrsquos length Then pressits handle against a desk a door a filing cabinetand other objects What do you hear Why

152 Section Review

physicsppcomself_check_quiz

Figure 15-20 Noise iscomposed of several frequenciesand involves random changes infrequency and amplitude

Sound Reproduction and NoiseHow often do you listen to music produced directly by a human voice

or musical instrument Most of the time the music has been recorded andplayed through electronic systems To reproduce the sound faithfully thesystem must accommodate all frequencies equally A good stereo systemkeeps the amplitudes of all frequencies between 20 and 20000 Hz thesame to within 3 dB

A telephone system on the other hand needs only to transmit the infor-mation in spoken language Frequencies between 300 and 3000 Hz are sufficient Reducing the number of frequencies present helps reduce thenoise A noise wave is shown in Figure 15-20 Many frequencies are present with approximately the same amplitude While noise is not help-ful in a telephone system some people claim that listening to noise has a calming effect For this reason some dentists use noise to help theirpatients relax

Am

plit

ud

e

T2

T1

TimeTbeat

Figure 15-19 Beats occur as aresult of the superposition of twosound waves of slightly differentfrequencies

411-419 CH15-S2-845813 33004 712 AM Page 419

Speed of SoundIf a vibrating tuning fork is held above a closed pipe of the proper length the airin the pipe will vibrate at the same frequency f as the tuning fork By placing aglass tube in a large water-filled graduated cylinder the length of the glass tubecan be changed by raising or lowering it in the water The shortest column of airthat will resonate occurs when the tube is one-fourth of a wavelength long Thisresonance will produce the loudest sound and the wavelength at this resonance is described by 4L where L is the length from the water to the open end ofthe pipe In this lab you will determine L calculate and calculate the speed of sound

QUESTIONHow can you use a closed-pipe resonator to determine the speed of sound

Alternate CBL instructionscan be found on the Web site

physicsppcom

Collect and organize data to obtain resonantpoints in a closed pipe

Measure the length of a closed-pipe resonator Analyze the data to determine the speed of

sound

Immediately wipe up any spilled liquids Glass is fragile Handle with care

three tuning forks of known frequenciesgraduated cylinder (1000-mL)watertuning fork mallet

metric rulerthermometer (non-mercury)glass tube (approximately 40 cm in length and 35 cm in diameter)

1 Put on your safety goggles Fill the graduatedcylinder nearly to the top with water

2 Measure the room air temperature and recordit in Data Table 1

3 Select a tuning fork and record its frequency in Data Tables 2 and 3

4 Measure and record the diameter of the glasstube in Data Table 2

5 Carefully place the glass tube into the water-filled graduated cylinder

6 Hold the tuning fork by the base Swiftly strikeit on the side with the tuning fork mallet Donot strike the tuning fork on the laboratorytable or other hard surface

7 Hold the vibrating fork over the open end ofthe glass tube and slowly raise the tube andthe fork until the loudest sound is heard Oncethis point is located move the tube up anddown slightly to determine the exact point ofresonance Measure the distance from thewater to the top of the glass tube and recordthis distance in Data Table 2

8 Repeat steps 3 6 and 7 for two additional tuningforks and record your results as trials 2 and 3The three tuning forks that you test should resonate at three different frequencies

9 Empty the water from the graduated cylinder

Procedure

Materials

Safety Precautions

Objectives

420

Hor

izon

s C

ompa

nies

420-421 CH15-LAB-845813 33004 720 AM Page 420

421

To find out more about the properties of soundwaves visit the Web site physicsppcom

Data Table 1Trial Temperature

(degC)Accepted Speed of Sound (ms)

ExperimentalSpeed of Sound

(ms)

1

2

3

Data Table 2Trial Tuning Fork

Frequency (Hz)

Diameter (m)

Length of Tube AboveWater (m)

CalculatedWavelength

(m)

1

2

3

1 Calculate the accepted speed of soundusing the relationship v 331 ms 060T where v is the speed of sound attemperature T and T is the air tempera-ture in degrees Celsius Record this asthe accepted speed of sound in DataTables 1 and 3 for all the trials

2 Since the first resonant point is located whenthe tube is one-fourth of a wavelength abovethe water use the measured length of the tubeto determine the calculated wavelength foreach trial Record the calculated wavelengths in Data Table 2

3 Multiply the values in Data Table 2 of wavelengthand frequency to determine the experimentalspeed of sound and record this in Data Table 1for each of the trials

4 Error Analysis For each trial in Data Table 1 determine the relative error between the experimental and accepted speed of sound

Relative error

100

5 Critique To improve the accuracy of your calculations the tube diameter must be takeninto consideration The following relationshipprovides a more accurate calculation of wavelength 4(L 04d ) where is thewavelength L is the length of the tube abovethe water and d is the inside diameter of thetube Using the values in Data Table 1 forlength and diameter recalculate and recordit in Data Table 3 as the corrected wavelengthCalculate the corrected experimental speed ofsound by multiplying the tuning fork frequencyand corrected wavelength and record the newvalue for the corrected experimental speed ofsound in Data Table 3

Accepted valueExperimental value

Accepted value

Analyze

6 Error Analysis For each trial in Data Table 3determine the relative error between the cor-rected experimental speed and the acceptedspeed of sound Use the same formula that youused in step 4 above

1 Infer In general the first resonant pointoccurs when the tube length 4 What are the next two lengths where resonance will occur

2 Think Critically If you had a longer tubewould it be possible to locate another positionwhere resonance occurs Explain your answer

Which result produced the more accurate speed ofsound

Explain the relationship between the size of organpipes and their resonant frequencies

Real-World Physics

Going Further

Conclude and Apply

Data Table 3Trial Tuning Fork

Frequency(Hz)

AcceptedSpeed of

Sound (ms)

CorrectedCalculated

Wavelength(m)

CorrectedExperimental

Speed ofSound (ms)

1

2

3

420-421 CH15-LAB-845813 61004 840 PM Page 421

422 Extreme Physics

Sound Waves in the SunThe study of wave oscillations in the Sun is called helioseismology Naturally occurringsound waves (p waves) gravity waves and surface gravity waves all occur in the Sun All of these waves are composed of oscillating particles but different forces cause theoscillations

For sound waves pressure differencescause the particles to oscillate In theSun sound waves travel through the convective zone which is just under the surface or photosphere The soundwaves do not travel in a straight line as shown in the image

Ringing like a Bell The soundwaves in the Sun cause the surface of the Sun to vibrate in the radial direc-tion much like a ringing bell vibratesWhen a bell is rung a clapper hits thebell in one place and standing waves arecreated The surface of the Sun does havestanding waves but they are not causedby one large event Instead scientistshypothesize that many smaller disrup-tions in the convective zone start mostof the sound waves in the Sun Just likeboiling water in a pot can be noisy bubblesthat are larger than the state of Texas form onthe surface of the Sun and start sound waves

Unlike a pot of boiling water the sound coming from the Sun is much too low for us tohear The A above middle C on the piano has aperiod of 000227 s (f 440 Hz) The middlemode of oscillation of the waves in the Sun hasa period of 5 min (f 0003 Hz)

Because we cannot hear the sound wavesfrom the Sun scientists measure the motion ofthe surface of the Sun to learn about its soundwaves Because a sound wave takes 2 h to travelfrom one side of the Sun to the other the Sunmust be observed for long time periods Thisnecessity makes observations from Earth difficult because the Sun is not visible duringthe night In 1995 the Solar and HeliosphericObservatory (SOHO) was launched by NASAThis satellite orbits Earth such that it alwayscan observe the Sun

The motion of the surface of the Sun ismeasured by observing Doppler shifts in sunlight The measured vibrations are a com-plicated pattern that equals the sum of all ofthe standing waves present in the Sun Just likea ringing bell many overtones are present inthe Sun Through careful analysis the individualstanding waves in the Sun and their intensitiescan be calculated

Results Because composition temperatureand density affect the propagation of soundwaves the Sunrsquos wave oscillations provideinformation about its interior SOHO resultshave given insight into the rotation rate of theSun as a function of latitude and depth as wellas the density and temperature of the Sun Theseresults are compared to theoretical calculationsto improve our understanding of the Sun

Photosphere

ChromosphereConvective zone

Radiativezone

Core

p

1 Hypothesize How do scientists sepa-rate the surface motion due to soundwaves from the motion due to the rotation of the Sun

2 Critical Thinking Would sound wavesin another star similar to the Sun butdifferent in size have the same wave-length as sound waves in the Sun

Going Further

Sound waves (p waves) travel through the Sunrsquos convective zone

422 CH15-FEATURE-845813 33004 723 AM Page 422

151 Properties and Detection of Sound

Vocabularybull sound wave (p 404)

bull pitch (p 406)

bull loudness (p 406)

bull sound level (p 406)

bull decibel (p 406)

bull Doppler effect (p 407)

152 The Physics of Music

Vocabularybull closed-pipe resonator

(p 412)

bull open-pipe resonator (p 412)

bull fundamental (p 417)

bull harmonics (p 417)

bull dissonance (p 418)

bull consonance (p 418)

bull beat (p 418)

Key Conceptsbull Sound is a pressure variation transmitted through matter as a longitudinal

wave

bull A sound wave has frequency wavelength speed and amplitude Sound wavesreflect and interfere

bull The speed of sound in air at room temperature (20degC) is 343 ms The speedincreases roughly 06 ms with each 1degC increase in temperature

bull Sound detectors convert the energy carried by a sound wave into anotherform of energy The human ear is a highly efficient and sensitive detector of sound waves

bull The frequency of a sound wave is heard as its pitch

bull The pressure amplitude of a sound wave can be measured in decibels (dB)

bull The loudness of sound as perceived by the ear and brain depends mainly on its amplitude

bull The Doppler effect is the change in frequency of sound caused by the motion of either the source or the detector It can be calculated with thefollowing equation

fd fsvv

vvd

s

Key Conceptsbull Sound is produced by a vibrating object in a material medium

bull Most sounds are complex waves that are composed of more than onefrequency

bull An air column can resonate with a sound source thereby increasing theamplitude of its resonant frequency

bull A closed pipe resonates when its length is 4 34 54 and so on Its resonant frequencies are odd-numbered multiples of the fundamental

bull An open pipe resonates when its length is 2 22 32 and so on Its resonant frequencies are whole-number multiples of the fundamental

bull A clamped string has a node at each end and resonates when its length is2 22 32 and so on just as with an open pipe The stringrsquos resonantfrequencies are also whole-number multiples of the fundamental

bull The frequencies and intensities of the complex waves produced by a musicalinstrument determine the timbre that is characteristic of that instrument

bull The fundamental frequency and harmonics can be described in terms ofresonance

bull Notes on a musical scale differ in frequency by small whole-number ratiosAn octave has a frequency ratio of 12

bull Two waves with almost the same frequency interfere to produce beats

423physicsppcomvocabulary_puzzlemaker

423-429 CH15-SG CA STP-845813 33004 728 AM Page 423

151 Properties and Detection of Sound

Vocabularybull sound wave (p 404)

bull pitch (p 406)

bull loudness (p 406)

bull sound level (p 406)

bull decibel (p 406)

bull Doppler effect (p 407)

152 The Physics of Music

Vocabularybull closed-pipe resonator

(p 412)

bull open-pipe resonator (p 412)

bull fundamental (p 417)

bull harmonics (p 417)

bull dissonance (p 418)

bull consonance (p 418)

bull beat (p 418)

Key Conceptsbull Sound is a pressure variation transmitted through matter as a longitudinal

wave

bull A sound wave has frequency wavelength speed and amplitude Sound wavesreflect and interfere

bull The speed of sound in air at room temperature (20degC) is 343 ms The speedincreases roughly 06 ms with each 1degC increase in temperature

bull Sound detectors convert the energy carried by a sound wave into anotherform of energy The human ear is a highly efficient and sensitive detector of sound waves

bull The frequency of a sound wave is heard as its pitch

bull The pressure amplitude of a sound wave can be measured in decibels (dB)

bull The loudness of sound as perceived by the ear and brain depends mainly on its amplitude

bull The Doppler effect is the change in frequency of sound caused by the motion of either the source or the detector It can be calculated with thefollowing equation

fd fsvv

vvd

s

Key Conceptsbull Sound is produced by a vibrating object in a material medium

bull Most sounds are complex waves that are composed of more than onefrequency

bull An air column can resonate with a sound source thereby increasing theamplitude of its resonant frequency

bull A closed pipe resonates when its length is 4 34 54 and so on Its resonant frequencies are odd-numbered multiples of the fundamental

bull An open pipe resonates when its length is 2 22 32 and so on Its resonant frequencies are whole-number multiples of the fundamental

bull A clamped string has a node at each end and resonates when its length is2 22 32 and so on just as with an open pipe The stringrsquos resonantfrequencies are also whole-number multiples of the fundamental

bull The frequencies and intensities of the complex waves produced by a musicalinstrument determine the timbre that is characteristic of that instrument

bull The fundamental frequency and harmonics can be described in terms ofresonance

bull Notes on a musical scale differ in frequency by small whole-number ratiosAn octave has a frequency ratio of 12

bull Two waves with almost the same frequency interfere to produce beats

423physicsppcomvocabulary_puzzlemaker

423-429 CH15-SG CA STP-845813 33004 728 AM Page 423

Sound

properties

frequency loudness

30 Complete the concept map below using thefollowing terms amplitude perception pitch speed

Mastering Concepts31 What are the physical characteristics of sound

waves (151)

32 When timing the 100-m run officials at the finishline are instructed to start their stopwatches at thesight of smoke from the starterrsquos pistol and not atthe sound of its firing Explain What would happento the times for the runners if the timing startedwhen sound was heard (151)

33 Name two types of perception of sound and thephysical characteristics of sound waves thatcorrespond to them (151)

34 Does the Doppler shift occur for only some types of waves or for all types of waves (151)

35 Sound waves with frequencies higher than can beheard by humans called ultrasound can betransmitted through the human body How couldultrasound be used to measure the speed of bloodflowing in veins or arteries Explain how the waveschange to make this measurement possible (151)

36 What is necessary for the production andtransmission of sound (152)

37 Singing How can a certain note sung by an operasinger cause a crystal glass to shatter (152)

38 Marching In the military as marching soldiersapproach a bridge the command ldquoroute steprdquo isgiven The soldiers then walk out-of-step with eachother as they cross the bridge Explain (152)

39 Musical Instruments Why donrsquot most musicalinstruments sound like tuning forks (152)

40 Musical Instruments What property distinguishesnotes played on both a trumpet and a clarinet ifthey have the same pitch and loudness (152)

41 Trombones Explain how the slide of a tromboneshown in Figure 15-21 changes the pitch of thesound in terms of a trombone being a resonance tube (152)

Applying Concepts42 Estimation To estimate the distance in kilometers

between you and a lightning flash count theseconds between the flash and the thunder anddivide by 3 Explain how this rule works Devise a similar rule for miles

43 The speed of sound increases by about 06 ms foreach degree Celsius when the air temperature risesFor a given sound as the temperature increaseswhat happens to the followinga the frequencyb the wavelength

44 Movies In a science-fiction movie a satellite blowsup The crew of a nearby ship immediately hearsand sees the explosion If you had been hired as anadvisor what two physics errors would you havenoticed and corrected

45 The Redshift Astronomers have observed that thelight coming from distant galaxies appears redderthan light coming from nearer galaxies With thehelp of Figure 15-22 which shows the visiblespectrum explain why astronomers conclude thatdistant galaxies are moving away from Earth

46 Does a sound of 40 dB have a factor of 100 (102)times greater pressure variation than the thresholdof hearing or a factor of 40 times greater

4107 m 5107 m 6107 m 7107 m

Concept Mapping

424 Chapter 15 Sound For more problems go to Additional Problems Appendix B

Figure 15-22

Figure 15-21

(t)Tim Fuller (b)Carolina BiologicalVisuals Unlimited

423-429 CH15-SG CA STP-845813 33004 729 AM Page 424

47 If the pitch of sound is increased what are thechanges in the followinga the frequencyb the wavelengthc the wave velocityd the amplitude of the wave

48 The speed of sound increases with temperatureWould the pitch of a closed pipe increase ordecrease when the temperature of the air risesAssume that the length of the pipe does not change

49 Marching Bands Two flutists are tuning up If theconductor hears the beat frequency increasing arethe two flute frequencies getting closer together orfarther apart

50 Musical Instruments A covered organ pipe plays a certain note If the cover is removed to make it an open pipe is the pitch increased or decreased

51 Stringed Instruments On a harp Figure 15-23along strings produce low notes and short stringsproduce high notes On a guitar Figure 15-23bthe strings are all the same length How can theyproduce notes of different pitches

Mastering Problems151 Properties and Detection of Sound52 You hear the sound of the firing of a distant cannon

50 s after seeing the flash How far are you fromthe cannon

53 If you shout across a canyon and hear an echo 30 slater how wide is the canyon

54 A sound wave has a frequency of 4700 Hz andtravels along a steel rod If the distance betweencompressions or regions of high pressure is 11 mwhat is the speed of the wave

55 Bats The sound emitted by bats has a wavelengthof 35 mm What is the soundrsquos frequency in air

56 Photography As shown in Figure 15-24 somecameras determine the distance to the subject bysending out a sound wave and measuring the timeneeded for the echo to return to the camera Howlong would it take the sound wave to return to sucha camera if the subject were 300 m away

57 Sound with a frequency of 2616 Hz travels throughwater at 25degC Find the soundrsquos wavelength in waterDo not confuse sound waves moving through waterwith surface waves moving through water

58 If the wavelength of a 440102-Hz sound infreshwater is 330 m what is the speed of sound in freshwater

59 Sound with a frequency of 442 Hz travels throughan iron beam Find the wavelength of the sound in iron

60 Aircraft Adam an airport employee is workingnear a jet plane taking off He experiences a soundlevel of 150 dBa If Adam wears ear protectors that reduce the

sound level to that of a typical rock concert what decrease in dB is provided

b If Adam then hears something that sounds like a barely audible whisper what will a person notwearing the ear protectors hear

61 Rock Music A rock band plays at an 80-dB soundlevel How many times greater is the sound pressurefrom another rock band playing at each of thefollowing sound levelsa 100 dBb 120 dB

62 A coiled-spring toy is shaken at a frequency of 40 Hz such that standing waves are observed with a wavelength of 050 m What is the speed of propagation of the wave

63 A baseball fan on a warm summer day (30degC) sitsin the bleachers 152 m away from home plate a What is the speed of sound in air at 30degCb How long after seeing the ball hit the bat does

the fan hear the crack of the bat

300 m

Chapter 15 Assessment 425physicsppcomchapter_test

Figure 15-23

Figure 15-24

a b

Getty Images

423-429 CH15-SG CA STP-845813 33004 730 AM Page 425

47 If the pitch of sound is increased what are thechanges in the followinga the frequencyb the wavelengthc the wave velocityd the amplitude of the wave

48 The speed of sound increases with temperatureWould the pitch of a closed pipe increase ordecrease when the temperature of the air risesAssume that the length of the pipe does not change

49 Marching Bands Two flutists are tuning up If theconductor hears the beat frequency increasing arethe two flute frequencies getting closer together orfarther apart

50 Musical Instruments A covered organ pipe plays a certain note If the cover is removed to make it an open pipe is the pitch increased or decreased

51 Stringed Instruments On a harp Figure 15-23along strings produce low notes and short stringsproduce high notes On a guitar Figure 15-23bthe strings are all the same length How can theyproduce notes of different pitches

Mastering Problems151 Properties and Detection of Sound52 You hear the sound of the firing of a distant cannon

50 s after seeing the flash How far are you fromthe cannon

53 If you shout across a canyon and hear an echo 30 slater how wide is the canyon

54 A sound wave has a frequency of 4700 Hz andtravels along a steel rod If the distance betweencompressions or regions of high pressure is 11 mwhat is the speed of the wave

55 Bats The sound emitted by bats has a wavelengthof 35 mm What is the soundrsquos frequency in air

56 Photography As shown in Figure 15-24 somecameras determine the distance to the subject bysending out a sound wave and measuring the timeneeded for the echo to return to the camera Howlong would it take the sound wave to return to sucha camera if the subject were 300 m away

57 Sound with a frequency of 2616 Hz travels throughwater at 25degC Find the soundrsquos wavelength in waterDo not confuse sound waves moving through waterwith surface waves moving through water

58 If the wavelength of a 440102-Hz sound infreshwater is 330 m what is the speed of sound in freshwater

59 Sound with a frequency of 442 Hz travels throughan iron beam Find the wavelength of the sound in iron

60 Aircraft Adam an airport employee is workingnear a jet plane taking off He experiences a soundlevel of 150 dBa If Adam wears ear protectors that reduce the

sound level to that of a typical rock concert what decrease in dB is provided

b If Adam then hears something that sounds like a barely audible whisper what will a person notwearing the ear protectors hear

61 Rock Music A rock band plays at an 80-dB soundlevel How many times greater is the sound pressurefrom another rock band playing at each of thefollowing sound levelsa 100 dBb 120 dB

62 A coiled-spring toy is shaken at a frequency of 40 Hz such that standing waves are observed with a wavelength of 050 m What is the speed of propagation of the wave

63 A baseball fan on a warm summer day (30degC) sitsin the bleachers 152 m away from home plate a What is the speed of sound in air at 30degCb How long after seeing the ball hit the bat does

the fan hear the crack of the bat

300 m

Chapter 15 Assessment 425physicsppcomchapter_test

Figure 15-23

Figure 15-24

a b

Getty Images

423-429 CH15-SG CA STP-845813 33004 730 AM Page 425

64 On a day when the temperature is 15degC a personstands some distance d as shown in Figure 15-25from a cliff and claps his hands The echo returns in20 s How far away is the cliff

65 Medical Imaging Ultrasound with a frequency of425 MHz can be used to produce images of thehuman body If the speed of sound in the body isthe same as in salt water 150 kms what is thelength of a 425-MHz pressure wave in the body

66 Sonar A ship surveying the ocean bottom sendssonar waves straight down into the seawater fromthe surface As illustrated in Figure 15-26 the firstreflection off of the mud at the sea floor is received174 s after it was sent The second reflection fromthe bedrock beneath the mud returns after 236 sThe seawater is at a temperature of 25degC and thespeed of sound in mud is 1875 msa How deep is the waterb How thick is the mud

67 Determine the variation in sound pressure of aconversation being held at a sound level of 60 dB

68 A fire truck is moving at 35 ms and a car in frontof the truck is moving in the same direction at 15 ms If a 327-Hz siren blares from the truckwhat frequency is heard by the driver of the car

69 A train moving toward a sound detector at 310 msblows a 305-Hz whistle What frequency is detected on each of the followinga a stationary trainb a train moving toward the first train at 210 ms

70 The train in the previous problem is moving awayfrom the detector What frequency is now detectedon each of the followinga a stationary trainb a train moving away from the first train at a

speed of 21 ms

152 The Physics of Music71 A vertical tube with a tap at the base is filled with

water and a tuning fork vibrates over its mouth Asthe water level is lowered in the tube resonance isheard when the water level has dropped 17 cm and again after 49 cm of distance exists from the waterto the top of the tube What is the frequency of thetuning fork

72 Human Hearing The auditory canal leading to the eardrum is a closed pipe that is 30 cm longFind the approximate value (ignoring endcorrection) of the lowest resonance frequency

73 If you hold a 12-m aluminum rod in the centerand hit one end with a hammer it will oscillate likean open pipe Antinodes of pressure correspond tonodes of molecular motion so there is a pressureantinode in the center of the bar The speed ofsound in aluminum is 5150 ms What would bethe barrsquos lowest frequency of oscillation

74 One tuning fork has a 445-Hz pitch When a secondfork is struck beat notes occur with a frequency of 3 Hz What are the two possible frequencies of thesecond fork

75 Flutes A flute acts as an open pipe If a flute soundsa note with a 370-Hz pitch what are the frequenciesof the second third and fourth harmonics of thispitch

76 Clarinets A clarinet sounds the same note with apitch of 370 Hz as in the previous problem Theclarinet however acts as a closed pipe What are thefrequencies of the lowest three harmonics producedby this instrument

77 String Instruments A guitar string is 650 cm long and is tuned to produce a lowest frequency of 196 Hza What is the speed of the wave on the stringb What are the next two higher resonant

frequencies for this string

t 174 s

Mud

Seawater

Bedrock

t 236 s

d

426 Chapter 15 Sound For more problems go to Additional Problems Appendix B

Figure 15-25 (Not to scale)

Figure 15-26 (Not to scale)

423-429 CH15-SG CA STP-845813 33004 730 AM Page 426

78 Musical Instruments The lowest note on an organis 164 Hza What is the shortest open organ pipe that will

resonate at this frequencyb What is the pitch if the same organ pipe is closed

79 Musical Instruments Two instruments are playingmusical A (4400 Hz) A beat note with a frequency of 25 Hz is heard Assuming that one instrument is playing the correct pitch what is the frequency of the pitch played by the second instrument

80 A flexible corrugated plastic tube shown in Figure15-27 is 085 m long When it is swung around it creates a tone that is the lowest pitch for an open pipe of this length What is the frequency

81 The tube from the previous problem is swung fasterproducing a higher pitch What is the new frequency

82 During normal conversation the amplitude of apressure wave is 0020 Paa If the area of an eardrum is 052 cm2 what is the

force on the eardrumb The mechanical advantage of the three bones

in the middle ear is 15 If the force in part a istransmitted undiminished to the bones whatforce do the bones exert on the oval window themembrane to which the third bone is attached

c The area of the oval window is 0026 cm2 Whatis the pressure increase transmitted to the liquidin the cochlea

83 Musical Instruments One open organ pipe has alength of 836 mm A second open pipe should havea pitch that is one major third higher How longshould the second pipe be

84 As shown in Figure 15-28 a music box contains aset of steel fingers clamped at one end and pluckedon the other end by pins on a rotating drum Whatis the speed of a wave on a finger that is 24 cmlong and plays a note of 1760 Hz

Mixed Review85 An open organ pipe is 165 m long What

fundamental frequency note will it produce if it is played in helium at 0degC

86 If you drop a stone into a well that is 1225 m deepas illustrated in Figure 15-29 how soon after youdrop the stone will you hear it hit the bottom of the well

87 A bird on a newly discovered planet flies toward a surprised astronaut at a speed of 195 ms whilesinging at a pitch of 945 Hz The astronaut hears a tone of 985 Hz What is the speed of sound in the atmosphere of this planet

88 In North America one of the hottest outdoortemperatures ever recorded is 57degC and one of thecoldest is 62degC What are the speeds of sound atthose two temperatures

89 A shiprsquos sonar uses a frequency of 225 kHz Thespeed of sound in seawater is 1533 ms What is thefrequency received on the ship that was reflectedfrom a whale traveling at 415 ms away from theship Assume that the ship is at rest

90 When a wet finger is rubbed around the rim of aglass a loud tone of frequency 2100 Hz is producedIf the glass has a diameter of 62 cm and thevibration contains one wavelength around its rimwhat is the speed of the wave in the glass

91 History of Science In 1845 Dutch scientistChristoph Buys-Ballot developed a test of theDoppler effect He had a trumpet player sound an Anote at 440 Hz while riding on a flatcar pulled by alocomotive At the same time a stationary trumpeterplayed the same note Buys-Ballot heard 30 beats persecond How fast was the train moving toward him

1225 m

Steel fingers

085 m

Chapter 15 Assessment 427physicsppcomchapter_test

Figure 15-27

Figure 15-28

Figure 15-29 (Not to scale)

Horizons Companies

423-429 CH15-SG CA STP-845813 6404 956 AM Page 427

78 Musical Instruments The lowest note on an organis 164 Hza What is the shortest open organ pipe that will

resonate at this frequencyb What is the pitch if the same organ pipe is closed

79 Musical Instruments Two instruments are playingmusical A (4400 Hz) A beat note with a frequency of 25 Hz is heard Assuming that one instrument is playing the correct pitch what is the frequency of the pitch played by the second instrument

80 A flexible corrugated plastic tube shown in Figure15-27 is 085 m long When it is swung around it creates a tone that is the lowest pitch for an open pipe of this length What is the frequency

81 The tube from the previous problem is swung fasterproducing a higher pitch What is the new frequency

82 During normal conversation the amplitude of apressure wave is 0020 Paa If the area of an eardrum is 052 cm2 what is the

force on the eardrumb The mechanical advantage of the three bones

in the middle ear is 15 If the force in part a istransmitted undiminished to the bones whatforce do the bones exert on the oval window themembrane to which the third bone is attached

c The area of the oval window is 0026 cm2 Whatis the pressure increase transmitted to the liquidin the cochlea

83 Musical Instruments One open organ pipe has alength of 836 mm A second open pipe should havea pitch that is one major third higher How longshould the second pipe be

84 As shown in Figure 15-28 a music box contains aset of steel fingers clamped at one end and pluckedon the other end by pins on a rotating drum Whatis the speed of a wave on a finger that is 24 cmlong and plays a note of 1760 Hz

Mixed Review85 An open organ pipe is 165 m long What

fundamental frequency note will it produce if it is played in helium at 0degC

86 If you drop a stone into a well that is 1225 m deepas illustrated in Figure 15-29 how soon after youdrop the stone will you hear it hit the bottom of the well

87 A bird on a newly discovered planet flies toward a surprised astronaut at a speed of 195 ms whilesinging at a pitch of 945 Hz The astronaut hears a tone of 985 Hz What is the speed of sound in the atmosphere of this planet

88 In North America one of the hottest outdoortemperatures ever recorded is 57degC and one of thecoldest is 62degC What are the speeds of sound atthose two temperatures

89 A shiprsquos sonar uses a frequency of 225 kHz Thespeed of sound in seawater is 1533 ms What is thefrequency received on the ship that was reflectedfrom a whale traveling at 415 ms away from theship Assume that the ship is at rest

90 When a wet finger is rubbed around the rim of aglass a loud tone of frequency 2100 Hz is producedIf the glass has a diameter of 62 cm and thevibration contains one wavelength around its rimwhat is the speed of the wave in the glass

91 History of Science In 1845 Dutch scientistChristoph Buys-Ballot developed a test of theDoppler effect He had a trumpet player sound an Anote at 440 Hz while riding on a flatcar pulled by alocomotive At the same time a stationary trumpeterplayed the same note Buys-Ballot heard 30 beats persecond How fast was the train moving toward him

1225 m

Steel fingers

085 m

Chapter 15 Assessment 427physicsppcomchapter_test

Figure 15-27

Figure 15-28

Figure 15-29 (Not to scale)

Horizons Companies

423-429 CH15-SG CA STP-845813 6404 956 AM Page 427

92 You try to repeat Buys-Ballotrsquos experiment from theprevious problem You plan to have a trumpetplayed in a rapidly moving car Rather than listeningfor beat notes however you want to have the carmove fast enough so that the moving trumpetsounds one major third above a stationary trumpeta How fast would the car have to moveb Should you try the experiment Explain

93 Guitar Strings The equation for the speed of a wave

on a string is v F

T where FT is the tension in the

string and is the mass per unit length of the stringA guitar string has a mass of 32 g and is 65 cm longWhat must be the tension in the string to produce a note whose fundamental frequency is 147 Hz

94 A train speeding toward a tunnel at 375 mssounds its horn at 327 Hz The sound bounces offthe tunnel mouth What is the frequency of thereflected sound heard on the train Hint Solve theproblem in two parts First assume that the tunnel is a stationary observer and find the frequency Thenassume that the tunnel is a stationary source and find the frequency measured on the train

Thinking Critically95 Make and Use Graphs The wavelengths of the

sound waves produced by a set of tuning forks withgiven frequencies are shown in Table 15-2 below

a Plot a graph of wavelength versus the frequency(controlled variable) What type of relationshipdoes the graph show

b Plot a graph of wavelength versus the inverse ofthe frequency (1f) What kind of graph is thisDetermine the speed of sound from this graph

96 Make Graphs Suppose that the frequency of a carhorn is 300 Hz when it is stationary What wouldthe graph of the frequency versus time look like asthe car approached and then moved past youComplete a rough sketch

97 Analyze and Conclude Describe how you coulduse a stopwatch to estimate the speed of sound ifyou were near the green on a 200-m golf hole asanother group of golfers hit their tee shots Wouldyour estimate of the speed of sound be too large ortoo small

98 Apply Concepts A light wave coming from apoint on the left edge of the Sun is found byastronomers to have a slightly higher frequencythan light from the right side What do thesemeasurements tell you about the Sunrsquos motion

99 Design an Experiment Design an experiment thatcould test the formula for the speed of a wave on a string Explain what measurements you wouldmake how you would make them and how youwould use them to test the formula

Writing in Physics100 Research the construction of a musical instrument

such as a violin or French horn What factors mustbe considered besides the length of the strings ortube What is the difference between a qualityinstrument and a cheaper one How are theytested for tone quality

101 Research the use of the Doppler effect in the studyof astronomy What is its role in the big bangtheory How is it used to detect planets aroundother stars To study the motions of galaxies

Cumulative Review102 Ball A rolling west at 30 ms has a mass of 10 kg

Ball B has a mass of 20 kg and is stationary Aftercolliding with ball B ball A moves south at 20 ms(Chapter 9)

a Sketch the system showing the velocities andmomenta before and after the collision

b Calculate the momentum and velocity of ball Bafter the collision

103 Chris carries a 10-N carton of milk along a levelhall to the kitchen a distance of 35 m How muchwork does Chris do (Chapter 10)

104 A movie stunt person jumps from a five-storybuilding (22 m high) onto a large pillow atground level The pillow cushions her fall so thatshe feels a deceleration of no more than 30 ms2If she weighs 480 N how much energy does thepillow have to absorb How much force does thepillow exert on her (Chapter 11)

Table 15-2Tuning Forks

Frequency (Hz) Wavelength (m)

131

147

165

196

220

247

262

233

208

175

156

139

428 Chapter 15 Sound For more problems go to Additional Problems Appendix B

423-429 CH15-SG CA STP-845813 71304 111 PM Page 428

1 How does sound travel from its source to your ear

by changes in air pressure

by vibrations in wires or strings

by electromagnetic waves

by infrared waves

2 Paulo is listening to classical music in thespeakers installed in his swimming pool A notewith a frequency of 327 Hz reaches his earswhile he is under water What is the wavelengthof the sound that reaches Paulorsquos ears Use1493 ms for the speed of sound in water

219 nm 219101 m

488105 m 457 m

3 The sound from a trumpet travels at 351 ms inair If the frequency of the note is 298 Hz whatis the wavelength of the sound wave

993104 m 118 m

0849 m 105105 m

4 The horn of a car attracts the attention of astationary observer If the car is approaching the observer at 600 kmh and the horn has a frequency of 512 Hz what is the frequency of the sound perceived by the observer Use 343 ms for the speed of sound in air

488 Hz 538 Hz

512 Hz 600 Hz

5 As shown in the diagram below a car isreceding at 72 kmh from a stationary siren If the siren is wailing at 657 Hz what is thefrequency of the sound perceived by the driverUse 343 ms for the speed of sound

543 Hz 647 Hz

620 Hz 698 Hz

6 Reba hears 20 beats in 50 s when she playstwo notes on her piano She is certain thatone note has a frequency of 262 Hz What arethe possible frequencies of the second note

242 Hz or 282 Hz

258 Hz or 266 Hz

260 Hz or 264 Hz

270 Hz or 278 Hz

7 Which of the following pairs of instrumentshave resonant frequencies at each whole-number multiple of the lowest frequency

a clamped string and a closed pipe

a clamped string and an open pipe

an open pipe and a closed pipe

an open pipe and a reed instrument

Extended Answer8 The figure below shows the first resonance

length of a closed air column If the frequencyof the sound is 488 Hz what is the speed ofthe sound

L 168 cm

vcar

vsound

x

Multiple Choice

Write It Down

Most tests ask you a large number of questions in asmall amount of time Write down your work wheneverpossible Do math on paper not in your headUnderline and reread important facts in passages and diagramsmdashdo not try to memorize them

Chapter 15 Standardized Test Practice 429physicsppcomstandardized_test

423-429 CH15-SG CA STP-845813 33004 733 AM Page 429

1 How does sound travel from its source to your ear

by changes in air pressure

by vibrations in wires or strings

by electromagnetic waves

by infrared waves

2 Paulo is listening to classical music in thespeakers installed in his swimming pool A notewith a frequency of 327 Hz reaches his earswhile he is under water What is the wavelengthof the sound that reaches Paulorsquos ears Use1493 ms for the speed of sound in water

219 nm 219101 m

488105 m 457 m

3 The sound from a trumpet travels at 351 ms inair If the frequency of the note is 298 Hz whatis the wavelength of the sound wave

993104 m 118 m

0849 m 105105 m

4 The horn of a car attracts the attention of astationary observer If the car is approaching the observer at 600 kmh and the horn has a frequency of 512 Hz what is the frequency of the sound perceived by the observer Use 343 ms for the speed of sound in air

488 Hz 538 Hz

512 Hz 600 Hz

5 As shown in the diagram below a car isreceding at 72 kmh from a stationary siren If the siren is wailing at 657 Hz what is thefrequency of the sound perceived by the driverUse 343 ms for the speed of sound

543 Hz 647 Hz

620 Hz 698 Hz

6 Reba hears 20 beats in 50 s when she playstwo notes on her piano She is certain thatone note has a frequency of 262 Hz What arethe possible frequencies of the second note

242 Hz or 282 Hz

258 Hz or 266 Hz

260 Hz or 264 Hz

270 Hz or 278 Hz

7 Which of the following pairs of instrumentshave resonant frequencies at each whole-number multiple of the lowest frequency

a clamped string and a closed pipe

a clamped string and an open pipe

an open pipe and a closed pipe

an open pipe and a reed instrument

Extended Answer8 The figure below shows the first resonance

length of a closed air column If the frequencyof the sound is 488 Hz what is the speed ofthe sound

L 168 cm

vcar

vsound

x

Multiple Choice

Write It Down

Most tests ask you a large number of questions in asmall amount of time Write down your work wheneverpossible Do math on paper not in your headUnderline and reread important facts in passages and diagramsmdashdo not try to memorize them

Chapter 15 Standardized Test Practice 429physicsppcomstandardized_test

423-429 CH15-SG CA STP-845813 33004 733 AM Page 429