sound solid

7/29/2019 Sound Solid

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C HA PT ER 5

SoundWaves inSolids,

Liquids, andGases

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.2 Sound Velocity Along a Solid Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.3 Rigorous Derivation of Velocity of Sound in a Solid Rod . . . . . . . . . . . . 875.4 Sound Waves in Liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.5 Sound Waves in Gases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.6 Intensity of Sound Waves in Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.7 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

5.1 Introduction

Longitudinal wavesinanelasticbody (or medium) aregenerally calledsoundwaves. The most familiar sound waves arethose that propagatein air. How-ever, sound waves can propagate even in solids or liquids. Sound waves areassociated with the compressional and rarefactional motion of molecules inthedirectionthatthewavepropagates.Thisissimilartothelongitudinal wavesthatpropagatealongamass-springtransmissionlinethatwasdiscussedinthepreviouschapter.Earthquakesgenerally producebothlongitudinal wavesandtransversewaves, thelatter propagatingslower thantheformer. Whenwearehit by an earthquake, wefirst feel ahorizontal motion arising fromthe lon-gitudinal waves, andsometimelater, atumbling vertical movementfromthetransversewaves. In this chapter, westudy theproperties of thelongitudinalsound waves in solids, liquids, and gases.

5.2

Sound Velocity Along a Solid RodInthepreviouschapter welearnedthat thepropagation velocitiesof mechan-ical waves arein general given by

cw =

elastic modulus

massdensity(5.1)

83


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84 Sound Waves in Solids, Liquids, and Gases

Cross sectionA

Clamp

l

l + l

ForceF

Force per unit area

isF/A

FIGURE 5--1

Relativeelongation of asolid rodl/ l is proportional totheforceper unit areaF/A (Hookes law).

Theelastic modulus is aconstant that relates thestress to thestrain as

stress= elastic modulus strain (5.2)For thecaseof acontinuousspring havingauniformly distributed mass den-sity, wehave

F = Kll

(5.3)

wherethestress is theforceitself (N) and thestrain is l/ l (dimensionless).TheproportionalitybetweenthestressandthestrainisknownasHookeslaw.For thecaseof aspring, Hookenotedthat after removingafew coilsfromthespring, thespring would becomestiffer.

Hookes law for an elastic body (rather than the spring) takes a slightlydifferent form. Consider a solid rod with a natural length l(m) and a cross-sectional areaA(m2) (Figure51). If atensionF(N) is appliedalongtherod,therod will beelongated by alengthl, and wemay write

F = constantll

(5.4)

However, if the cross section is increased, a larger forcemust be applied inorder toobtainthesamelongitudinal deformationl. Thereforetheconstantin the preceding equation is still size dependent and is not a real material

constant. If wedivideEq. (5.4) by thecross sectionA and writeF

A= Y l

l(5.5)

amoregeneral relationship canbefound. TheconstantY(N/m2) is calledtheYoungsmodulusandit isamaterial constant. Equation(5.5) mayberegardedas themicroscopic formof Hookes law.


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5.2 Sound Velocity Along a Solid Rod 85

The stress in this case is thus given by the force per unit area N/m2 andtheelastic modulusY has thesamedimensions. Themass density of therodshould be the volume mass density, v (kg/m

3), rather than the linear massdensity as in thecaseof thespringand string. Thevelocityof propagationofthelongitudinal waves in therod is thus given by

cw =

elastic modulus

massdensity=Y

v(5.6)

It should be remembered that weare dealing with a slender rod rather thanan unbounded volume of the solid. Strictly speaking, the expression for thewavevelocity Eq. (5.6) is valid only for arod, along whichpurelongitudinalwavespropagate. ForthevelocityformulainEq.(5.6) tobeapplicable, theroddiameter must bemuchsmaller thanthewavelengthof thepropagatingwave.

Thevelocity of longitudinal spherical waves in an unbounded solid containsan additional factor in the elastic modulus (shear modulus) and it is largerthan that given by Eq. (5.6). Liquids and gases, on the other hand, cannotsupportshear stress, andthustransversewaves, andsuchacomplicationdoesnotoccur. Transversewavesonastringdiscussedinthepreviouschapter maybealternatively called shear waves.

Any longitudinal mechanical wave can appropriately be called a soundwave. The most familiar formof a sound wave is of course is the one thatpropagatesinair.Theair moleculesmovebackandforthinthesamedirectionasthevelocityof propagationof thesoundwave. Themoleculesintheroddothesamething. Theyundergodisplacementsinthesamedirectionasthewavevelocity. Whenever molecules are displaced, an increase or decrease in thedensity corresponding toacompressionor rarefactionoccurs. This issimilarto thelongitudinal waves that propagatein themass-springsystem.

Theenergy and momentumdensities associated with thesound waves insolidscansimilarly befoundby just replacingthelinear massdensityl withthevolumemassdensity v in Eqs. (4.28) and (4.43). Wewrite

energy density = 12

v220 (J/m

3) (5.7)

momentumdensity=

1

2

v220

cw(N

sec/m3) (5.8)

for asinusoidal wavewithadisplacementamplitude0 (m). Thepower flowneedssomefurther explanation.If therodhasacrosssectionA(m2), thewaveenergy per unit lengthalong therod (seeFigure52) is

1

2v

220A (J /m) (5.9)


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t = 0

t = 1 sec

A

cw

12rvw2x0

2A(J/m)

FIGURE 5--2

In 1 second, theenergy 12

v 220Acw is

transferred throughthecross sectionof therod.

In onesecond, theenergy of an amount

1

2

v220Acw (J) (5.10)

goes through thecross sectional areaA (m2) at an arbitrary location. Hence,thepower density (energy flow rateacross aunit areain unit time) is

1

2v

220cw (W/m2) (5.11)

This quantity is called theintensityof thewave.

EXAMPLE 5.1

Steel has aYoungs modulus of 2

1011 N/m2 and avolumemass densityof 7800 kg/m3.Assumingasinusoidal longitudinal wavewith adisplacementamplitudeof1.0 106 mmandafrequency = 5 kHz in asteel rod, find(a) thewavevelocity,(b) thewaveintensity.

Solution

(a) Thevelocity of thewaveis

cw =Y

v=

2 10117.8

103

= 5.1 103 m/sec

(b) Theintensity of thewavecan becomputed fromEq. (5.11),

intensity = 12

v 220cw

= 12 7800 (2 5 103)2 (109)2 5.1 103 = 1.96 102 W/m2.


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5.3 Rigorous Derivation of Velocity of Sound in a Solid Rod 87

In the table, Youngs moduli, densities, and sound velocities in rods oftypical materials at roomtemperatureareshown.

Material Y(1010N/m2) v (kg/m3) cw(10

3m/sec)

Aluminum 6.9 2,700 5.0Cast iron 19 7,200 5.1Copper 11 8,900 3.5Lead 1.6 11,340 1.2Steel 20 7,800 5.1Glass 5.4 2,300 5.0Brass (70% Cu, 30% Zn) 10.5 8,600 3.5

5.3 Rigorous Derivation of Velocity of Sound in aSolid Rod

Infindingthepropagationvelocityof soundwavesinsolidrods, wehavejustused thegeneral expressionthat was obtained in Chapter 4,

cw =

elastic modulus

massdensity(5.12)

Here, for redundancy, wederiveEq. (5.6) directly fromHookes law and theequation of motion. Theprocedure, however, is almost exactly the same asweemployed in Chapter 4.

Consider alonguniformrodwithacrosssectionA (m2) havinganelasticor Youngs modulusY(N/m2) andamassdensityv (kg/m

3). Weselect athinslice of thickness x located at a distancex (m) from one end of the rod(Figure53).

When a sound wave is excited along the rod, the thin slice will moveabout its original location. At the same time, the slice is deformed since theforces that act onthe cross sections Ax and Ax+x will be different. Let thedisplacementsof thecrosssectionsofAx and Ax+x be(x) and(x+x),

Ax Ax+ x

x

x+ x

A

FIGURE 5--3A solid rodin equilibrium. Thesmall element,havingavolumeofAx, experiences nonetforce.


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Ax Ax+ x

x+x(x)

F(x) F(x+ x)

x+ x+x(x+ x)

A

FIGURE 5--4

With thepresenceof thewave, thevolumeAx is displaced. At thesametime, thevolumeis deformed.

respectively (Figure54). Then thenet deformation is

(x+ x) (x) x

x (5.13)

wherewehaveused the Taylor series expansionfor (x+ x) and retainedonly thelowest-order terms

(x+ x) (x)+ x

x (5.14)

Then Hookeslaw,

stress = Y strain (5.15)becomes

F

A= Y (/x)x

x= Y

x(5.16)

Note that x is the original thickness of the slice that corresponds to l inEq (5.5), and (/x)x correspondsto l. Next, theequationof motion,

mass acceleration= net force (5.17)can bewritten as

vxA2

t2= F (x+ x) F (x) F

xx (5.18)

wherewehaveexpandedF(x+x) usingtheTaylor seriesexpansion. Sub-stitutingEq. (5.16) into Eq. (5.18), wefind

2

t2= Y

v

2

x2(5.19)

and thevelocity of propagation can immediately befound tobe

cw =Y

v(5.20)

Onenotesthatanyphysicalquantityassociatedwiththesoundwaveshouldobey the same equation. If wedifferentiateEq. (5.19) with respect tox, we


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5.3 Rigorous Derivation of Velocity of Sound in a Solid Rod 89

obtain

3

xt2= Y

v

3

x3(5.21)

But thepressureor forcewavesatisfies

F

A= Y

x

Hence we find that the force F(x, t) obeys the same differential equation.Similarly, thevelocity defined by

v = t

should also obey the same equation. However, the forceF and the velocityv arenot of thesamefunctional formas thedisplacement . For example, if

weassume a sinusoidal wavegiven by (x, t) = 0 sin(kx t), the forcebecomes

F (x, t) = AY x

= AYk0 cos(kx t) (5.22)

whichis 90 out of phasewithrespecttothedisplacement asshownin Fig-ure55. Thevelocity is also90 outof phasewithrespecttothedisplacement. Sincetheworkis givenby force displacement, wecaneasily seethattherodis, ontheaverage, notdoingany work orgainingany energy. Theaverageof thefunction

sin(kx t) cos(kx t) (5.23)is exactly zero. The rod is transferring energy fromleft to right but it is notdissipatingorcreatingenergy.Therodcanaccommodateenergythatismovingor propagatingin the rodand only acts as a mediumthat accommodates thewave.

F(x, t = 0)

x(x, t = 0)

x0 x0 sinkx

AYk x0 coskx

x

x

FIGURE 5--5

Theforcewaveis 90 out of phasewithrespect to thedisplacement wave. Both

waves areshownat t = 0.


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Infact, goodwavemediaall havethisnondissipationproperty. Bygooditismeantherethat themediumdoesnot dissipateenergy. Otherwise, thewaveenergy is gradually absorbed or dissipated by the medium and by the timethewavepropagates totheother end, littleenergy would remain. In practice,

however, any wavethat propagates in material media actually does undergodamping. Inour studies,however, weassumethat thedissipation issmall andcanbeneglectedexcept for onecaseinthechapter onelectromagnetic waveswhere weexamine wavepropagation in metals and describe skin effects. Inotherwords,themediawestudyareall reactivemedia.Theconceptof reactivemedia will later becomeclear in thechapters onelectromagnetic waves.

5.4 Sound Waves in Liquids

Sound waves require a compressive (and thus rarefactive) medium. Soundwaves in solids can exist because solids are elastically compressive. Wateris compressive also and there is a relationship between the forceF and thechangein volumeof thewater V, just as in Hookes law for solids.

Consider a liquid in a rigid cylinder occupying a volumeV (m3) in theabsence of a compression force (Figure 56). If the cylinder has a cross-sectional area A(m2), the stress due to the force F(N) is F/A (N/m2) andthe strain isl/ l wherel is essentially the change in the volume VsinceV = Al andthecross-sectional areadoesnot change. Thestressstrainrelation for liquidsis written as

FA= MB l

l= MB V

V(5.24)

Piston

Volume V=Al

ForceF

l

V=A(l + l)

=V+ V

Cross

section

A

Rigid cylinder wall

l + l

l< 0

V< 0

FIGURE 5--6

Liquidssuchas water can becompressed by an external force.


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5.4 Sound Waves in Liquids 91

whereMB (N/m2) playsthe role of Youngs modulusin solids and is called

thebulk modulusof liquids. For acompressiveforceasshowninFigure56,the change in the volume V is negative or V < 0. The bulk modulus isamaterial constantandhas thesamedimensions as Youngs modulus. Sincethemass density is inversely proportional to thevolume

vV = constant, (5.25)

thedecreaseinthevolumecausesanincreaseinthemassdensity andwemaywriteEq. (5.24) in an alternativeform,

F

A= MB

v

v, v > 0 ifV < 0 (5.26)

Oncewe find the elastic modulus (MB in this case), the velocity of the

longitudinal (sound) waves is immediately found to be

cw =

MB

v(5.27)

in analogy with Eq. (5.6). Sinceliquids cannot support shear stress, this ex-pressionis notsubjectto thegeometrical constraint as sound waves in solidsare. Thevelocityof soundwavesinisotropic freesolids(without boundaries)is given by

cs =K+43G

v (5.28)

whereK is thebulk elastic modulus andG is theshear elastic modulus. Thedifferencefromthesound speed in rods,cs =

Y/v should benoted.

EXAMPLE 5.2

Water has amassdensity ofv = 103 kg/m2 and abulk modulusofMB = 2.1 109 N/m2. Find thevelocityof sound waves in water.

Solution

cw =

MB

v=

2.1 109 N/m2103 kg

= 1.45 103 m/sec


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EXAMPLE 5.3

DeriveEq. (5.27) directly fromNewtons equationof motion for asmall differentialvolumein aliquid. Follow theprocedureemployed in Section 5.3 for thesound waves insolids.

S1 S2

F(x) F(x+x)

x+x+x(x+x)

Cylindrical wall

Filled

with

liquid

x+x(x)

(x x)

Cylindrical

Filled

wit

iqui

FIGURE 5--7

Displacementof twosurfaces S1 and S2, originally separated byadistancex.

Solution

Weconsider apipefilled withaliquid in which asound waveis excited. A small elementwithathickness xhas amassv Ax, wherev is theunperturbed liquid massdensityand A is thecross sectionof thepipe. Let thedisplacementsof surfacesS1 andS2 be(x)and (x+ x), respectively (Figure57). Then thenet changein thevolumefromtheoriginal volumeAx is

A[x+ (x+ x) (x)] Ax= A[(x+ x) (x)] Ax

x

wherewehaveused thefirst two terms of aTaylor seriesexpansionfor thequantity(x+ x),(x+ x) (x)+

xx

This changein thevolumemust becaused by theinternal forceF acting onthesurfaceoftheelement. FromEq. (5.24), wethen find

F

A= MB

1

AxAx

x= MB

x(A)

However, theforces that actonthesurfaceS1 andonS2 must bedifferent inorder tocausethedisplacementof thewhole volumeAx. Thenet forcedirected to theright is

F (x) F (x+ x) Fx

x (B)

wherewehaveagain Taylor series-expandedF (x+x). Then theequationof motion forthesegmentcan bewritten as

v Ax 2

t2= F

xx (C)


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5.5 Sound Waves in Gases 93

After substituting theforcein Eq. (A) into Eq. (C), wefinally obtain

2

t2= MB

v

2

x2(D)

which immediately yields thesound velocity, Eq. (5.27).

5.5 Sound Waves in Gases

Sound waves in air areprobably themost familiar wavephenomena. Humanears can detect sound waves with frequencies ranging fromabout 20 Hz to20kHz(theaudiofrequencyrange). Someanimals(dogs, bats)apparentlycandetectsoundwaveswithhigherfrequencies,calledtheultrasonic frequencies.Frequencies belowtheaudiblelimit arecalled infrasonic frequencies. Earth-quakes are usually accompanied by infrasonic waves in the air in additionto the physical waves in the ground. Acoustics is a scientific branch that isdevoted to thestudies of sound waves andtheir applications.

Sound waves can be created with physical objects that are oscillating orvibrating. When wespeak, our vocal cordsvibratetocreatecompressiveandrarefactive motion of the molecules in air. At roomtemperature 20C, thesecompressive and rarefactive air molecule perturbations propagateat a speedof approximately 340 m/sec. Since sound waves in air (or gases, in general)fall into the class of waves called mechanical waves, wecan find the sound

velocity fromthegeneral formula

cw =

elastic modulus

massdensity

To find the mass density of air, let us recall that 1 mole of gas occupies22.4litresof volumeinthestandardcondition, 0C andatmosphericpressure.Since1 mole of air hasamassof 29g(about 80%of nitrogenwithmolecularmolar massof 28gand20% of oxygenwith32gof molar mass), wefindthemass density

v =0.029 kg

22.4 103 m3 = 1.29 kg/m3

Theelasticmodulusof gasesisdefinedin thesamemanner asthatfor liquids.As before(seeFigure58), weconsider acylinder withacross-sectional

area A(m2) and alength l filled withagas havingapressureP. If thepistonis pushed by an external force, F (N), the pressurerises (P, the change inthepressure, is positive), butthevolumedecreases byAl. If thevolumeof


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PPGas pressureP

volume V=AlPiston

Force

F=AP

l

P= P

V= V

A

Cylinder wall

l + l

l< 0

V< 0

P> 0

FIGURE 5--8

When agas is compressed, thepressureincreases duetotheincreasein bothmolecular density andtemperature.

the gas is large compared with the wavelength of sound, the compression is

adiabatic andthe gas pressureP and the volume occupied by the gasV arerelated throughtheadiabaticequationofstate:

PV = constant (5.29)where is theratio of specific heats. Wefind after differentiating Eq. (5.29)

PV + PV1V = 0 (5.30)Therefore

P = P VV

(5.31)

SinceV = Al, V = Al, wefinally obtain

P = P ll

(5.32)

Thisisidentical withEq. (5.24) (stressstrainrelationfor solids) providedwereplaceF/A (whichhas thedimensions of pressure, N/m2) by P, and MBby P. Thereforewemay definethebulk modulusof agashavingapressureP and aratio of specific heats by

MB = P (N/m2) (5.33)

Denoting the volume mass density of the gas with v (kg/m

3

), we find thesound velocity is given as

cw =

P

v(m/sec) (5.34)

The origin of the additional factor (compared with the sound velocityinliquids) stemsfromthefact that gasestendtobeheatedwhencompressed.


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Thepressureof agas is given by

P = nkBT (N/m2) (5.35)wheren is the number density of gas molecules (m3),kB is theBoltzmann

constant,kB = 1.38 1023J /K, andT is the absolutetemperature (K). Ifthegasiscompressed, themolecular densityobviously increases.Atthesametime, thegas is heated if thegas is thermallyinsulated fromexternal agents.Thisprocessis calledadiabaticcompression. Thenthetotal pressureincreaseis given by

P = kB(nT + nT) = P

n

n+ T

T

(5.36)

Recalling

n

n= V

V(5.37)

Eq. (5.36) becomes

P = P VV

+ P TT

(5.38)

which is obviously larger than the pressure change due to the density (orvolume) changealone. Thecoefficient is defined from

P VV

+ P TT

P VV

(5.39)

andisalwayslargerthanorequal to1.Itsnumerical valuedependsontheactualmolecules that the gas contains. For gases withmonatomicmolecules such

as helium(He) and argon(Ar), = 5/3. For gases withdiatomicmoleculessuchas oxygen (O2), nitrogen (N2), and air (mixtureof oxygen and nitrogengases), = 7/5. Theratio of specific heats is given by = ( f+ 2)/ f,where f is thenumber of degrees of freedomof molecular energy partition.For monatomic gases, f = 3 (or = 5/3) corresponding to three possiblekinetic energies, 1

2mv2x,

12mv2y,

12mv2z, in the x, y, z directions, respectively.

In adiatomic gas, thereareadditional two degrees of freedom(Figure59).These are associated with rotational energies about the two axes of boundatoms as shown. Therotation about thezaxis has negligible energy becauseof thesmall moment of inertiaof theatom. Youmay wonder why oneshould

notincludethevibrational energy along thezaxis. Only quantummechanicscan answer this question. As the energy of the electron in a hydrogen atomis not arbitrary but quantized (or discrete), so is the energy of this vibrationin the diatomic molecule. That is, the vibration energy cannot increase withtemperature and this energy does not contribute to the number of degreesof freedom. (A bound vibration is not free in a sense.) A brief introductionto quantummechanics is given in Chapter 13. It is interesting that quantum


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Moment of inertia

aboutz axis is

negligible

Atom

Atom

Spring to simulate

molecular binding force

Translational

motion of the

whole molecule

Vibrational

motionRotational

motion

z

x

y

vz

vy

vx

FIGURE 5--9

Translational, rotational, and vibrationalenergies of adiatomic molecule.

effects appear even in sound waves, which havebeen an important physicalsubject sinceNewtonsera.

EXAMPLE 5.4

Find thevelocityof sound in (a) air and (b) heliumgas at atmospheric pressureand atemperatureof 0C.

Solution

(a) Sinceair is largely composed of nitrogen and oxygen gases withboth componentspossessingdiatomic molecules, wemay choose = 7/5. ThepressureP is1.013 105 N/m2 andthemassdensity v is 1.293 kg/m3 at 0C, 1 atmospherepressure. Then

cw =

P

v=

1.4 1.013 1051.293

= 331 m/sec

(b) Sinceheliumgas is monatomic, wechoose = 5/3. Themassdensity v can befound as follows. Onemole of agas occupies avolumeof 22.4litres (L) atatmospheric pressureat 0C and contains6 1023 (Avogadros number) molecules.Sincethemolar mass of Heis 4 g, wefind

v = 0.004 kg22.4 103 m3 = 0.179 kg/m

3

Thevelocity of sound in Hegas in thestandard condition is

cw =

P

v=

5/3 1.013 105 N/m20.179 kg/m3

= 972 m/sec


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The reader may have experienced the demonstration of the elevation ofpitchof anindividual who inhaledheliumandstartedtospeak fromthesameindividual who was breathing normal air and speaking. This is related to thedifferences in thevelocities of sound becauseof thedifferenceof themasses

between heliumandair.Theexpressionfor thevelocity of soundwavesingasesEq. (5.34) canbe

rewritten in terms of microscopic quantities. Since the pressure P is nkBTEq. (5.35) and themass density is

v = nm(kg/m3) (5.40)

wheremis themass of onemolecule, Eq. (5.39) becomes

cw =

kBTm

(5.41)

Observethat thevelocity of sound in gases isactually independent of thegasdensityn andis determined by the gas temperature andthe molecular mass.Froman investigationof thekinetic theory of gases, onefinds that thesoundvelocity givenby Eq. (5.41) is verysimilar to therandommolecular velocity,

3kBT

m(5.42)

Theonlydifferenceisinthenumerical factors, and3.Thecloseresemblanceisactually duetothepropagationmechanismof soundwavesor theorigin ofthebulk modulus (elasticity) of gases.

If we further multiply both numerator and denominator in Eq. (5.42) byAvogadrosnumber N = 6.01023/mole, weobtainanother way toexpressthesoundvelocity.

cw =

NkBT

Nm(5.43)

Here, NkB = 8.3J/(K mol) RJ /(K mol) isknownasthegasconstant, andNm= Mmol is themass of onemole. Then, using these, wehave

cw =

RT

Mmol(5.44)


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EXAMPLE 5.5

Findthevelocityof sound in air at 20C. Onemoleof air has amass of 29 g.

Solution

Substituting = 7/5, R= 8.3J/Kmol,T = 273+ 20= 293 K, andMmol = 0.029 kg,wefind

cw =

(7/5) 8.3 2930.029

= 343 m/sec

5.6 Intensity of Sound Waves in Gases

In Section 5.2 we derived expressions for the average energy density andpower density or energy flux for sinusoidal sound waves in solids. Thoseexpressions can also be applied for sound waves in gases. The quantity ofpractical importanceisthepower density(W/m2), whichindicateshow muchenergy (J) passes through aunit area(1 m2) per unit time(1 sec). Thepowerdensity is alternatively called theintensity of sound waves.

If asinusoidal wavetrain is described with thedisplacement profile

(x, t) = 0 sin(kx t),

theintensity I is given by Eq. (5.11)

I = 12

v220cw (W/m

2) (5.45)

The human ear is avery sensitive organ. At the same time, thehuman ear isveryflexibleandcanstandatremendouslywiderangeof intensities.Thelowerlimit of theaudibleintensity is of theorder of 1012 W/m2 andthemaximumsafety limit is of theorder of 1 W/m2. Theratio between thesetwo values is1012! The intensity of ordinary conversation is of the order of 106 W/m2,street traffic is 105 W/m2 and jet planes is 102 W/m2.

EXAMPLE 5.6

Sinusoidal sound waves in air havean intensity1.0 106 W/m2 andafrequency of2 kHz. Assuming aroomtemperature(20C) and1 atmospherepressure, findtheamplitude0 of thedisplacement waves. What is theamplitudeof thevelocitywave?What is theamplitudeof thepressurewave?


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5.6 Intensity of Sound Waves in Gases 99

Solution

At 20C thevelocity of sound is 343 m/sec (Example 5.5). In Eq. (5.45),

I = 1.0 106 W/m, v = 1.29 kg/m3 (273 K/293 K)= 1.20 kg/m3, = 2 2 103 rad/sec, andcw = 343m/sec. Then

0 =1

2I

vcw= 1

2 2 103

2 1061.2 343 = 5.55 10

9 (m)

Theamplitudeof thevelocity waveis 0 since

v = t

= 0 cos(kx t)

Then

0=

2

2

103

5.55

109

=7.0

105 m/sec

Theamplitudeof thepressurewavecan befound from

P = P x

in analogy with sound waves in liquids, as in Eq. (A) in Example 5.3. Then

P = Pk0 cos(kx t)and theamplitudeof thepressurewaveis

Pk0

where thewavenumberk can befound fromthedispersion relation

k= cw or k=

cw

Substituting = 7/5, P = 1.013 105 N/m2, 0 = 5.55 109 m, andk= 2 2 103/343= 36.6 rad/m, wefind

P0 = 2.85 102 N/m2

Asthisexampleindicates,theair moleculeshardly move(they moveonly5.6

109 m!) for theintensity of 106 W/m2, whichis thetypical intensity

of human conversation. Thepressureperturbation relativeto theequilibriumpressureis only

2.85 1021.0 105 = 2.85 10

7

Weindeed realizehowsensitiveour ears actually are.


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Becausetheaudibleintensity rangeissowide(1012!) andthepressureontheeardrummayhavedeleteriouseffects,itiscommontospecifytheintensityintermsof aratio of theactual intensitydividedby astandardintensityusinga logarithmic expression. The intensity I is proportional to the square ofthe amplitude A of the wave. The decibel (deci = ten, bel after Bell),abbreviated as dB is defined as

dB = 10log10I

I0

= 10log10

A2

A20

= 20log10

A

A0

(5.46)

Thestandardsoundintensity I0 is chosento be1.01012 W/m2, this beingapproximately thethreshold of human hearing. For example, theintensity of1.0 106 W/m2 is equivalent to 60 dB since

10log10(106/1012)

=10log10 10

6

=60

Apparently, the human feeling for sound intensity (loudness of sound) is inlogarithmic formand notin alinear form. In this respect, theintroduction ofdecibelsisrather natural. Thedecibel representationis also usedin electricalengineering to express a power relative to a standard power. The citation of3 dB, whichoften appears in electrical engineering, indicates adifferenceofafactor of 2 between two powers. (Recall that log10 2 0.3.)

5.7 Problems

1. Icehasadensity of 920kg/m3 andYoungs modulusof 11010 N/m2 at0C. Estimatethespeedof soundalong an icerod.

2. Longitudinal earthquake waves typically havea ve-locity of 5 103 m/sec. Assumingtheaverageearthdensity is 1500 kg/m3, estimatethe elastic modulus

of theearth.

3. A longsteel rodhavingadiameterof5cmisforcedtotransmit longitudinal waves excited with amechani-

cal oscillator connected totheend. Theamplitudeof

thedisplacementwaves is10

5 mandthefrequency

is 400 Hz. Find

(a) The expression to describe the displacementwave.

(b) Theaverageenergy per unit length of therod.(c) Theaveragepower transfer throughacross sec-

tion of therod.

(d) Thepower delivered bytheoscillator.

4. In Problem3, find theexpression for(a) Thevelocity wave.(b) Theforcewave.(c) Theenergy wave.Assume that the displacement waveis described by

0 sin(kx t).5. Computethevelocityof soundin

(a) A hydrogen (H2) gas.

(b) An argon (Ar) gas. Both gases areat 0C.Theatomic massof hydrogen atomis 1.0andthat of

argonis 40. Be careful withyour choiceof thevalue

of.6. Show that the change in the sound velocity cw

caused by a small change in the temperature T is

given by

cw =1

2

T

Tcw

wherecw =

RT/Mmol is theoriginal velocity.


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5.7 Problems 101

7. An observer detects the sound in air caused by anexplosion on a lake 2 sec after hedetects the sound

in water caused by the same explosion. How far is

the explosion point from the observer? Assume the

sound velocity in water is 1500 m/sec and in air is340 m/sec.

8. In a thunderstorm, an individual detectsthe thunderthree seconds after seeing the lightning flash. How

faraway wasthis particularlightningstrokefromthe

individual?

9. A sinusoidal sound wave with a frequency of =400Hz in air (20C, 1atmpressure) hasan intensityof 1 107 W/m2. What is(a) Theamplitudeof thedisplacementwave?(b) Theamplitudeof thepressurewave?

(c) Theintensity of thewaveexpressed in dB?10. Theair moleculedisplacementassociatedwithahar-

monic (sinusoidal) sound wave train in air (20C,1 atmospherepressure) is described by

(x, t) = 1.0 108 sin(kx t)(m)

where = 2 103 rad/sec.(a) What is thevalueof thewavenumberk?

(b) How intense is the wave? Answer in terms ofW/m2 anddB (decibels).

(c) Whatistheamplitudeof thepressurewave?Howdoesthiscomparewiththeequilibriumpressure?

11. Twosoundwaves,oneinairandoneinwater,havethe

same frequency and the same intensity. What is the

ratio between molecular displacement amplitudes?

AssumeT = 20C.12. Thedisplacementwaveof aharmonic soundwavein

air is given by

(x, t) = 0 sin(kx t),

k= cs

Show that the ratio between the pressure wave

P x

and the velocity wave t

is constant and

given by

Pv, where v is the volume mass

density. This quantity is called the impedance for

sound wave.


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sound solid

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