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Physics140
Sound
Chapter12
Soundwaves
Physics140,Prof.M.Nikolic 2
Soundiscomposedoflongitudinalpressurewaves.
wavepropagaBon
Compression
RarefacBon
Compression
RarefacBon
CompressionèwhenparBclescometogether
RarefacBonèwhenparBclesarefarthestapart
Soundwaves
Physics140,Prof.M.Nikolic 3
Describedbythegaugepressure–differencebetweenthepressureatagivenpointandaveragepressure
Thespeedofsoundwaves
Physics140,Prof.M.Nikolic 4
Let’srememberChapter11andthespeedofstringwaves v = Tµ
Thespeedofsoundwaves
ρBv = Bisthebulkmodulusofthemedium
andρitsdensity.
AllotherequaBonsfromChapter11sBllapply.
v = f λλπ2
=k
Thespeedofsoundwavesinthinrods v = Y
ρYistheYoung’smodulusofthemediumandρitsdensity.
GiventheinformaBoninthetablebelow,whichsubstancewouldhavethehighestspeedofsound?
A. aluminumB. brassC. copperD. mercury
Substance BulkModulus(GPa)
Density(kg/m3)
Speed(m/s)
Aluminum 70 2700
Brass 61 8400
Copper 14 8900
Mercury 27 13600
ConceptualquesBon–Speedofsound Q1
ρBv =
Physics140,Prof.M.Nikolic 5
GiventheinformaBoninthetablebelow,whichsubstancewouldhavethehighestspeedofsound?
A. aluminumB. brassC. copperD. mercury
Substance BulkModulus(GPa)
Density(kg/m3)
Speed(m/s)
Aluminum 70 2700 5.1x103
Brass 61 8400 2.7x103
Copper 14 8900 1.25x103
Mercury 27 13600 1.4x103
ConceptualquesBon–Speedofsound Q1
ρBv =
Physics140,Prof.M.Nikolic 6
Exercise:Speedofsoundingold
Physics140,Prof.M.Nikolic 7
What’sthespeedofsoundinagoldingot?Densityofgoldisρ=19,360kg/m3anditsbulkmodulusis220GPa.
ρBv =
ρBv =
m/s 1038.3kg/m 19300
Pa 10220 33
9
×=×
=v
What’sgiven:ρ=19,360kg/m3
B=220GPa=220x109Pa
CharacterisBcsofsound
Physics140,Prof.M.Nikolic 8
Soundcantravelthroughanykindofma`er,butnotthroughavacuum.
Thespeedofsoundisdifferentindifferentmaterials;ingeneral,itisslowestingases,fasterinliquids,andfastestinsolids.
Thespeeddependssomewhatontemperature,especiallyforgases.
00 TTvv = V0istheabsolutespeedat
absolutetemperatureT0.e.g.v0(air)=343m/satT0=200C.
Thespeedofsoundwaves
Physics140,Prof.M.Nikolic 9
00 TTvv =Speedofsoundinidealgases:
Toconverttemperaturefrom0CtoKelvins(SIunitoftemperature)youjustneedtoadd273K.
T (in K) = TC (in C)+ 273
SIunitsoftemperature:[K]
Forthespeedofsoundinair(andonlyinair)wecanuseanapproximateformula:
( )m/s 606.0331 CTv +=
whereTcistheairtemperaturein0C.
Physics140,Prof.M.Nikolic 10
Exercise:AlightningflashAlightningflashisseenintheskyand8.2secondslatertheboomofthunderisheard.Thetemperatureoftheairis12.00C.
(a)Whatisthespeedofsoundinairatthattemperature?
Whatisgiven:t=8.2sTC=120C
Sincewearelookingforthespeedofsoundinair,wecanusetheapproximateformula:
( )m/s 606.0331 CTv +=
( ) m/s 338m/s C 12606.0331 0 =⋅+=v
(b)Howfarawayisthelightningstrike?
vtd = ( )( ) km 2.8m 2800s 2.8m/s 338 ===d
(c)Thespeedoflightis3.00x108m/s.Howlongdoesittakethelightsignaltoreachtheobserver?
vdt = s 103.9
km/s 103.0m 2800 6
8−×=
×=t
Theamplitudeandintensity
Physics140,Prof.M.Nikolic 11
TheintensityofawaveistheenergytransportedperunitBmeacrossaunitarea.
ItcanbegivenasafuncBonofapressureamplitude(P0)
I = P02
2ρv
OrbylookingattheamplitudeofthedisplacementofparBcles
P0 =ωvρsmax
Unitsofintensity:W/m2
ωistheangularfrequencyofsoundvisthespeedofsoundρisthedensityofthemediumsmaxisthemaximumdisplacement
Loudnessanddecibels
Physics140,Prof.M.Nikolic 12
Loudnessofasoundismeasuredbythelogarithmoftheintensity.
Soundlevel(intensitylevel)ismeasuredindecibels(dB)andisdefined:
( )0
logdB 10II
=β
I0istakentobethethresholdofhearing:
I0=1.0×10−12W/m2
Soundlevelisnotthesameastheintensityofthesoundwave!
Physics140,Prof.M.Nikolic 13
Exercise:DecibelsfromaspeakerThesoundlevel25mfromaloudspeakeris71dB.a)Whatistherateatwhichsoundenergyisbeingproducedbytheloudspeaker,assumingittobeanisotropicsource?Whatisgiven:r=25mβ=71dB
( )0
logdB 10II
=βI0=10-12W/m2
Wecansolvefortheintensityofthesoundwave:
( )0
logdB10II
=β log II0=
β10dB
II0=10
β10dB
25dB 10dB 71
212 W/m103.110 W/m10 −− ×=⋅=I
Physics140,Prof.M.Nikolic 14
Exercise:DecibelsfromaspeakerThesoundlevel25mfromaloudspeakeris71dB.a)Whatistherateatwhichsoundenergyisbeingproducedbytheloudspeaker,assumingittobeanisotropicsource?
Whatisgiven:r=25mβ=71dBAndwecalculated:I=1.3x10-5W/m2
Weneedtofindtherateatwhichsoundenergyisbeingproduced–that’spower,butweknowintensityofsoundanddistance.
OK,let’sgobacktoChapter11andusetheequaBonforintensityoftheisotropicsource:
I = P4πr2
24 rIP π=
( ) W10.0m 254)W/m103.1( 225 =×= − πP
Physics140,Prof.M.Nikolic 15
Exercise:DecibelsfromaspeakerThesoundlevel25mfromaloudspeakeris71dB.b)Whatisthepressureamplitudeofasoundwave?Densityofairis1.2kg/m3andthespeedofsoundisabout343m/s.Whatisgiven:r=25mβ=71dBρ=1.2kg/m3v=340m/sAndwecalculated:I=1.3x10-5W/m2
I = P02
2ρv vIP ρ20 ⋅=
Pa 1.0m/s 343kg/m 2.12 W/m103.1 3250 =⋅⋅⋅×= −P
Standingwavesofsound
Physics140,Prof.M.Nikolic 16
AllofthesemusicalinstrumentsareeffecBvelysealedoffononeendandopenontheother.
TheairmoleculesrightnexttotheclosedendareessenBallynotmovingèdisplacementnodes
Themoleculesattheopenendareexposedtotheatmosphere.ThisgivesthemthegreatestfreedomofmovementèdisplacementanBnodes
Standingwavesofsound-Tubeclosedatoneend-
Physics140,Prof.M.Nikolic 17
L
Firstharmonic-fundamental
L = λ14⇒ λ1 = 4L
Secondharmonic
L = 3λ14⇒ λ1 =
43L
Thirdharmonic
L = 5λ14⇒ λ1 =
45L
Physics140,Prof.M.Nikolic 18
Standingwavesofsound-Tubeclosedatoneend-
Thegeneralresultforstandingwavesinatubeopenatoneendandclosedattheotheris
nL
n4
=λwheren=1,3,5,…(oddvaluesonly!!)
Orλn ' =
4L2n '−1 wheren’=1,2,3,…
fn =vλn
=nv4L
= nf1f1isthefundamentalfrequency.
FrequencieshaveonlyoddmulBplesofthefundamentaloneèEvenharmonicsarenotpresent
Physics140,Prof.M.Nikolic 19
Exercise:AbugleWhatarethefirst3harmonicfrequenciesemi`edbyabuglethathasatube50cmlongareritwasleroutinthecold,coldnight(assumethetemperatureis-12oC)?
Whatisgiven:L=50cm=0.5mTC=-120C
fn =(2n−1)v4L
= (2n−1) f1
Sincewearelookingforthespeedofsoundinair,wecanusetheapproximateformula:
( )m/s 606.0331 CTv += ( )( ) m/s 323.7m/s C 12606.0331 0 =−⋅+=v
n=1-fundamentalfrequency: f1 =(2n−1)v4L
=v4L Hz 8.161
m 5.04m/s 7.323
1 =⋅
=f
n=2 f2 = (2 ⋅2−1) f1 = 3 f1 Hz 6.4852 =f
n=3 113 5)132( fff =−⋅= Hz 3.8093 =f
Physics140,Prof.M.Nikolic 20
PressurevariaBons-Tubeclosedatoneend-
DisplacementnodeattheclosedendcorrespondstothepressureanBnodeèmoleculesatthatlocaBonaretheonesgesngmaximallycompressed
DisplacementanBnodeattheopenendcorrespondstothepressurenodeèmoleculesatthatlocaBonaretheonesgesngmaximallystretchedout
Physics140,Prof.M.Nikolic 21
Standingwavesofsound-Tubeopenatbothends-
L
Firstharmonic-fundamental
L = λ12⇒ λ1 = 2L
Secondharmonic
L = 2λ12
⇒ λ1 = L
Thirdharmonic
L = 3λ12⇒ λ1 =
23L
Physics140,Prof.M.Nikolic 22
Standingwavesofsound-Tubeopenedatbothends-
Thegeneralresultforstandingwavesinatubeopenatoneendandclosedattheotheris
λn =2Ln
fn =vλn
=nv2L
= nf1f1isthefundamentalfrequency.
Sincebothendsareopen,airmoleculesateachendhavemaximumpossibledisplacementsèdisplacementanBnodes
Unlikethepipeopenatonlyoneend,allharmonicsarepresent.
Physics140,Prof.M.Nikolic 23
PressurevariaBons-Tubeopenedatbothends-
DisplacementanBnodesattheopenendscorrespondtothepressurenodesèmoleculesatthatlocaBonaretheonesgesngmaximallystretchedout
Physics140,Prof.M.Nikolic 24
Exercise:AnorganpipeAnorganpipethatisopenatbothendshasafundamentalfrequencyof382Hzat00C.a)Whatisthefundamentalfrequencyforthispipeat200C?
Whatisgiven:f1,0=382HzT0=00CT20=200C
fn =vλn
=nv2L
= nf1 f1 =v2L
Thespeedofsoundinairat:T0=00Cisv0=331m/s T20=200Cisv20=343m/s
Thefundamentalfrequencyat00C:Lvf20
0,1 =
Thefundamentalfrequencyat200C:Lvf220
20,1 =
WecanfindtheraBo:
0
20
0
20
0,1
20,1
2
2vv
LvLv
ff
==
Hz 396Hz 382m/s 331m/s 343
20,1 ==f
Notethatincaseyouneedtofindthespeedofsoundatanyothertemperaturenotprovidedinthetable,youwouldneedtousetheeqn.forthespeedofsound(Slides8and9).
Physics140,Prof.M.Nikolic 25
Exercise:AnorganpipeAnorganpipethatisopenatbothendshasafundamentalfrequencyof382Hzat00C.b)Howlongisthepipe?
Whatisgiven:f1,0=382HzT0=00CT20=200C
Lvf20
0,1 =0,1
02fvL =
m 43.0m/s 3822
Hz 331=
⋅=L
YoucouldalsochoosetheotherequaBon,forf1,20andv20andsBllgetthesameanswer.
Timbre
Physics140,Prof.M.Nikolic 27
StandingwaveisasuperposiBonofmanystandingwavepa`ernswithdifferentfrequencies
• Fundamentalfrequency(1stharmonic)• Overtones(higherharmonics)
Thesamenotesoundsdifferentlyondifferentinstrumentsè fundamentalfrequencymightbethesamebut
theovertonesappearindifferentintensiBes
Timbre(tonequality)
Beats
Physics140,Prof.M.Nikolic 28
Let’sconsidertwodifferentwaves,eachwiththesamewavespeedbutwithslightlydifferentfrequencies(and,hence,wavelengths).
ThesuperposiBonofthewaveswillproduceapulsaBoncalledbeats.
Beats
Physics140,Prof.M.Nikolic 29
Beatfrequencyis 21 fff −=Δ Ifthebeatfrequencyexceedsabout15Hz,theearwillperceivetwodifferenttonesinsteadofbeats.
Example:AviolinistunedbyadjusBngthetensioninthestrings.Brian’sAstringistunedtoaslightlylowerfrequencythanJennifer’s,whichiscorrectlytunedto440Hz.WhatisthefrequencyofBrian’sstringifbeatsof2Hzareheardwhentheybowthestringstogether?
21 fff −=Δ f2 = f1 −Δf = 438Hz
WhatwouldhappenifthetemperaturewhereBrianseatsdropsfrom250Cto200C?
ThefrequencyofBrian’sviolinwouldchange: v = v0TT0
f = f0TT0
Hz 442)27320()27325(438 =
+
+=f
TheDopplereffect
30
TheDopplereffectoccurswhenasourceofsoundismovingwithrespecttoanobserver.
• Whensourceismovingtowardanobserverthesoundhasahigherfrequencyandshorterwavelength
• Whenasourceismovingawayfromanobserverthesoundhasalowerfrequencyandlongerwavelength
TheDopplereffect
Physics140,Prof.M.Nikolic 31
Crest
vsound = f λ =1Tλ
λ = vsoundT
InthesameBmeT,thesourcewillmovedistancedtowardtheobserver:
Theobserverwillhearasoundwavewithadifferentwavelength:
d = vsourceT
λ0 = λ − d = λ − vsourceT
λ0 = λ − vsourceλ
vsoundλ0 = λ 1−
vsourcevsound
"
#$
%
&'
TheDopplereffect
Physics140,Prof.M.Nikolic 32
Ingeneral,whenboththeobserverandthesourcearemoving:
f0 =vsound ± vobservervsound ± vsource
!
"#
$
%& f
Whentheobserverismovingtowardsthesourceuse
vsound+vobserver
Whenthesourceismovingtowardstheobserveruse
vsound–vsource
Whentheobserverismovingawayfromthesourceuse
vsound-vobserver
Whenthesourceismovingawayfromtheobserveruse
vsound+vsource
Note,thatiftheobserverorthesourcearestaBonary,justsettheirspeedstozero.
Exercise:Rockedsleds
Physics140,Prof.M.Nikolic 33
Amaniaconarocketsledismovingalongatraintrack.Shehearsatrainwhistleatafrequencyof1200Hz.Themaniac,beingabigfanoftrains,knowsthatatraintravelsatanaveragespeedof30m/sandthewhistleisnormallyatafrequencyof720Hz.Whatisthespeedofthemaniacifshestartssleddingtowardsthetrain?Assumethatthespeedofsoundis340m/s.
f0 =vsound ± vobservervsound ± vsource
!
"#
$
%& f
30m/s340m/s
720Hz1200Hz
Sincetheobserver(maniac)ismovingtowardsthesource(train): vsound + vobserverSincethesource(train)ismovingtowardstheobserver(maniac): vsound − vsource
f0 =vsound + vobservervsound − vsource
"
#$
%
&' f
Exercise:Rockedsleds
Physics140,Prof.M.Nikolic 34
Amaniaconarocketsledismovingalongatraintrack.Shehearsatrainwhistleatafrequencyof1200Hz.Themaniac,beingabigfanoftrains,knowsthatatraintravelsatanaveragespeedof30m/sandthewhistleisnormallyatafrequencyof720Hz.Whatisthespeedofthemaniacifshestartssleddingtowardsthetrain?
f0 =vsound + vobservervsound − vsource
"
#$
%
&' f vsound + vobserver
vsound − vsource=f0f
m/s 67.176=observerv
( )sourcesoundobserversound vvffvv −=+ 0
( ) m/s 67.516m/s 30m/s 340Hz 720Hz 1200m/s 340 =−=+ observerv
EcholocaBon
Physics140,Prof.M.Nikolic 35
Soundwavescanbesentoutfromatransmi`erofsomesort;theywillreflectoffanyobjectstheyencounterandcanbereceivedbackattheirsource.TheBmeintervalbetweenemissionandrecepBoncanbeusedtobuildupapictureofthescene.
IfthesoundtakesBmettogofromthesource(bat,dolphin,…)totheobjectandbackthenittravelsthedistance:
d = vsound ⋅t2
AndallequaBonsforcalculaBngthespeedofsoundsBllapply!
Physics140,Prof.M.Nikolic 36
Exercise:AsonarAboatisusingsonartodetectthebo`omofafreshwaterlake.Iftheechofromasonarsignalisheard0.540sareritisemi`ed,howdeepisthelake?Assumethelake’stemperatureisuniformandat250C.Whatisgiven:t=0.54sT=250C
d = vsound ⋅t2
Fromtable12.1inourtextbook→thespeedofsoundinfreshwateris1493m/s.
m 1.403s 0.27m/s 1493 =⋅=d
Physics140,Prof.M.Nikolic 37
( )0
logdB10II
=β
HowdoIsolveforthelogarithm:
log II0=
β10dB
II0=10
β10dB
Generalrule:
logB A =C A = BC
Summary
Physics140,Prof.M.Nikolic 38
Thespeedofsoundwaves
ρBv =
Thespeedofsoundwavesinthinrods v = Y
ρ
Bisthebulkmodulusofthemediumandρitsdensity.
YistheYoungmodulusofthemediumandρitsdensity.
v = f λλπ2
=k
Thespeeddependstemperature-gases 0
0 TTvv = V0istheabsolutespeedat
absolutetemperatureT0.
Summary
Physics140,Prof.M.Nikolic 39
IntensityifasoundwaveasafuncBonofapressureamplitude I = P0
2
2ρv
P0 =ωvρsmaxSoundlevel(intensitylevel)
( )0
logdB10II
=βI0=1.0×10−12W/m2
T (in K) = TC (in C)+ 273
Beatfrequencyis 21 fff −=Δf0 =
vsound ± vobservervsound ± vsource
!
"#
$
%& f
TheDopplereffect
