defining'wave'fronts'and'rays. - physicshumanic/p1200_lecture24.pdf ·...
TRANSCRIPT
Wave%Fronts%and%Rays
Defining'wave'fronts'and'rays.
Consider)a)sound)wave)since)itis)easier)to)visualize.
Shown)is)a)hemispherical)view)of)a)sound)wave)emitted)by)a)pulsating)sphere.
The)rays are)perpendicularto'the'wave'fronts (e.g.)crests))which)are)separated)from)each)other)by)the)wavelength)of)the)wave,)!.
Wave%Fronts%and%Rays
The$positions$of$two$spherical$wave$fronts$are$shown$in$(a)$with$theirdiverging$rays.$
At$large$distances$from$the$source,$the$wave$fronts$become$lessand$less$curved$and$approach$the$limiting$case$of$a$plane&waveshown$in$(b).$A$plane$wave$has$flat$wave$fronts$and$rays$parallel$toeach$other.
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Reflection*of*string*pulses*at*boundaries*and*interfaces
}}
From&less&dense&mediumto&more&dense&medium
From&more&dense&mediumto&less&dense&medium
Reflection&inverted&&&&&&&&&Reflection¬&inverted
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Law$of$reflection$of$waves$at$a$boundary$or$interface:
θi =θr
Chapters)11)and)12
Sound&and&Standing&Waves
The$Nature$of$Sound$Waves
LONGITUDINAL*SOUND*WAVES
Speaker'makingsound'waves'ina'tube
The$Nature$of$Sound$Waves
The$distance$between$adjacent$condensations$is$equal$to$the$wavelength$of$the$sound$wave.
The$Nature$of$Sound$Waves
Individual)air)molecules)are)not)carried)along)with)the)wave.
When)the)sound)hits)your)ear)it)causes)your)eardrum)to)vibrate)and)your)braininterprets)the)vibration)as)thepitch and)loudness of)thesound.
The$Nature$of$Sound$Waves
THE$FREQUENCY$OF$A$SOUND$WAVE
The$frequency is$the$number$of$cyclesper$second.
A$sound$with$a$single$frequency$is$calleda$pure$tone.
The$brain$interprets$the$frequency$in$termsof$the$subjective$quality$called$pitch.
The$Nature$of$Sound$Waves
THE$PRESSURE$AMPLITUDE$OF$A$SOUND$WAVE
Loudness$is$an$attribute$ofa$sound$that$depends$primarily$on$the$pressure$amplitudeof$the$wave.
The$Speed$of$Sound
Sound&travels&through&gases,&liquids,&and&solids&at&considerablydifferent&speeds.
The$Speed$of$Sound
Conceptual$Example:$$Lightning,(Thunder,(and(a(Rule(of(Thumb
There%is%a%rule%of%thumb%for%estimating%how%far%away%a%thunderstorm%is.After%you%see%a%flash%of%lighting,%count%off%the%seconds%until%the%thunder%is%heard.%%Divide%the%number%of%seconds%by%five.%%The%result%gives%theapproximate%distance%(in%miles)%to%the%thunderstorm.%%Why%does%thisrule%work?
The$Speed$of$Soundvsound 20o C( ) = 343 m/s× 1 mile
1600 m= 0.214 miles/s
vlight = c = 3.00×108 m/s× 1 mile1600 m
=188, 000 miles/s
1) Time for a light flash to travel distance x in miles
tlight =xc=
x188, 000 miles/s
= (5.32×10−6 s/mile)x
light travels 1 mile in 5.32 µs ⇒ ≈ instantaneous
2) Distance thunder travels x in miles in t seconds
x = vsoundt = 0.214 miles/s( ) t = t4.67 s
miles ≈ t5 s
miles
Sound&Intensity
For$a$1000$Hz$tone,$the$smallest$sound$intensity$that$the$human$earcan$detect$is$about$1x10912$W/m2.$$This$intensity$is$called$the$thresholdof&hearing.$$
On$the$other$extreme,$continuous$exposure$to$intensities$greater$than$1$W/m2 can$be$painful.
As$we$saw$before,$if$the$source$emits$sound$uniformly*in*all*directions,*the$intensity$depends$on$the$distance$from$the$source$in$a$simple$way:
I = P4πr2 , P = power emitted from source
r = distance from source
Decibels
The$decibel (dB)$is$a$measurement$unit$used$when$comparing$two$soundintensities.$$
Because$of$the$way$in$which$the$human$hearing$mechanism$responds$tointensity,$it$is$appropriate$to$use$a$logarithmic$scale$called$the$intensitylevel:
( ) !!"
#$$%
&=
oII
logdB 10'
212 mW1000.1 !"=oI
Note$that$log(1)=0,$so$when$the$intensity$of$the$sound$is$equal$to$the$threshold$of$hearing,$the$intensity$level$is$zero.$
Decibels
Quick review of logarithms:
log x ≡ log10 x = y ⇒ 10y = x
For example, log1= 0 since 100 =1
logA− logB = log AB$
%&
'
() logA+ logB = log AB( )
log AN( ) = N logA
Decibels
( ) !!"
#$$%
&=
oII
logdB 10' 212 mW1000.1 !"=oI
Decibels
Example:..Comparing*Sound*Intensities
Audio&system&1&produces&a&sound&intensity&level&of&90.0&dB,&and&system2&produces&an&intensity&level&of&93.0&dB. Determine&the&ratio&of&intensities.
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logdB 10'
Decibels
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logdB 10'
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Transverse(Standing(Waves
In#reflecting#from#the#wall,#aforward3traveling#half3cyclebecomes#a#backward3travelinghalf3cycle#that#is#inverted.
Unless#the#timing#is#right,#thenewly#formed#and#reflected#cyclestend#to#offset#one#another.
Repeated#reinforcement#betweennewly#created#and#reflected#cyclescauses#a#large#amplitude#standingwave#to#develop.
Transverse(Standing(WavesTransverse(standing(wave(patterns(on(a(string.
One-half of the longest wavelength, λ1, can fit on the string of length L
⇒ L = λ1
2⇒ f1 =
vλ1
=v
2L
node%%%%%%anti)node
L
Transverse(Standing(Waves
!,4,3,2,1 2
=!"#
$%&= nLvnfnString(fixed(at(both(ends
4/5/2014 Note Frequencies
http://www.seventhstring.com/resources/notefrequencies.html 1/2
Note Frequencies Seventh String Contact
Transcribe! Overview Screen shots Reviews Download Buy Affiliates Video Pedals Lost Licenses History FAQ
Fake Book Index Search
Utilities Metronome Tuner Tuning Fork
Resources Transcription How to Transcribe How to Slow Down Note Frequencies Other
Here is a table giving the frequencies in Hz of musical pitches, covering the full range of allnormal musical instruments I know of and then some. It uses an even tempered scale with A= 440 Hz.
C C# D Eb E F F# G G# A Bb B0 16.35 17.32 18.35 19.45 20.60 21.83 23.12 24.50 25.96 27.50 29.14 30.871 32.70 34.65 36.71 38.89 41.20 43.65 46.25 49.00 51.91 55.00 58.27 61.742 65.41 69.30 73.42 77.78 82.41 87.31 92.50 98.00 103.8 110.0 116.5 123.53 130.8 138.6 146.8 155.6 164.8 174.6 185.0 196.0 207.7 220.0 233.1 246.94 261.6 277.2 293.7 311.1 329.6 349.2 370.0 392.0 415.3 440.0 466.2 493.95 523.3 554.4 587.3 622.3 659.3 698.5 740.0 784.0 830.6 880.0 932.3 987.86 1047 1109 1175 1245 1319 1397 1480 1568 1661 1760 1865 19767 2093 2217 2349 2489 2637 2794 2960 3136 3322 3520 3729 39518 4186 4435 4699 4978 5274 5588 5920 6272 6645 7040 7459 7902
The octave number is in the left column so to find the frequency of middle C which is C4,look down the "C" column til you get to the "4" row : so middle C is 261.6 Hz.
Some Specific Notes
Middle C is C4=261.6Hz
Standard tuning fork A is A4=440Hz
Piano range is A0=27.50Hz to C8=4186Hz
Guitar strings are E2=82.41Hz, A2=110Hz, D3=146.8Hz, G3=196Hz, B3=246.9Hz,E4=329.6Hz
Bass strings are (5th string) B0=30.87Hz, (4th string) E1=41.20Hz, A1=55Hz, D2=73.42Hz,G2=98Hz
Mandolin & violin strings are G3=196Hz, D4=293.7Hz, A4=440Hz, E5=659.3Hz
Viola & tenor banjo strings are C3=130.8Hz, G3=196Hz, D4=293.7Hz, A4=440Hz
Cello strings are C2=65.41Hz, G2=98Hz, D3=146.8Hz, A3=220Hz
Coda
Bear in mind that everything here is in relation to the even tempered (aka equal tempered)scale, where an octave is a frequency ratio of exactly two and a semitone is a frequency ratioof exactly the twelfth root of two. In the real world however many different temperamentsmay be used -‐‑ see en.wikipedia.org/wiki/Musical_temperament -‐‑ and octaves too can vary insize, see en.wikipedia.org/wiki/Stretched_octave.
Also we call middle C "C4" : this is the commonest octave numbering but some people callmiddle C "C3" or even "C5".
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Note%frequencies%(Hz)%of%the%chromatic%music%scale
Example:)The$A$string$on$a$violin$has$a$fundamental$frequency$of$440$Hz.$The$length$of$the$string$is$32.0$cm$and$it$has$a$mass$of$0.450$g.$Under$what$tension$must$the$string$be$placed?
v = FTm L
⇒ FT = v2 mL
f1 =v
2L⇒ v = 2Lf1
∴FT = 2Lf1( )2 mL= 4Lf1
2m = 4 0.320( ) 440( )2 0.450×10−3( ) =112 N
Transverse(Standing(Waves
!,4,3,2,1 2
=!"#
$%&= nLvnfn
Changing'the'pitch'of'a'guitar'string'by'fingering'it:'the'smaller'you'makeL,'the'higher'the'pitch.
Longitudinal+Standing+Waves
A"longitudinal"standing"wave"pattern"on"a"slinky.
Longitudinal+Standing+Waves
!,4,3,2,1 2
=!"#
$%&= nLvnfnTube+open+at+both+ends
Standing(sound(waves(in(a(tube(open(at(both(ends
The$anti)nodes$occur$at$the$open$ends$of$the$tube.
One-half of the longest wavelength, λ1, can fit in the tube of length L
⇒ L = λ1
2⇒ f1 =
vλ1
=v
2L
L
Note$that$the$string$fixed$at$both$ends$and$the$tube$open$at$both$endshave$the$same$equation$for$the$standing$wave$frequencies$!
Longitudinal+Standing+Waves
Example:++Playing(a(Flute
When(all(the(holes(are(closed(on(one(type(offlute,(the(lowest(note(it(can(sound(is(middleC((261.6(Hz).((If(the(speed(of(sound(is(343(m/s,and(the(flute(is(assumed(to(be(a(cylinder(openat(both(ends,(determine(the(distance(L.
Longitudinal+Standing+Waves
!,4,3,2,1 2
=!"#
$%&= nLvnfn
( )( ) m 656.0
Hz 261.62sm3431
2 ===
nfnvL
Longitudinal+Standing+Waves
!,5,3,1 4
=!"#
$%&= nLvnfnTube+open+at+one+end
One-quarter of the longest wavelength, λ1, can fit in the tube of length L
⇒ L = λ1
4⇒ f1 =
vλ1
=v
4L
Standing(sound(waves(in(a(tube(open(at(one(end
L
1st harmonic 3rd harmonic,/1st overtone
Example:)The$fundamental$frequency$of$an$open$organ$pipe$corresponds$to$the$note$D2$(f1 =$73.42$Hz$on$the$chromatic$musical$scale).$The$third$harmonic$of$another$organ$pipe$that$is$closed$at$one$end$has$the$same$frequency.$Compare$the$lengths$of$these$two$pipes.$
!,4,3,2,1 2
=!"#
$%&= nLvnfn
!fn = !n v4 !L"
#$
%
&' !n =1,3, 5,…
Tube$open$at$both$ends
Tube$closed$at$one$end
f1 =v
2L⇒ L = v
2 f1=
3432 73.42( )
= 2.34 m
!f3 = 3 v4 !L"
#$
%
&'= f1 ⇒ !L =
3v4 f1
=3 343( )
4 73.42( )= 3.50 m
Longitudinal+Standing+Waves
Tube+open+at+one+end
Conceptual+Example:+Why+does+inhaling+Helium+raise+the+pitch+of+a+voice?
!"Warning:"It"is"dangerous"to"inhale"Helium"!
!,5,3,1 4
=!"#
$%&= nLvnfn
Assume&that&the&mouth/larynx&system&acts&as&a&tube&open&at&one&end&oflength&0.25&m.&
In&air,&where&vsoundair&=&343&m/s,&the&fundamental&frequency&of&a&voice&is
In&Helium,&where&vsoundHelium =&965&m/s,&the&fundamental&frequency&of&a&voice&is
f1air =
vsoundair
4L=
3434 0.25( )
= 343 Hz ⇒ ≈ an octave 4 F note
f1Helium =
vsoundHelium
4L=
9654 0.25( )
= 965 Hz ⇒ ≈ an octave 5 B note
Complex(Sound(Waves
Complex(Sound(Waves
Fourier'spectrum
Complex(Sound(Waves