chapter 13 sound go. sound what is the sound of 1 hand clapping ? what is the sound of 1 hand...

Chapter 13Chapter 13SoundSound

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SoundSound

What is the sound of 1 hand clapping What is the sound of 1 hand clapping ??

Try itTry it

SoundSound

Sound is the propagation of Sound is the propagation of longitudinal waves whose frequency longitudinal waves whose frequency (and wavelength) are within the (and wavelength) are within the range of hearing (It’s a mechanical range of hearing (It’s a mechanical wave – needs a medium to travel in )wave – needs a medium to travel in )

Human hearing is in the approximate Human hearing is in the approximate range of 20 to 20,000 Hzrange of 20 to 20,000 Hz

SoundSound

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Infrasonicbelow 20 Hz Ultrasonic

above 20 000 Hz

SoundSound

(Ex) – Ultrasonic – waves “see” (Ex) – Ultrasonic – waves “see” object in body because they are object in body because they are about the same size or smaller than about the same size or smaller than objectobject

SoundSound

* Pitch ………the perceived highness * Pitch ………the perceived highness or lowness of sound depending on or lowness of sound depending on the frequency of the wavethe frequency of the wave

SoundSound

Speed sound depends on Speed sound depends on

Medium…solids highestgases lowest

TemperatureT increases – Speed increases

SoundSound

Sound, of course, spreads out in 3 Sound, of course, spreads out in 3 dimensions from a source……..dimensions from a source……..

SoundSound

front is the peak orcompression of wave

Doppler EffectDoppler Effect

DE…..frequency shift that is the DE…..frequency shift that is the result of relative motion between the result of relative motion between the source of waves and the observersource of waves and the observer

Page 507Page 507

1) Sound waves are longitudinal 1) Sound waves are longitudinal because air molecules vibrate in a because air molecules vibrate in a direction parallel to the direction of direction parallel to the direction of the wave motionthe wave motion

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2)2)

Dark compressions

White rarefactions

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3) Frequency is an objective measure 3) Frequency is an objective measure of the rate of particle vibrations. of the rate of particle vibrations. Pitch is the subjective quality that Pitch is the subjective quality that depends on the listener (based on depends on the listener (based on frequency).frequency).

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4)puppy dogs can hear higher 4)puppy dogs can hear higher frequenciesfrequencies

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5) Infrasonic is below 20 Hz5) Infrasonic is below 20 Hz Audible is from 20 to 20 000 HzAudible is from 20 to 20 000 Hz Ultrasonic is above 20 000 HzUltrasonic is above 20 000 Hz

Sound IntensitySound Intensity

Intensity……the rate at which energy Intensity……the rate at which energy flows through a unit area flows through a unit area perpendicular to the direction of the perpendicular to the direction of the wave motionwave motion

Sound IntensitySound Intensity

rate at which soundtravels wavefrontto wavefront….

IntensityIntensity

Intensity (I) = Δ E/Δ t / AreaIntensity (I) = Δ E/Δ t / Area = Power/Area= Power/Area = Power / 4πr= Power / 4πr22

distance from source of sound

Sample 13 ASample 13 A

* distance = 3.2 m * Power = 0.20 * distance = 3.2 m * Power = 0.20 WW

Intensity = P/4πrIntensity = P/4πr22

I = 0.20 W/ 4π (3.2)I = 0.20 W/ 4π (3.2)22

I = 1.6 x 10I = 1.6 x 10-3 -3 W/mW/m22

P 415P 415

1a) I = P/ 41a) I = P/ 4ππ r r22

= .25W/ 4= .25W/ 4ππ (5.0) (5.0)22

I = 8.0 x 10I = 8.0 x 10-4-4 W/m W/m22

P 415P 415

1b) I = P/41b) I = P/4ππrr22

= .50 W/ 4= .50 W/ 4ππrr22

I = 1.6 x 10I = 1.6 x 10-3-3 W/ m W/ m22

p415p415

3) I = P/ 43) I = P/ 4ππrr22

I (4I (4ππrr22) = P) = P

(4.6 x 10(4.6 x 10-7-7)(4)(4ππ2222) = ) =

2.3 x 102.3 x 10-5-5 W = P W = P

p415p415

5) I = P/45) I = P/4ππrr22

I/P = 1/4I/P = 1/4ππrr22

(I)(4(I)(4ππ)/ P = 1/r)/ P = 1/r22

P/ (I)(4P/ (I)(4ππ) = r) = r22

[P/ (I)(4[P/ (I)(4ππ)])]1/21/2 = r = r [.35/ (1.2x10[.35/ (1.2x10-3-3)(4)(4ππ)])]1/2 1/2 = 4.8 m = r= 4.8 m = r

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4) The 24) The 2ndnd tuning fork will pick up the tuning fork will pick up the vibration of the first fork and cause a vibration of the first fork and cause a faint sound due to sympathetic faint sound due to sympathetic vibration or resonance if they are the vibration or resonance if they are the same frequencysame frequency

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7) Doppler effect: there has to be 7) Doppler effect: there has to be relative motion of source and/or relative motion of source and/or observer so itsobserver so its

e) all of thesee) all of these

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10) 10) λλ goes down by factor of 2 so goes down by factor of 2 so frequency goes up by factor of 2 or frequency goes up by factor of 2 or doubles as speed is constantdoubles as speed is constant

v = fv = fλλ

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1) level goes 40 to 60 dB….1) level goes 40 to 60 dB….

40 to 50 is 2x as loud40 to 50 is 2x as loud so 40 to 60 is 4x as loudso 40 to 60 is 4x as loud

and I goes up by 2 steps or by a and I goes up by 2 steps or by a factor of 100 !factor of 100 !

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* Sound gets louder * Sound gets louder so these go up:so these go up: a) intensitya) intensity d) decibel level (rel inten)d) decibel level (rel inten) f) amplitudef) amplitude Sound goes to higher pitchSound goes to higher pitch c) freq goes upc) freq goes up d) wavelength goes downd) wavelength goes down

What can we hear ?What can we hear ?

* Intensity and frequency (range) * Intensity and frequency (range) determine what we can hear……..determine what we can hear……..

What can we hear ?What can we hear ?

IntensityIntensity th of pain

thr hearing

200 Freq Hz

music

Measures of SoundMeasures of Sound

IntensityIntensity

Relative Intensity

Decibel Level………relative intensity determined by intensityto intensity of threshold of hearing

Measure of SoundMeasure of Sound

Conversion Intensity to decibel levelConversion Intensity to decibel level Intensity dB ExampleIntensity dB Example 1.o x 10 -12 0 threshold hear

1.0 x 10-11 10 rustling leaves

1.0 x 10-10 20 quiet whisper

1.0 x 10-9 30 whisper

multiply by 10

add 10

Sympatheic Vib & ResonanceSympatheic Vib & Resonance

* Sympathetic vibration……..the * Sympathetic vibration……..the vibration of one object transfers its vibration of one object transfers its vibration (energy) to another objectvibration (energy) to another object

ex) the body of a stringed instrument ex) the body of a stringed instrument vibrates (sympathetically or forced) vibrates (sympathetically or forced) due to the vibration of the stringsdue to the vibration of the strings

* Resonance ……* Resonance ……

SV & ResonSV & Reson

String vibrationString vibration

body vibration

air vibration ( sound)

Standing WavesStanding Waves

(mostly music) (mostly music)

StringInstruments

Air Column Instruments

..usually several SWfrom string vibrations

..vibration of aircolumns

SWSW

Harmonic seriesHarmonic series

Strings Air columnand openair closed air

Harmonic SeriesHarmonic Series

Strings and open airStrings and open air

L of string

N N

A

Fundamental freqor 1st Harmonic

½ λ = Lor λ = 2L

Harm SeriesHarm Series

Second Harmonic (1Second Harmonic (1stst overtone) overtone)

NN N

A A λ = Lf2 = 2f1

L

Harmonic Ser (String/Open)Harmonic Ser (String/Open)

33rdrd Harmonic – 2 Harmonic – 2ndnd Overtone Overtone

N N N N

A A A

L

3/2 λ = Lλ = 2/3 L

f3 = 3f1

Harmonic Ser (Str/Open)Harmonic Ser (Str/Open)for all harmonics for all harmonics

Then the frequency f is found Then the frequency f is found by: by:

ffnn = n (v/2L) n = 1,2,3… = n (v/2L) n = 1,2,3…

harmonic no.

FreFre qq

……… ……….or the frequency may be .or the frequency may be found by found by

ff11 (fundamental or 1 (fundamental or 1stst harmonic) harmonic)

x 2 = 2x 2 = 2ndnd harmonic harmonic

x 3 = 3x 3 = 3rdrd harmonic harmonic

x 4 = 4x 4 = 4thth harmonic….etc harmonic….etc

exex

What is the frequency of the 2What is the frequency of the 2ndnd harmonic of a .50 m long instrument harmonic of a .50 m long instrument string when the speed of sound is string when the speed of sound is 340 m/s ?340 m/s ?

exex

ff22 = n (v/ 2L) = n (v/ 2L)

= 2 (340/ 2(.5))= 2 (340/ 2(.5))

ff22 = 680 Hz = 680 Hz

Harmonic Series (Open)Harmonic Series (Open)

… …it looks like this: it looks like this:

HS (open)HS (open)

……the air column may look like this:the air column may look like this:

A

A

N

1st harmonic

N

N

A

2nd harmonic

Harm Series (Open air)Harm Series (Open air)

We find freq same way as in strings:We find freq same way as in strings:

ffn = n = n(v/2L) n = 1,2,3…n(v/2L) n = 1,2,3…

harmonic no.

Harmonic Series (Closed air)Harmonic Series (Closed air)

Closed air columnClosed air column

Harm Series (Closed air)Harm Series (Closed air)

..or like this:..or like this:

N

A

L

1/4λ =Lλ = 4L

Fundamental1st Harmonic

N

A

N

A

3/4λ =Lλ = 4/3L

3rd Harmonic

Harmon (Closed air col)Harmon (Closed air col)

Only odd harmonics in a closed air Only odd harmonics in a closed air column !column ! Freq for closed air col:Freq for closed air col:

ffn n == n(v/4L) n = 1,3,5….(odds)n(v/4L) n = 1,3,5….(odds)

harmonic no.

Harmonic series…lets do it:Harmonic series…lets do it:

SAMPLE 13 BSAMPLE 13 B

L = 2.45 m long open organ pipeL = 2.45 m long open organ pipe * 1* 1stst Harmonic = fundamental Harmonic = fundamental = f= f1 1 = (1)(345/2(2.45)= (1)(345/2(2.45) = 70.4 Hz= 70.4 Hz

HSHS

* 2* 2ndnd Harmonic is: Harmonic is: use formula or justuse formula or just ff2 2 = 2f= 2f11 = 2(70.4) = 2(70.4) = 141 Hz= 141 Hz

* 3* 3rdrd Harmonic is: Harmonic is: ff3 3 = 3f= 3f1 1 = 3(70.4)= 3(70.4)

= 211 Hz= 211 Hz

HarmonicsHarmonics

PRACTICE 13BPRACTICE 13B 1) L = 0.20 m long closed pipe1) L = 0.20 m long closed pipe

* Fundamental = n(v/4L)* Fundamental = n(v/4L) = 1(352/4(0.20)= 1(352/4(0.20) = 440 Hz= 440 Hz

HarmonicsHarmonics

PRAC 13 BPRAC 13 B #3#3 a) L = .700 m stringa) L = .700 m string fundam = n (v/2L)fundam = n (v/2L) = 1 (115/2(.700)= 1 (115/2(.700) 82.1 Hz82.1 Hz

HarmonicsHarmonics

What is the 3What is the 3rdrd harmonic or 2 harmonic or 2ndnd overtone of this string ?overtone of this string ?

f3 = 3f1f3 = 3f1 = 3(82.1) = 3(82.1) f3 = 246 Hzf3 = 246 Hz

Harmon & freqHarmon & freq

Ex: Middle C = 261.63 HzEx: Middle C = 261.63 Hz Major 2Major 2ndnd or or full step above = 9/8 ratiofull step above = 9/8 ratio = 1.125(261.63)= 1.125(261.63) = 294.33 Hz= 294.33 Hz

Freq .Freq .

* 5* 5thth above C = 3/2 ratio above C = 3/2 ratio = 1.5(261.63) = 392.44 Hz= 1.5(261.63) = 392.44 Hz * Octave = 2/1 ration* Octave = 2/1 ration = 2(261.63) = 523.26 Hz C = 2(261.63) = 523.26 Hz C

aboveabove

FreqFreq

…………so 2so 2ndnd harmonics are an octave harmonics are an octave above the fundamental for strings above the fundamental for strings and open air columns……and open air columns……

p499p499

1) f1) f11 = n (v/ 4L) = n (v/ 4L)

= 1( 352/ 4(.20))= 1( 352/ 4(.20))

ff11 = 440 Hz = 440 Hz ff22 = 2 x 440 = 880 Hz = 2 x 440 = 880 Hz ff33 = 3 x 440 = 1320 Hz = 3 x 440 = 1320 Hz

TimbreTimbre

* Timbre……..* Timbre……..

* Beats…..interference of waves of * Beats…..interference of waves of slightly different frequencies slightly different frequencies traveling in the same direction, traveling in the same direction, perceived as variations of loudnessperceived as variations of loudness

BeatsBeats

So beats are difference between 2 So beats are difference between 2 freq..freq..

ex) Frq 1 = 200 Hzex) Frq 1 = 200 Hz Frq 2 = 205 HzFrq 2 = 205 Hz …… …….so the beat is difference =.so the beat is difference = 5 Hz5 Hz

BeatsBeats

page 503 #3page 503 #3

Beat = 4 /sec Beat = 4 /sec Frq 1 = 392 HzFrq 1 = 392 Hz

…………so frq 2 is either 396 Hz or 388 so frq 2 is either 396 Hz or 388 HzHz

HW IntensityHW Intensity

P 508P 508 27) * r = d = 5.0 m * P = 3.1 x 1027) * r = d = 5.0 m * P = 3.1 x 10-3 -3 WW Rel Int (dB) = ?Rel Int (dB) = ? I = P/AI = P/A = 3.1 x 10= 3.1 x 10-3-3/ 4/ 4ππ (5) (5)22

I = 1.0 x 10I = 1.0 x 10-5-5 W/m W/m22

so its 70 dB on tableso its 70 dB on table

HW HarmonHW Harmon

P 508P 508 30) a) this is ½ 30) a) this is ½ λλ…so ½ …so ½ λλ = L = L and then and then λλ = 2L so = 2(2.0m) = 4.0 m = 2L so = 2(2.0m) = 4.0 m

B) this is 1 B) this is 1 λλ… … λλ = L = 2,0 m = L = 2,0 m

HarHar

30c) this is 1 ½ or 3/2 of 30c) this is 1 ½ or 3/2 of λλ…so…so 3/2 3/2 λλ = L or = L or λλ = 2/3 L = 2/3 L so so λλ = 2/3(2.0m) = 4/3 = 1.3 m = 2/3(2.0m) = 4/3 = 1.3 m

D) this is 2 D) this is 2 λλ so …..2 so …..2λλ = L or = L or λλ = 1/2L = 1/2L then then λλ = ½(2.0m) = 1.0 m = ½(2.0m) = 1.0 m

HW harmHW harm

32) The instruments have different 32) The instruments have different harmonics present at various harmonics present at various intensities…intensities…

HW HarmonHW Harmon

39) It’s a string so….f39) It’s a string so….fnn = n (v/2L) = n (v/2L)

Lets get the fundamental then the Lets get the fundamental then the other harmonics….other harmonics….

HW HarmonicsHW Harmonics

f 1 = 1((274.4)/ 2 (.31m))f 1 = 1((274.4)/ 2 (.31m)) f1 = fundamental = 443 Hzf1 = fundamental = 443 Hz f2 = 2f2 = 2ndnd harmonic = 2 x 443 = 886 harmonic = 2 x 443 = 886

HzHz f3 = 3f3 = 3rdrd harmonic = 3 x 443 = 1330 harmonic = 3 x 443 = 1330

HzHz

HW HarmHW Harm

41) Open pipe so…f41) Open pipe so…fnn = n(v/2L) = n(v/2L) a) solve for L right ?a) solve for L right ?

f = n(v/2L)f = n(v/2L) nv/ 2f = Lnv/ 2f = L

(1)(331)/ 2(320) = .52 m = L(1)(331)/ 2(320) = .52 m = L

HW HarmHW Harm

41b) f1 fundam = 320 Hz41b) f1 fundam = 320 Hz f2 = 2f2 = 2ndnd harmonic = 2 x 320 = 640 harmonic = 2 x 320 = 640

HzHz F3 = 3F3 = 3rdrd harmonic = 3 x 320 = 960 harmonic = 3 x 320 = 960

HzHz

HWHarmHWHarm

41c)41c) f1 = fundam = n(v/2L)f1 = fundam = n(v/2L) = (1)((367)/ 2(.52m))= (1)((367)/ 2(.52m)) f1 = 353 Hzf1 = 353 Hz

HW beatsHW beats

44) Beats is just the difference so its 44) Beats is just the difference so its /132 – 137/ = 5 Hz/132 – 137/ = 5 Hz