PHYSICS 220

PHYSICS 220

Lecture 21SoundSound

Textbook Sections 13.1 – 13.7

Lecture 21 Purdue University, Physics 220 1

Overview• Last Lecture

– Interference and Diffraction• Constructive, destructive• Diffraction: bending of waves around obstaclesDiffraction: bending of waves around obstacles

– Coherence• Same frequency and constant phase difference

– Reflection and Refraction• fixed point inverts wave• sinθ1/sinθ2 = v1/v2sinθ1/sinθ2 v1/v2

– Standing Waves (fixed ends)• λn = 2L/n• fn = n v / 2L

• TodayS d W


– Sound Waves

Quiz1) A violin string of length L is fixed at both ends. Which one

of these is not a wavelength of a standing wave on the string?string?

A) 5L/3B) L/2 λ =

2LC) L

D) 2L

λn = nL

2) Two strings, one thick and the other thin, are connected to form one long string. A wave travels along the string andform one long string. A wave travels along the string and passes the point where the two strings are connected. Which of the following does not change after crossing that point:

A) frequencyB) wavelength


B) wavelengthC) propagation speed

Sound Waves


Speed of Sound• Recall for pulse on string: v = sqrt(F / μ)

For fluids: v sqrt(B/ )• For fluids: v = sqrt(B/ρ)– B = bulk modulus:

ΔP = −BΔVV

• For ideal gas: the speed of sound is proportional to the sq are root of the absol te temperat re b t independent

V

square root of the absolute temperature, but independent of pressure and density Medium Speed (m/s)B Tρ∝ Air 343

Helium 972T

B Tρ∝

Water 1500

Steel 5600 v = v0 T0

T i K l iLecture 21 Purdue University, Physics 220 5

Steel 5600T in KelvinRoom T = 20 0C or 293 K

Speed of Sound


Speed of Sound• The general relationship for a wave, v = ƒ λ holds• The speed of the sound wave is a function of the

medium and is independent of the frequency and amplitude– The frequency and amplitude depend on how the wave is

generated• Frequency range of normal human hearing is 20 Hz q y g g

to 20 kHz• Sound waves below 20 Hz are called infrasonicSound waves below 20 Hz are called infrasonic• Sound waves above 20 kHz are called ultrasonic

Lecture 21 7Purdue University, Physics 220

Intensity and Loudness• Intensity is the power per unit area

– I = P / A– Units: Watts/m2

• For Sound Waves– I = p0

2 / (2 ρ v) (po is the pressure amplitude)– Proportional to p0

2 (note: Energy goes as A2)• Loudness (Decibels)

L d ti i l ith i– Loudness perception is logarithmic– Threshold for hearing I0 = 10-12 W/m2

– β = (10 dB) log10 ( I / I0)β (10 dB) log10 ( I / I0)– β2 – β1 = (10 dB) log10(I2/I1)

• Pitch


– Perception of frequency

Log10 Review

• Log10(1) = 0• Log (10) = 1

log10(10x ) = x• Log10(10) = 1• Log 10(100) = 2

L (1 000) 3logx2 = 2 log x

• Log10(1,000) = 3

• Log10(10 000 000 000) = 10Log10(10,000,000,000) = 10


Decibels• The sensitivity of the ear depends on the frequency

f th dof the sound• In the most favorable frequency range, the ear can

detect intensities as small as about 10-12 W/m2

• A more convenient unit, the decibel (dB), is often , ( ),used to look at sound intensity

• The sound intensity level β is measured inThe sound intensity level, β, is measured in decibels


Decibels• Sound intensity level β

– β = (10 dB) log10 ( I / I0)– Units: Bels

I 10−5 W/ 2

A ratio of 107 indicates a sound intensity of 7

II0

=10 5 W/m2

10−12 W/m2 = 107

– A ratio of 107 indicates a sound intensity of 7 bels or 70 decibels (dB)

• An intensity of 0 dB corresponds to the y phearing threshold– Intensity increase by factor 10 → intensity

Alexander Graham Bell(1847-1922)

level adds 10 dB– Adding 3 dB to the intensity level doubles the

intensity (log102=0.3)


te s ty ( og10 0 3)

Intensity Level of Some Sounds


Question

If 1 person can shout with loudness 50 dBIf 1 person can shout with loudness 50 dB. How loud will it be when 100 people shout?

A) 52 dB B) 70 dB C) 150 dB

β100 – β1 = (10 dB) log10(I100/I1)β100 = 50 + (10 dB) log10(100/1)β100 = 50 + 20


β100 50 20

Standing Sound Waves

λ2

LPipe open at both ends

λn = nL


Standing Sound Waves

λ4

LPipe open at one end

λn = nL


Standing Waves in PipesOpen at both ends: Open at one end:

Pressure node at endλ = 2 L / n n=1,2,3, ...

Pressure antinode at closed end: λ = 4 L / n, , ,

n=1,3,5, ...


Standing Wave Summary• The closed end of a pipe is a pressure antinode for

th t dithe standing wave• The open end of a pipe is a pressure node for the

standing wave• A pressure node is always a displacement p y p

antinode, and a pressure antinode is always a displacement nodep

• These three points will allow you to predict the allowed frequencies of standing sound waves inallowed frequencies of standing sound waves in pipes


Organ Pipe Example

A 0 9 m organ pipe (open at both ends) isA 0.9 m organ pipe (open at both ends) is measured to have it’s n=2 harmonic at a frequency of 382 Hz What is the speed of sound in the pipe?of 382 Hz. What is the speed of sound in the pipe?

Pressure node at each end.

λ = 2 L / n n=1,2,3.., ,

λ = L for n=2 harmonic f = v / λf = v / λ

v = f λ = (382 s-1 ) (0.9 m)


= 343 m/s

Resonance ILQWhat happens to the fundamental frequency of pp q ya closed pipe, if the air (v=343 m/s) is replaced by helium (v=972 m/s)?y ( )

A) Increases B) Same C) DecreasesA) Increases B) Same C) Decreases

f = v/λλ1=4Lf = v/(4L)


Musical Tones• A sound wave described by a single frequency is

ll d tcalled a pure tone• Most sounds are combinations of many pure tones• Many sound are a combination of frequencies that

are harmonically relatedy• One way to characterize a combination tone is by a

property called pitchproperty called pitch• Generally, the pitch of a pure tone is its frequency

With l d th it h i i t d• With more complex sounds, the pitch is associated with the fundamental frequency


Real Musical Tones• Pipes are the basis for many musical instruments

Th t di i th i t t i• The standing wave in the instrument is a combination of standing waves with different frequenciesfrequencies

• These combination tones are often said to have a property called timbreproperty called timbre– Also known as tone color

The timbre depends on the mix of frequencies• The timbre depends on the mix of frequencies• The timbre is different for different instruments and

how the instrument is playedhow the instrument is played


TimbreFundamental + OvertonesFourier decomposition


Superposition & Interference• Consider two harmonic waves A and B meeting at x=0.

– Same amplitudes, but ω2 = 1.15 x ω1.

• The displacement versus time for each is shown below:

A(ω1t)

B(ω2t)

Wh t d C(t) A(t) + B(t) l k lik ?


What does C(t) = A(t) + B(t) look like?

Superposition & Interference• Consider two harmonic waves A and B meeting at x=0.

– Same amplitudes, but ω2 = 1.15 x ω1.

• The displacement versus time for each is shown below:

A(ω1t)

B(ω2t)

DESTRUCTIVEINTERFERENCEC(t) = A(t) + B(t)


CONSTRUCTIVEINTERFERENCE

C(t) A(t) + B(t)

Beats• If two tones are played after one another, you will

b bl t d t t i i f diffbe able to detect a minimum frequency difference, ∆ƒ

• For most people, ∆ƒ ≈ 0.01 ƒ– This is the smallest detectable frequency difference for

tones that are played consecutively• If two tones are played simultaneously, you can p y y y

detect much smaller frequency differences• When two tones are played together, theWhen two tones are played together, the

superposition principle indicates the sound pressures will addpressures will add


Beat Frequency• The amplitude of this

combination pressure wave oscillates

• These amplitude oscillations are called beats

• The frequency of the oscillations is the beatoscillations is the beat frequency, ƒbeat– ƒ = | ƒ – ƒ |– ƒbeat = | ƒ1 – ƒ2 |

• Beats can be used to tune musical instrumentsmusical instruments


BeatsAdd two cosines and remember the identity:

where and

( )1 2cos( ) cos( ) 2 cos cos( )L HA t A t A t tω ω ω ω+ =

( )1ω ω ω= − ( )1ω ω ω ω= + ≈where and( )1 22Lω ω ω= ( )1 22Hω ω ω ω= + ≈

cos(ωLt)( L )

Beat Frequency: 2beat L beator f fω ω ω= = Δ = Δ


Doppler Effect Moving Source


Doppler Effect Moving Source• When source is coming toward you (vs > 0)

– Distance between waves decreases– Frequency increases

• When source is going away from you (vs < 0)g g y y ( s )– Distance between waves increases– Frequency decreasesq y

• f = f / (1 v /v)• fo = fs / (1- vs/v)


iClickerA police car passes you with its siren on. Th f f th d h fThe frequency of the sound you hear from its siren after the car passes

A) Increases B) Decreases C) Same) ) )


Doppler Effect Moving Observer


Doppler Effect Moving Observer• When moving toward source (vo < 0)

– Distance between waves decreases– Frequency increases

• When away from source (vo > 0)y ( o )– Distance between waves increases– Frequency decreasesq y

• f = f (1 v /v)

Combine: f = f (1-v /v) / (1-v /v)

• fo = fs (1- vo/v)


Combine: fo = fs (1-vo/v) / (1-vs/v)

Moving Observer and Source

Combine: fo = fs (1-vo/v) / (1-vs/v)

A: You are driving along the highway at 65 mph, and behind you a police

Combine: fo fs (1 vo/v) / (1 vs/v)

g g g y p y pcar, also traveling at 65 mph, has its siren turned on.B: You and the police car have both pulled over to the side of the road, but the siren is still turned onthe siren is still turned on.In which case does the frequency of the siren seem higher to you?A) Case A B) Case B C) same) ) )

f f ’v

v v


vs vo

Velocity ILQA sound wave having frequency f0, speed v0 and wavelength λ0, is traveling through air when it encounters a large helium-filled balloon. Inside the balloon the frequency of the wave is f1, its speed is v1, and its wavelength is λ1.C th d f th dCompare the speed of the sound wave inside and outside the balloonA)A) v1 < v0

B) v1 = v0

C) v1 > v0

V1=965m/sV0=343m/s


Frequency ILQA sound wave having frequency f0, speed v0 and wavelength λ0, is traveling through air when it encounters a large helium-filled balloon. Inside the balloon the frequency of the wave is f1, its speed is v1, and its wavelength is λ1.Compare the frequency of the soundwave inside and outside the balloonA) f1 < f0B) f1 = f0

ffC) f1 > f0f1f0

Time between wave peaks does


pnot change!

Wavelength ILQA sound wave having frequency f0, speed v0 and wavelength λ is traveling through air when it encounters a large heliumλ0, is traveling through air when it encounters a large helium-filled balloon. Inside the balloon the frequency of the wave is f1, its speed is v1, and its wavelength is λ1.1, p 1, g 1

Compare the wavelength of the sound wave inside and outside the balloonwave inside and outside the balloonA) λ1 < λ0

B) λ1 = λ0

λ0 λ1

B) λ1 λ0

C) λ1 > λ0

λ = v / f


Human Ear

• Audible range: 20 Hz - 20 kHzAudible range: 20 Hz 20 kHz• Your ear is sensitive to an amazing

range! 1dB – 100 dBrange! 1dB – 100 dB– 10-12 Watts/m2

1 Watt/m2– 1 Watt/m2

• Like a laptop that can run using all powerLike a laptop that can run using all power of– Battery– Battery– Entire Nuclear Power Plant


Hearing• When a sound wave reaches

h hyour ear, the pressure on the outside of your eardrum is the atmospheric pressure plusatmospheric pressure plus the oscillating sound pressure

• On the inside of yourOn the inside of your eardrum, the pressure is equal to the atmospheric pressure

• Generally, your eardrum is insensitive to atmospheric pressure, but detects the sound pressuresound pressure


Ear as a Pressure Detector• Intensity is proportional to the square of the

lit dpressure amplitude• There is a relationship between the threshold

intensity for human hearing Io and the smallest pressure oscillation that can be detected by the ear

• A sound intensity of Io = 1.00 x 10-12 W/m2

corresponds to a pressure amplitude of p p ppo = 2 x 10-5 Pa

• This is about 2 x 10-10 times smaller thanThis is about 2 x 10 times smaller than atmospheric pressure


Human Perception of Sound• The lowest line plots the

intensity of the minimum ydetectable sound as a function of the frequency

• Doubling the perceived loudness corresponds to

i f t limoving from one contour line to the next

• The highest contour curve has• The highest contour curve has a loudness at which the ear is damagedg– Occurs at β ≈ 120 dB

• The loudness contours depend on frequency


Ultrasound Images• Ultrasonic imaging uses sound

waves to obtain images inside a material– The most familiar use is to produce

views inside the human body• The depth of objects inside the

body is determined by the time it takes the reflected ray to return to the detector

• Some medical ultrasounds also use the Doppler effectuse the Doppler effect


Summary of Concepts• Speed of sound v = sqrt(B/ρ)

• Intensity β = (10 dB) log10 ( I / I0)

• Standing Wavesf /(2L) O t b th d 1 2 3– fn = n v/(2L) Open at both ends n=1,2,3, …

– fn = n v/(4L) Open at one end n=1,3,5, …

• Doppler Effect fo = fs (v-vo) / (v-vs)

• Beats ω L =

12ω1 −ω2( )


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