• result from periodic disturbancelwillia2/2b/2bch14.pdf• result from periodic disturbance •...

• result from periodic disturbance • same period (frequency) as source• Longitudinal or Transverse Waves• Characterized by

– amplitude (how far do the “bits” move from their equilibrium positions? Amplitude of MEDIUM)

– period or frequency (how long does it take for each “bit” to go through one cycle?)

– wavelength (over what distance does the cycle repeat in a freeze frame?)

– wave speed (how fast is the energy transferred?) v fλ=

1f =Τ

SoundString

Types of Waves

Sound is a Longitudinal Wave

Pulse

Tuning Fork

Guitar String

is a Pressure Wave

Spherical Waves

3Hz

5Hz

Wavelength and Frequency are Inversely related:vfλ

=The shorter the wavelength, the higher the frequency.The longer the wavelength, the lower the frequency.

Wave speed: Depends on Properties of the Medium: Temperature, Density, Elasticity, Tension, Relative Motion

v fλ=

• The speed of sound waves in a medium depends on the compressibility, density and temperature of the medium

• The compressibility can sometimes be expressed in terms of the elastic modulus of the material

• The speed of all mechanical waves follows a general form:

Bvρ

=

elastic propertyinertial property

v =

Yvρ

=

C(331 m/s) 1273 C

Tv = + o

Liquid or Gas:

Solid Rod:

Dependence on Temperature:

(1D string)Tvρ

=

v fλ=

343 m/s in Air @ 20 C °

5960 m/s in Steel @ 20 C °

1522 m/s in Ocean Water @ 20 C°

Speed of Sound in a Vacuum?

343 m/s soundv =

Problem: You see lightening flash and 10 seconds later you hearthe thunder clap. How far away was the lighting from your position?

d vt= (343 / )10m s s= 3.43km=

(Rule of thumb: divide time by 5 to get miles)~ 2 miles

Speed of wave depends on properties of the MEDIUM

Speed of particle in the Medium depends on

SOURCE: SHM

v fλ=

( ) sinv t A tω ω= −

v fλ=

(1D string)

/ (linear mass density)

Fv

m Lμ

μ

=

=

Waves on Strings

Problem:

The displacement of a vibrating string vs position along the string is shown. The wave speed is 10cm/s.

If the linear density of the string is .01kg/m, what is the tension of the string?

v fλ=

Problem:


If the linear density of the string is .01kg/m, what is the tension of the string?

2 ( / )F v m L=2 5(.1 ) (.01 / ) 10F m kg m N−= =

/Fv

m L=

Problem:


If the the tension doubles, how does the wave speed change?

/Fv

m L=

22 /

Fvm L

=2 2

/F v

m L= =

Wave speed increases by a factor of 2

Wave PULSE:

• traveling disturbance• transfers energy and momentum• no bulk motion of the medium • comes in two flavors

• LONGitudinal• TRANSverse

Reflected PULSE:

Free End Bound End

If the end is bound, the pulse undergoes an inversion upon reflection: “a 180 degree phase shift”

If it is unbound, it is not shifted upon reflection.

Reflection of a Wave Pulse

Pulsed Interference

SuperpositionWaves interfere temporarily.

Reflection of a Traveling Wave

SuperpositionWaves ADD in space.

Simply add them point by point.

Superposition of Sinusoidal Waves

• Case 1: Identical, same direction, with phase difference (Interference)

• Case 2: Identical, opposite direction (standing waves)

• Case 3: Slightly different frequencies (Beats)

InterferenceWaves ADD: Constructive Interference.

Waves SUBTRACT: Destructive Interference.

In Phase Out of Phase

Interference of Sound Waves• Sound waves interfere

– Constructive interference occurs when the path difference between two waves’ motion is zero or some integer multiple of wavelengths

• path difference = nλ

– Destructive interference occurs when the path difference between two waves’ motion is an odd half wavelength

• path difference = (n + ½)λ

Interference

Chapter 14 Problem 26

Interference: Beats

beats frequency = difference in frequencies

Interference: BeatsAmplitude ~ Power ~ Loudness

Interference: Beats2 1

2 1

2

B

ave

f f ff ff

= −

+=

BeatsTwo tuning forks have frequencies of 440Hz and

438 Hz. What average frequency will you hear and what is the beat frequency?

BeatsTwo tuning forks have frequencies of 440Hz and

438 Hz. What average frequency will you hear and what is the beat frequency?

2 12 1

2B avef ff f f f +

= − =

2 1 440 - 438 2Bf f f Hz Hz Hz= − = =

2 1 440 438 4392 2ave

f f Hz Hzf Hz+ += = =

Standing Waves: Boundary Conditions

Node & Antinodes

• A node occurs at a point of zero amplitude

• An antinode occurs at a point of maximum displacement, 2A

0,1,2

nx nλ= = K

1,3,4

nx nλ= = K

Transverse Standing WaveProduced by the superposition of two identical

waves moving in opposite directions.

Standing WavesStanding waves form in certain MODES based on the length of the string or tube or the shape of drum or wire. Not all frequencies are permitted!

Standing Waves on a String Harmonics

Standing Waves on a String

1 2Lλ =

2 Lλ =

323Lλ =

Standing Waves on a String2

nLn

λ =

/n nf v λ=

2nvf nL

=

Standing Wave on a String Problem

A string with a mass of 8.00 g and a length of 5.00 m has one end attached to a wall; the other end is draped over a pulley and attached to a hanging object with a mass of 4.00 kg. If the string is plucked, what is the fundamental frequency of vibration?

You Try Standing Waves

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the fundamental frequency?

a. 316 Hzb. 632 Hzc. 158 Hzd. 215 Hze. 79 Hz

The possible frequency and energy states of a wave on a string are quantized.

1 2v vf

lλ= =

Standing Waves on a String Harmonics

1nf nf=

2nvf nl

=

Multiple Harmonics can be present at the same time.

Which harmonics (modes) are present on the string?

The Fundamental and third harmonic.

The possible frequency and energy states of an electron in an atomic orbit or of a wave on a string are quantized.

2vf nl

=

Strings & Atoms are Quantized

34

, n= 0,1,2,3,...

6.626 10nE nhf

h x Js−

=

=

Standing Waves in a Tube

1nf nf=

2nvf nL

=

Resonant Frequencies:

Open at both ends.

2Lλ =

Lλ =

Same as a string fixed at both ends.

Standing Waves in a Tube

4n

odd

Ln

λ =4n oddvf nL

=

Open at one end.

Standing Waves in a TubeWhat is the length of a tube open at both ends that has a fundamental frequency of 176Hz and a first overtone of

352 Hz if the speed of sound is 343m/s?

2nvf nL

=2 n

vL nf

=

343 /12(176 )

m sHz

=

.974m=

Problem 33

Fig. P14.33, p.384

is a Longitudinal Wave

is a What you Hear

The Pressure Wave sets the Ear Drum into Vibration.

The ear converts sound energy to mechanical energy to a nerve impulse which is transmitted to the brain.

Drum to Stirrup: Simple Machine Amplification

Since the pressure wave striking the large area of the eardrum is concentrated into the smaller area of the stirrup, the force of the vibrating stirrup is nearly 15 times larger than that of the eardrum. This feature enhances our ability of hear the faintest of sounds.

Resonance of the Cilia NervesThe inner surface of the cochlea is lined with over

20 000 hair-like cilia connected to nerve cells, each differing in length by minuscule amounts. Each hair cell has a natural sensitivity to a particular frequency of vibration. When the frequency of the sound wave matches the natural frequency of the nerve cell, that nerve cell will resonate with a larger amplitude of vibration which induces the cell to release an electrical impulse along the auditory nerve towards the brain.

Sound GenerationEnergy is transmitted as a pressure wave.

There is no net motion of the medium.The medium oscillates in simple harmonic motion.

The frequency of the wave is the same as the vibrating source.

VibratingString

Spherically SymmetricSound Source (bell).

Reflect

ECHO

Echo vs ReverberationA reverberation is perceived when the reflected sound wave reaches your ear in less than 0.1 second after the original sound wave. Since the original sound wave is still held in memory, there is no time delay between the perception of the reflected sound wave and the original sound wave. The two sound waves tend to combine as one very prolonged sound wave.

DiffractWe can hear around corners.

Why can’t we see around corners?

If the size of the wave (wavelength) is close in size to the object (door way) then the wave will diffract (bend).

RefractSound waves refract (bend) when moving between

mediums in which it travels at different speeds.

2 2

W 4 m

PIrπ

⎡ ⎤= ⎢ ⎥⎣ ⎦

The intensity of a wave, the power per unit area, is the rate atwhich energy is being transported by the wave through a unit area A perpendicular to the direction of travel of the wave:

2 2

W 4 m

PIrπ

⎡ ⎤= ⎢ ⎥⎣ ⎦The power transmitted by a wave is proportional to the amplitude of the wave.

A bell produces sound energy at a rate of 4.0 × 10−3 W and radiates it uniformly in all directions. What is the intensity of

the wave 10 m from the bell?

24Power PIArea rπ

= =

3

2

4.0 104 (10 )

x WImπ

−

= 6 23.2 10 /x W m−=

Cochlear Cilia Nerve Damage

Normal Ear Damaged Ear

Excessive exposure to loud sound can damage your cilia.

12 20Threshold of hearing : 10 /I W m−=

4 2Bursting of eardrums : 10 /I W m=

6 2Normal Conversation: 10 /I W m−=

10 2Whisper: 10 /I W m−=

2

0

10WhisperI

I=

0

log 2WII

= 2 bels

10 1decibels bel= 20 decibels

0 dB

20 dB

60 dB

160 dB

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

Decibel Index:

12 20Threshold of hearing : 10 /I W m−=

Whisper: 20dbConversation: 60dbLoud Music: 120 dbJet: 140 dBRocket: 250dB

OSHA Safety StandardsOSHA - Occupational Safety and Health Act - The OSHA criteria document reevaluates and reaffirms the Recommended Exposure Limit (REL) for occupational noise exposure established by the National Institute for Occupational Safety and Health (NIOSH) in 1972.

The REL is 85 decibels, A-weighted, as an 8-hr time-weighted average (85 dBA as an 8-hr TWA). Exposures at or above this level are hazardous.

11

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

If a sound is twice as intense, how much greater is the sound level, in db?

22

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

2 12 1

0 0

10 log 10 logI IdB dBI I

β β⎛ ⎞ ⎛ ⎞

− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 12 1

0 0

10 log /I IdBI I

β β⎛ ⎞ ⎛ ⎞

− = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1

10 log IdBI

⎛ ⎞= ⎜ ⎟

⎝ ⎠

2 1 10 log 2dBβ β− = 3.01dB=

53 dB is twice as intense as 50dB. Log Scale!!

11

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

The decibel level of a jackhammer is 130 dB relative to the threshold of hearing. Determine the sound intensity produced by the jackhammer.

1

0

130 10 log IdB dBI

⎛ ⎞= ⎜ ⎟

⎝ ⎠1

0

13 log II

⎛ ⎞= ⎜ ⎟

⎝ ⎠1

0log

1310 10II

⎛ ⎞⎜ ⎟⎝ ⎠=

13 1

0

10 II

=

131 010I I= 13 1210 10−= 210 /W m=

Intensity

A point source emits sound with a power output of 100 watts. What is the intensity (in W/m2) at a distance of 10.0 m from the source? What is it in dB?

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

2 2

W 4 m

PIrπ

⎡ ⎤= ⎢ ⎥⎣ ⎦

You TryCalculate the intensity level in dB of a sound

wave that has an intensity of 15 × 10–4 W/m2.a. 20b. 200c. 92d. 9e. 10

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

2 2

W 4 m

PIrπ

⎡ ⎤= ⎢ ⎥⎣ ⎦

You Try

By what factor will an intensity change when the corresponding sound level increases by 3 dB?

a. 3b. 0.5c. 2d. 4e. 0.3

0

10 log IdBI

β⎛ ⎞

= ⎜ ⎟⎝ ⎠

2 2

W 4 m

PIrπ

⎡ ⎤= ⎢ ⎥⎣ ⎦

Loudness Perception: PhonsPerception of Loudness depends on Frequency & Intensity

Sonic: 20 Hz 20 kHzINFRAsonic: 20HzULTRAsonic: 20kHz

ff

−<>

A middle C vibrates 252 times per second.

Sound Frequencies

Ultrasound:Pulverizing Tumors

5 2

~ 23~ 10 /

f kHzI W m

Deep Heat

3 2

~ 1~ 10 /

f MHzI W m

UltrasoundIntensity of reflected sound wave (echo) is

related to change in density in target.Ultrasound beam:

-2

7 1 detail~ 10MHz mm

I W→

William@10 Weeks

"A Womb With a View" and "Fetal Fotos” “Peek in the Pod”

Hi Cost Hi-Definition UltrasoundAre there RISKS?

"We do know in animal studies, certain levels of ultrasound can cause damages in growing bones, in developing bones," said Dr. Dan Schultz of the Food and Drug Administration.

Ultrasound QuestionHow far apart are two layers of tissue that produce echoes having round-trip times that differ by 0.750μs? What minimum frequency must the ultrasound have to see detail this small?

The speed of sound in human tissue is 1540m/s.

( )( )64w

1540 m s 0.750 10 s5.78 10 m

2 2v td

−−

×ΔΔ = = = ×

v f f vw

w 6m s m

2.67 10 Hz= ⇒ = =×

= ×−λλ

1540 578 10 4.

2 1 s 2 s 1 s / 2d d d v t v t v tΔ = − = − = Δ

Freaky QuestionWhich travels further, high or low frequencies?

Why?

Low frequency waves travel further because high frequency waves are absorbed by molecules in

the medium. All dat gets thru da wall is da boom boom Bass!

Animal Perception of Sound

•domestic cats •100-32,000 Hz•domestic dogs •40-46,000 Hz•African elephants •16-12,000 Hz

•bats •1000-150,000 Hz

•rodents •70-150,000 HzHuman: 20-20,00Hz

http://images.google.com/imgres?imgurl=http://www.reject.org/fatman/cats/cute_kitty.jpg&imgrefurl=http://www.pnwquizzing.org/bboard/index.cgi%3Faction%3Drecent&h=496&w=400&sz=55&tbnid=z4fH5hGdX_8J:&tbnh=126&tbnw=102&start=7&prev=/images%3Fq%3Dcats%26hl%3Den%26lr%3D

Infrasonic Contact Calls

Female African elephants use "contact calls" to communicate with other elephants in their bands (usually a family group). These infrasonic calls, with a frequency of about 21 Hz and a normal duration of 4-5 seconds, carry for long distances (several kilometers), and help elephants to determine the location of other Elephants. Calls vary among individual elephants, so that others respond differently to familiar calls than to unfamiliar calls. Perhaps elephants can recognize the identity of the caller.

Scientists first detected infrasound in 1883, when the eruption of the Krakatoa volcano in Indonesia sent inaudible sound waves careening around the world, affecting barometric readings.

The eruption of the Fuego volcano in Guatemala last year generated high-amplitude infrasound, mostly below 10 hertz. The pressure readings show that the strength of these sound waves can reach the equivalent of 120 decibels.

Infrasonic: < 20Hz

Echolocation: Sonic Vision

Dolphins produce high frequency (100kHz) clicks that pass through the melon. These sound waves bounce off objects in the water andreturn to the dolphin in the form of an echo. The brain receives the sound waves in the form of nerve impulses. By this complex system of echolocation, dolphins can determine size, shape, speed, distance, direction, and even some of the internal structure of objects in the water.

Dolphin Vocalization

SOund Fixing And Ranging

Acoustic Thermometry of Ocean ClimateATOC: 70 Hertz, with a sound pressure level of 195 dB

Dolphin, pinniped species sensitive to high frequencies (above 10,000 Hz) Baleen whales sensitive to low-frequencies (below 100 Hertz)

SOFAR Channel

Low Frequency Active Sonar

The LFAS system consists of a 35-ton block of 18 huge underwater speakers and dozens of microphones. The speakers emit a consistent low-frequency tone, between 100 and 500 Hertz, at 240dB, which travels out into the water at a depth of several hundred meters. The low frequency permits the sound to travel tremendous distances, detecting objects many hundreds of miles away by echolocation.

Physical Effect on Marine Life“At a 1 mile radius from the ship the noise only dissipates to 180 db which causes a bubbling effect in marine mammals' blood stream which creates embolisms. At 100 mile radius from the ship the noise only drops to 160 db which causes shearing of the tissues in the air sack behind whales' and dolphins' brain. This air sack is highlysensitive since it is used in echolocation. This shearing of tissue then causes hemorrhaging in their brains. Fish have little hairs in their ears that transmit sound waves from their ear canals to their central nervous system. The 160 db level shears these hair right off. Granted they grow back in 2 weeks, but they are deaf and are more likely to be picked off by predators and can't find food. Any fish or marine mammals caught in this "death zone" would have to swim 100 miles to escape the noise and pain.”

http://www.usagainstwhaling.org/soundkills.htm

Novermber 28, 2004“Sound bombing" of ocean floors to test for oil and gas for National

Security?

More than 100 whales and dolphins died in two separate beachings in 24 hours on remote Australian islands

http://story.news.yahoo.com/news?g=events/sc/113004whalebeach&a=&tmpl=sl&ns=&l=1&e=7&a=0&printer=

Deadly Sonar: NRDChttp://www.nrdc.org/wildlife/marine/sonar.asp

http://www.nrdc.org/wildlife/marine/sonar.asp

Gray whales migrating off the coast of Southern California

Sea Quakes produce powerful pressure waves that rupture the sinuses and middle ear of whales and dolphins.

Sound Weapons

Atomic Blast WaveA fraction of a second after a nuclear explosion, the heat from the fireball causes a high-pressure wave to develop and move outward producing the blast effect. The front of the blast wave, i.e., the shock front, travels rapidly away from the fireball, a moving wall of highly compressed air. The blast wind may exceed several hundred km/h. The range for blast effects increases with the explosive yield of the weapon and also depends on the burst altitude.

Which is traveling at subsonic, sonic, or supersonic speeds?

a)

b)

c)

Subsonic

Sonic

Supersonic

RADAR: RAdio Detecting And Ranging

•Cosmological Redshift: Expanding Universe•Stellar Motions: Rotations and Radial Motions•Solar Physics: Surface Studies and Rotations•Gravitational Redshift: Black Holes & Lensing•Extra-solar Planets via Doppler Wobbler

Case 1: Moving Source Stationary Observer 0Ov =

Sv

Observer Reference Frame

Sv

Observer Reference Frame


S O ?v′ =Speed of a wave is determined by the properties of the Medium!


What is the speed of sound to the observer?

wavev v=

S O

Speed of a wave is determined by the properties of the Medium!


What is the speed of sound to the observer?

v v′ =

wavev v=

S O

f fλ λ′ <′ >


v v′ =

Sv

source moves in time a distance Svτ τ


Sv

emits another wavelength


Sv

travels a distance and emits again...Sv τ


Sv

and so on...


Sv

bunching up the wavecrests by Sv τ


Sv

is sho rtened by = Svλλ λ τ′ −


= (1 )Svvτλτ

−

= (1 )Svv

λ −

= (1 )Sv τλλ

−

= ( )sv vv

λ λ −′

Sv

?f ′ =Case 1: Moving Source Stationary Observer 0Ov =

Use v v′ =

= sv vv

λ λ −′

( )sf f f v v

v

λ λλ λ

′ = =−′

'f fλ λ′ =

S

vf fv v

′ =−

Sv

Case 1: Source moving TOWARD (-) and AWAY (+) from Observer

What if ?Sv v=

= sv vv

λ λ ±′

S

vf fv v

′ =±

1

(1 )Sf f v

v

′ =±

If Sv v=1

(1 1)f=

−

10

=

1

(1 )Sf f v

v

′ =±

Source Moving:

1

(1 )Sf f v

v

′ =±

If Sv v=1

(1 )Sf f v

v

′ =−

1(1 1)

f=−

10

=

= Mach #Svv

1

(1 )Sf f v

v

′ =−

When the duck speed is equal or greater thanthe speed of waves in water, the waves form a bow wave.

Case 2: Observer Moving & Stationary Source

SOv

Observer Moving TOWARD (+) and AWAY (-) from Source

???

vf

λ′ =′ =′ =

Case 2: Observer Moving & Stationary Source

SOv

Observer Moving TOWARD (+) and AWAY (-) from Source

ov vf fv±′ =

ov v vλ λ′ =′ = ±

0(1 )vf fv

′ = ±

A siren , mounted on the tower, emits a sound with a frequency of 2140 Hz. What is the difference in the

frequency heard by the driver travelling away from the towerat 27 m/s between the directed and reflected sound of the siren?

Take the speed of sound to be 343 m/s.

(1 )Ovf fv

′ = ±

:2140 ,

343 / , 27 /O

Givenf Hzv m s v m s== =

Direct (1 )Ovf fv

′ = − Reflected (1 )Ovf fv

′ = +

Doppler Shift Problem

:2140 ,

343 / , 27 /O

Givenf Hzv m s v m s== =

Direct (1 )Ovf fv

′ = − Reflected (1 )Ovf fv

′ = +

Direct (1 )Ovf fv

′ = −

Reflected (1 )Ovf fv

′ = + 2310Hz=

1970Hz=

Doppler Shift Problem

If both Source and Observer are moving…..

o

s

v vf fv v+′ =−

+ : Moving Towards each other- : Moving Away from each other

0(1 )vf fv

′ = ±1

(1 )Sf f v

v

′ =±

Source Moving: Observer Moving:

• result from periodic disturbancelwillia2/2b/2bch14.pdf• result from periodic disturbance •...

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