chap17 physics unit 1 smh-waves i - waves ii

7/31/2019 chap17 Physics Unit 1 SMH-WAVES I - WAVES II

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4: Sound Waves

(Chapter 16)Phys130, A04

Dr. Robert MacDonald

On SoundSound is any longitudinal (compression) wave in atangible medium, like air, wood, rock, the sun, etc.

At least some aspects of an earthquake areessentially huge sounds.

The audible range of sound for humans is about2020,000 Hz. Sounds below this range are calledinfrasonic; above this range theyre called ultrasonic.

Elephants use low frequency (1335 Hz) sounds tocommunicate with each other through the earth,over distances of more than 2.5 km. They can findeach other, converse, even recognize each other.

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http://dsc.discovery.com/news/2008/09/30/elephant-communication.html

Wavefunction of SoundSound waves generally travel out in all directions froma source. But for now, well focus on sound wavestravelling in a straight line in one direction, which well

call the positivexaxis (since we can).

This is the same wave function weve been workingwith up till now:y(x, t) =A cos(kx t). (The phaseconstant 0 isnt important for now, so well set thatto zero to keep things tidy.)

Here Im usingyto represent the displacement ofthe particles along the direction of wave motion. Thisis different than on a string, but the principles are

the same.3

Displacement vs PressureIn addition to talking about sound as a wave of particledisplacements, we can talk about pressure waves,describing how the air pressure changes as the sound

wave passes and the molecules bunch or spread.This is a useful description since its how we hear. Theeardrum has air on both sides.

Inside your head the eardrum is vented by theEustachian tube, so its always at atmospheric pressure(unless the tube is plugged!).

A sound wave changes the pressure up and down onthe outside of the eardrum. The difference moves the

eardrum back and forth.4


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Since what the wave causes is these variations inpressure, thats what well use to describe sound thedifference from atmospheric pressure.

Recall how a high-then-low longitudinal displacementgraph results in particles bunching and spreading. Lets

look at the sinusoidal version of that and figure out howto describe what the pressures doing.

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http://www.animations.physics.unsw.edu.au/jw/sound-pressure-density.htm

x

Ox x+x

Undisturbed cylinderof air:

Surface area S

When a sound wave passed through, each end of thiscylinder will be displacedaccording to the wavefunctiony.The displacement of the left end at time t will bey(x, t);the displacement of the right end will bey(x+x, t).

Ox x+x

Ox x+x

y(x, t) y(x+x, t)

>x


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That last version becomes a derivative ifxis very small(i.e. the limit x0):

The change in volume results in a change in pressure(youre compressing or expanding the gas). The amountof pressure change depends on the bulk modulus B,defined as

If we define p(x,t) as the difference between the pressure

in our cylinder and atmospheric pressure (i.e. p = P),then

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See Section 12.7 formore details about B.

Since we know the wavefunction,y(x, t) =A cos(kx - t), wecan evaluate y/xand determine a formula for p(x,t):

where k is the wavenumber, B is the bulk modulus,A is the

displacement amplitude, and is the angular frequency.

More amplitude means more pressure change.

More bulk modulus means its harder to compress thegas, so for a given amplitude you get more pressurechange.

Shorter wavelength means more pressure change (!).

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Difference from atmospheric

pressure in a sinusoidal soundwave.

Maxdifference from atmospheric

pressure in a sinusoidal sound wave.

http://www.animations.physics.unsw.edu.au/jw/sound-pressure-density.htm12

Perception of SoundLoudness depends on the amplitude of the soundwave. But the perceived loudness varies from personto person!

We dont have a uniform frequency response some frequencies sound louder than others, evenat the same amplitude. The details vary fromperson to person.

We lose sensitivity to sound over time, especiallyat the higher frequencies.

Loud sounds damage hearing, too look atrock musicians and orchestra members.


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Pitch is of course mainly dependent on the(fundamental) frequency of the sound wave. But thereare a number of auditory illusions that make a pitchsound higher or lower. For example, if you hear twotones at the same frequency but different amplitudes, thelouder one will tend to sound a little lower.

Timbre (aka, tone colour or quality), the characteristicsound of an instrument, comes from the particularcombination of sine waves of various frequencies andamplitudes that make up its sound that is, its harmoniccontent. These waves come from the air or string as wellas from the instrument body.

The human voice works basically the same way.13

http://phet.colorado.edu/simulations/sims.php?sim=Fourier_Making_Waves

From The Physics Hypertextbookhttp://physics.info/music/

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Noise is a combination of a range of frequencies, notjust harmonics.

White noise is a uniform combination of allfrequencies of sound (within some range, of course).

This is why instruments that are out of tune witheach other sound so bad: the combination isntharmonic but noisy.

We have an expression for the speed of a wave on astring:

What does the speed of a wave (sound) in a fluiddepend on?

We can expect it has something to do with howdifficult it is to compress the fluid; this is describedby the bulk modulus.

Its probably related to how hard it is to get themolecules of gas moving their mass, or (similarto the string) their mass density.

We can use the wave equation to find the speed.16

Speed of Sound in a Fluid


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Well start by looking at the speed of sound through apipe.

This is a relevant example, since it describes windinstruments and the human voice.

Well ignore any effects of the walls of the pipe (like

friction); as long as the pipe is reasonably big it shouldnthave much effect. (Sound through a small tube, it turnsout, is slowed down by the walls. This is called the tubeeffect.)

All we care about is that the fluid cant expand laterally(i.e. cant squish out the sides of the pipe).

Remember our three ingredients for the wave equation:a restoring force, Newton II (F = ma), and linearization.

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is the mass density "the mass per unit volume(usually measured in kg/m3). Its how heavy the air orwater etc is.

We can describe the sound by the way it changes thefluids pressure as it passes through, like before. Thepressure wave p(x,t) gives the amount that the pressureat pointxis increased or decreased by the sound wave,at time t.

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p(x,t) p(x+x,t)

x A = areaA = areaVolume of pipe segment:! V=AxMass of fluid in pipe segment:! m = V = A x

Greek letter rho

xaxis

pipe

Remember that pressure is a force distributed over anarea (force per unit area). For example, a largerparachute will slow your fall better than a smaller one.

So the force on each side of the segment of fluid is given

by the pressure there times the area. The force on theleft end (pushing right) is F(x,t) =A p(x,t), and on theright end (pushing left) is F(x+x,t) =A p(x+x,t).

The net force on the segment of fluid is just thedifference of these (remember right = positive):

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We can put this information into F = ma, as a steptowards relating what we know to the Wave Equation.

So we can cancel outA; the speed of sound doesntdepend on the size of the pipe!

We found previously thatwhich gives us:

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Net Force mass acceleration


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That last equation should look vaguely familiar from thederivation of wave speed on a string. Lets rearrange it abit:

The left hand side is just thex-derivative of the slope or the secondx-derivative of the position ifxisvery small. So we put that in:

Huzzah! The Wave Equation! This means that /B=1/v2,or:

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Speed of sound in a fluid.B is the bulk modulus.

is the mass density.

The speed of sound in a solid is more complex thanwere going to deal with here. Its complicated by theway the surrounding material keeps the wave from

squishing the material out sideways.It depends on the density, the bulk modulus (how hard itis to compress), and the shear modulus (how hard it is toshear the material), and sometimes some other things.

All this adds complexity without adding much morethan a few formulas. (If youre curious, though, its inyour textbook...) Well focus on fluids here.

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Speed of Sound in a Solid

Speed of Sound in a GasA gas is of course just a type of fluid, but its interestingto study specifically. For anyfluid the speed of sound isgiven by:

In a gas, though, the bulk modulus B depends on thepressure of the gas which can be changed quiteeasily. So, obviously, does the mass density .

Temperature and pressure in a gas are closely related.

So the speed of sound in a gas is very sensitive to the

pressure and the temperature.23

Here is the speed of sound in a few different media:

Air (0C):! 331 m/s Air (20C):! 343 m/s Helium (0C):! 965 m/s Water (20C):! 1482 m/s Lead! 1960 m/s Copper! 5010 m/s Glass (Pyrex)! 5640 m/s

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Speeds of Sound


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Compare the wavelength of sonar used by bats (in air,say at 20C) to that used by dolphins. Assume theyboth use a frequency of 100 kHz = 1.00x105 Hz. Anyrequired constants will be supplied.

Start by looking for a connection betweeninformation we have frequency "andinformation were looking for wavelengths. Thisis a simple one: v = ffor anywave, so well start

with that: = v/f.

Frequency is the same in both situations, but thespeed of soundchanges.

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Echolocation: Bat vs Dolphin Dolphins in water:

For water, the density is 1.00 g/cm3 = 1000 kg/m3, andthe textbook says B = 2.18x109 Pa. This givesv= 1476 m/s.

Then the wavelength is = v/f= (1476m/s)/(1.00x105Hz)or = 0.0148 m.

This is independent of temperature, aside from slightchanges in the density of water.

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Bats in air:

The bulk modulus of air at 20C is about 1.42x105 Pa.

The density of air at 20C is about 1.204 kg/m3

.

This gives v= 343 m/s, and = v/f= 0.00343 m.

Dolphins in water: = 1.48 cm

Bats in air: 0.34 cm

Bats can distinguish much smaller objects.(This makes sense they hunt bugs!)

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how the displacement of molecules in a sound waveresults in a change in pressure.

how to describe a sound wave in terms of eitherdisplacement or pressure.

what determines the speed of sound in a liquid or agas.

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So now you know:


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Wave IntensityIn one-dimensional waves (e.g. waves on a string), all ofthe energy that enters the string at one end makes it tothe other end (aside from energy lost to damping).

In three-dimensional waves (e.g. sound, light), the energyspreads out in more than one direction.

The intensity(I) of a (3D) wave is the average rate atwhich energy is transported across some unit of area.In other words, its the average power per unit area.

Written mathematically, intensity is defined as:! I = P/A.

Britains Listening Ears

http://www.ajg41.clara.co.uk/mirrors/dungeness.html

30 footsound mirror

200 footsound mirror

Consider a sphericalwave a wave which spreadsout evenly in all directions. This is the kind of waveyou get from a point source of light (bulb, candle,

star), sound, etc.

At a distance rfrom the source, the energy of thewave is spread over the surface of a sphere of area4r2. So the intensity is given by:

The way intensity drops with the square of distance iscalled the inverse square law.

Spherical Symmetry

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Power comingfrom the source

Surface area of a sphere

Intensity of sound inspherical waves.


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Inverse Square LawThe Inverse Square Lawdescribes how intensitychanges as a function of distance.

Consider two spheres of radius r1 and r2 centred on

the wave source. As long as the medium doesntabsorb any energy as the wave travels between thespheres (i.e. no damping), the total power Pgoingthrough each sphere should be the same. Then

Equate these and rearrange and we find:

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and

Inverse square lawfor intensity ofspherical waves.

Example: Solar PanelsConsider a 1 m2 solar panel in use at the equator (sowe dont have to worry about seasons).

The Earth is closest to the Sun in January, at a distance

ofr1 = 1.47x108 km.

The Earth is furthest from the Sun in July, at a distanceofr2 = 1.52x108 km.

If the solar panel generates 200 W of power in January(P1), how much power does it generate in July (P2)?

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200 W

1.47x108 km

1.52x108 km

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Intensity and AreaIntensity is defined aspower per unit area "thatis, the amount of energy passing through some surfaceevery second (or whatever).

For a spherical wave, at a distance rfrom the sourcethe original power is distributed over the surface of asphere of radius r. So the intensityof the wave at thisdistance is given by I = P 4r2.

If youre standing at a distance rand hold up a sheetwith areaA, the amount of power striking that sheet isgiven by I A the intensity times the area of thesheet. (Not the area of the imaginary sphere.)


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Not all waves are spherical! The formula for intensitydepends entirely on the shape of the waves. Theintensity of a laser, for example the amount of energygoing into that little dot of light per second is almostindependent of distance r.

So we need to know the shape of the wave in order tofigure out the intensity at a distance rfrom the source.

But if you want to know how much power is strikingyour photosensor, microphone, etc, it doesnt matterwhat the shape of the wave was. All that matters is theintensity of the wave and how big your sensor is. P = I Ais always true "its the definition of intensity!

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Shaping SoundYour voice can be reasonably modelled as a pointsource radiating in all directions. But if you cup yourhands, the sound that would have gone to the sidesalso gets reflected forward. The result is that the

sound doesn't spread out as much, so it decreasesmore slowly than 1/r2.

Reflections from walls mean that the inverse squarelaw is blown away indoors.

In fact, theatres (including home theatres!) andconcert halls should be carefully designed so youdont end up with weird dead spots or othereffects.

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Winspear Centre

The Winspear Centre downtown has fantasticacoustics "there are no bad seats! The whole placewas designed around how sound is transported,reflected, focussed, and mixed. No inverse square law!

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e.g. Intense ConcertDuring a particularly thrilling part of an EdmontonSymphony concert, the sound intensity reaching youreardrum is 0.80 W/m2.

Assume the eardrum is a circle of radius 0.4 mm.

What is the average rate at which energy is reachingyour ear?

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Examples of Sound Intensity

PowerIntensity at 1 m

away

Threshold of hearing 1x10-12 W/m2

Typical conversation 1x10-5 W 8x10-7 W/m2

Loud shout 3x10-2 W 2x10-3 W/m2

Threshold of pain 1 W/m2

I calculated intensity assuming spherical wavesand the inverse square law.

Intensity vs Amplitude

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We should expect the intensityof a sound wave to berelated to the displacement amplitude, or equivalently tothe pressure amplitude. Lets figure out what therelationship is.

Remember that Energy = (Force) x (Distance),which means that Power = (Force) x (Velocity).

(Power / unit area) = (Force / unit area) x (Velocity)

In other words, (Intensity) = (Pressure) x (Velocity).

The velocity depends on the displacement amplitude, sowe can use this to relate intensity to amplitude.

So if we want the velocity of the air, caused by thechange in pressure from the sound wave, look to thewave equation:y(x, t) =A cos(kx t). The derivativey/t gives the particle velocity were looking for.

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This is the instantaneous intensity.

This last equation, I(x,t), is the instantaneous intensity,the power per unit area at some place and at some time.Were almost never interested in this; rather, when wesay intensity we generally mean the time-averagedintensity.

Remember that the average of sin2 over a period is 1/2.

Then the time average ofI(x,t) = BkA2sin2(kxt) is

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rho

Average intensity of a

sinusoidal sound wavein a fluid.

= mass densityB = bulk modulus = angular frequency

A = displacement amplitude

wave speed!


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Remember that the maximum pressure (the pressureamplitude) is given by pmax"="BkA. ThenA = pmax/Bk.Use this in I!=!(1/2)BkA2:

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Intensity vs Pressure

Average intensity of asinusoidal sound wave

in a fluid.

pmax = pressure amplitude

= mass densityB = bulk modulusv= wave speed

I=p2max

2B

= Greek letter rho

= Greek letter omega

The Decibel ScaleHuman sight and hearing are logarithmic. Theperceived brightness or loudness of somethinggoes like the logarithm of the intensity.

So for sound they developed something called thesound intensity level (not to be confused withsound intensity!). Its represented by a (Greekletter beta), and is defined as:

Its measured in decibels or dB. One dB is 1/10 of abel. We pretty much always use decibels, though.

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sound intensity

reference intensity:10-12 W/m2base 10 logarithm

not a B!

The threshold of pain (1 W/m2) then corresponds to(10"dB)log(100/10-12)"="120"dB.A typical conversation (~10-6 W/m2) is around 65"dB.A whisper is more like 20"dB (~10-10 W/m2).Remember that the frequency response of the humanear isnt uniform, so 65 dB will sound louder at somefrequencies than at others. There are specialized soundscales and meters that adjust the numbers so that aparticular sound level will sound the same at allfrequencies.

The dBA scale is one of these. So 65 dBA soundsthe same at all frequencies, but high and lowfrequencies will carry a lot more power.

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Examples of Sound IntensityIntensity

at your ear

Intensity Level

at your ear

Ninja through dry leaves 0.5x10-12 W/m2 3 dB

Threshold of hearing 1x10-12 W/m2 0 dB

Typical conversation 3.2x10-6 W/m2 65 dB

Pirate conversation 3.0x10-2 W/m2 105 dB

Rock concert 1x100 W/m2 120 dB

Threshold of pain 1.0x101 W/m2 130 dB


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Its often useful to use other reference intensities forvarious purposes.

Amplifiers and attenuators will sometimes list how

much theyre changing the signal in dB. In this case the reference intensity is the input.

Stereos often display volume in negative decibels.

The stereo is attenuating(quietening) the signalfrom the CD or whatever.

0 dB on this scale is an unattenuated signal.(Positive dB means youre actually amplifying it.)

Decibels and Attenuation

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Consider an ideal bird "a point source. The birdssound will be radiating equally in all directions, in aspherical or hemispherical way, so it follows an inversesquare law.

If you go three times farther away from the bird, howmuch does the sound intensity levelof the birdsongchange?

How much does the pressure amplitude change?

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Example: Chirping Bird

Starting with the sound intensity level, first lets writedown the difference:

So its basically the usual definition of sound intensitylevel, but using the original intensity as the reference. (Inother words, decibels add!)

Now well apply the inverse square law:

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So moving three times farther away from the birdreduces the sound intensity level of the song by almost10"dB.What does this do to the pressure amplitude?

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Were looking for pressure amplitude, and we have thechange in sound intensity level. This is related to thesound intensity, which is related to the pressureamplitude by the formula:

Since were looking for the relative change in pressureamplitude we dont actually need to know the density"or the bulk modulus B. (Ill call the pressure amplitude phere instead ofpmaxto tidy it up a bit.)

These cancel out;its the same air!

A (very large!) electric spark jumps along a straight lineof length L = 10 m, and produces a noise with acousticpower ofPs = 1.6x104 W.

The bang travels radially outward from the spark; thewave looks like an expanding cylinder.

What is the intensity I of the sound at a distance ofr"="12"m from the spark?

At what rate Pd does sound energy reach an acousticdetector of areaAd = 2.0 cm2, aimed at the spark andlocated 12"m away from it?

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e.g. Cylindrical sound wave

Consider an imaginary cylinder of radius 12"m and height10 m, with no end caps.

All of the acoustic energy that leaves the spark must passthrough this cylinder, at the same (total!) rate.

So the total power passing through the cylinder is thesame as the total power coming from the source.

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That energy is uniformly distributedover the whole cylinder bysymmetry.

So the intensity at anypoint on thiscylinder is the total power divided bythe cylinders area (ignoring the ends!).

Total power passing through the cylinder = Ps

Total area of the cylinder sides = 2rL

Sound intensity at 12 m from the spark:

! I = P/A (definition of intensity!)I = Ps / (2rL)

! = (1.6x104 W) / (2(12 m)(10 m))! = 21.2 W/m2.

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So thats the intensity of sound at the surface of ouracoustic detector.

How much acoustic power is the detector receiving?

From the definition of intensity: P= IA.

Area of the detector =Ad = 2.0 cm2 = 2.0x104 m2.

So the detector receives a power of:! Pd = (21.2 W/m2)(2.0x104 m2) 4.2x103 Wor 4.2 mW.

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what intensity means (definition: I = P/A).

how the intensity of a wave can change withdistance, what the inverse square law is, and when

it applies.

how the intensity of a sound wave is related to itsdisplacement amplitude and pressure amplitude.

what the decibel scale is, and how it measuresrelative intensity.

That is, the intensity of a wave relative to somereference intensity.

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So now you know:

The Treachery of Images, Ren Magritte, c. 1928

When a wave reaches the end of a pipe, whether theend is open or closed, it reflects back into the pipe.

The wave may be inverted or not at the reflection,just like with a wave on a string with a loose orfixed end.

The incoming and reflected wave interfere (theirdisplacements add), and just as with strings you get astanding wave.

We can describe the standing wave in terms ofdisplacement or pressure, just like with other soundwaves.

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Standing Waves in a Pipe


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LongitudinalStanding Wave

Same wavefunction as a transverse standing wave!displacementyof the particle at positionx, at time t:

! y(x,t) =A sin(kx) sin(t)But now the displacementyis along the same directionas the wave motion.

Magnutude of pressure change p is largest wheredisplacementyis 0, and p = 0 where |y| is max.

displacement nodes = pressure antinodespressure nodes = displacement antinodes

So a good pressure wavefunction is:!p(x,t) = pmax cos(kx) sin(t)

Waves and Pipe Ends

Closed end of a pipe = displacement node(air has nowhere to go!)

Pressure can change! !pressure antinode

Open end of a pipe = displacement antinode

Pressure is equalized to atmospheric pressure!!"pressure node.

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WindInstruments

Usually have at least one end open, theother either open or closed.

Generate noise by buzzing lips or areed, or blowing air past a mouth(e.g. recorder, pipe organ).

Noise is a continuous combinationof many frequencies!

Frequencies matching the pipeharmonics resonate and are amplified.

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Davis Organ @ Winspear


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Standing Waves in a Pipe

Displacementenvelopes shown.

Two Open Ends

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This is basically the opposite ofthe clamped string, which haddisplacement nodes at each end.With two open ends we havedisplacement antinodes and

pressure nodes.

The longest wavelength that canproduce a standing wave is still1=2L, just like with a clampedstring.

The situation is basically the sameas before, so the formulas turnout the same. (Remember v= f.)

Allowed standing

waves in a pipewith two open

ends.

One Closed End (Stopped)A pipes open end has adisplacement antinode. The closedend has a displacement node. Thisis a little different, obviously.

The distance between a node andits nearest antinode is only /4. Sothe longest wavelength you can getin a stopped pipe is 1=4L twice as long as before (so twiceas low a note!).

The next harmonic will occurwhen weve squeezed in anotherhalf cycle, so = 4L/3 = 1/3.After that well get = 4L/5, etc...

only odd harmonics allowed!

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Allowed standingwaves in a pipe

with one openend.

Frequency and TemperatureThe wavelengths of standing waves possible in a pipe aredetermined by the length of the pipe.

Since v = f(for anywave, remember!) this means thefrequency will depend on the speed of sound in the pipe and that depends on air temperature!

As the temperature changes, the tuning of theinstrument changes. (As you may know if youve everplayed music outdoors.)

This can be a big problem for large pipe organs; oftensome pipes are warmer than others.

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Resonance and SoundWe studied forced oscillations previously. It worksthe same way with sound in a pipe as it does with asimple harmonic oscillator.

If you generate a sound with some frequency near thepipe, the molecules of air in the pipe will oscillate withthat frequency (just like the rest of the air). Thats allthere is to forced oscillations and sound.

Just like before, if you drive the molecules of air at afrequency theyd like to move at anyway, theoscillations build up, the amplitude increases, and youget resonance.

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The simple harmonic oscillator has only one resonantfrequency; if you let it go it will oscillate with thatfrequency.

But air in a pipe (or a string!) will be happy to oscillate inanyof its harmonics.

So if you apply sound with the same frequency as one(or more!) of the pipe's harmonics the air in the pipewill resonate.

As the sound reflects back and forth in the pipe, itsreinforced and added to by the incoming sound.

You may have already seen this in your labs.

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e.g.: Sound resonanceYou pick up a cardboard tube of length L = 67.0 cm andhold it near your ear. Random background sounds fromthe room set up a standing wave in the tube at its

fundamental frequency.

(Other standing waves, too, but the fundamentalfrequency is strongest.)

Assuming the speed of sound is v= 343 m/s, whatfundamental frequency do you hear from the tube?

If you jam your ear against one end of the tube, whatfundamental frequency do you hear from the tube?

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A 3 m stopped organ pipe (one end open) generatessound in its first harmonic. The speed of sound in thepipe is 350 m/s.

A nearby piece of guitar string is excited into its thirdharmonic by the sound. The string has linear massdensity 5 g/m and tension 50 N.

How long is the guitar string?

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Example: guitar and pipe


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So now you know: what standing sound waves are possible in a pipe,

and how that depends on whether the pipe ends areopen or closed.

How the wavelength and frequency of a standingsound wave are related to the sound speed and thepipe length (and pipe ends).

how sound waves can set up standing waves byresonance.

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Interference of WavesInterference is the term for what happens when twoor more waves overlap. Standing waves are a goodexample.

A different type of interference occurs when you havetwo or more similar waves travelling in the samedirection, or spreading out together in space.

The waves can add constructively, or destructively. Whichone you get depends on the relative phase of thetwo waves at the place youre looking.

(Remember, the principle of superposition is thestatement that when two waves overlap, they just add.)

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Combinedwaveform

wave 1

wave 2

Waves exactlyin phase = (2n), n = 0, 1, 2, ...

Waves exactlyout of phase = (2n+1), n = 0, 1, 2, ...

At the place where were looking, the two waves havedifferent phases (in general). = difference in phase.

depends on:

how the waves were created (different 0 values)

how far they travelled (difference inx/ (or kx))

what happened to them on the way (reflections etc)

can have any value! 75

Remember: phase = kxt + 0.

Some DemosSeveral animations showing superposition of 1-Dwaves, including beats, standing waves, and more:

http://www.kettering.edu/~drussell/Demos/superposition/superposition.html

Building snapshot & history graphs:

http://www.kettering.edu/~drussell/Demos/wave-x-t/wave-x-t.html

Two-source interference (shown later):

http://www.acoustics.salford.ac.uk/feschools/waves/super2.htm

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Phase DifferenceThink of a microphone placed near a setof two speakers, producing sound like this:

the same pure (single frequency) tone.

produced in phase (simultaneous

crests, etc).The principle of superposition says that thesound wave at the microphone will be thesum of the waves from each source.

Here we have identical waves, with

different phases when they reach themic.

How they add depends on the difference inphaseat the microphone.

In this case, depends only on thedifference in path length,L.

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vs LRemember that phase is kx t + 0.

Were told the speakers are emitting in phase, so

their wavefunctions have the same 0 values.

The mic is listening to both speakers at the sametime, so their wavefunctions have the same t values.

The waves travelled different distances to reach themic, so their wavefunctions have differentxvalues!

x1 = L1 andx2 = L2.

Use this to relate the phase difference to thepath differenceL.

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Difference in phaseat the microphone

Difference innumber of cycles

Radiansper cycle

Remember, this equation assumes:

the waves were created in phase, and that

nothing happened to them on the way.79

Example: Pirates vs NinjasTwo Ninjas board a pirates shipand head toward the mast whentheyre spotted by a Pirate . Theyboth attack with sound beam guns

at the same time. The frequencyof the beams is 940 Hz.

Does the Pirate experience thebeams as constructive or destructiveinterference, or somewhere inbetween?

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Pirate

Ninja Ninjamast

1.00 m 3.00 m

5.00 msound

beam

soundbeam

Speed of sound = 344 m/s.


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Beats

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Beats in sound are one particular case of superposition. When two

periodic waves of very close (but not equal) wavelengths overlap, thecombined wave will pulse, or beat.

Combinedwaveform

wave 1

wave 2

Consider two sound waves with fa > fb. Then Ta < Tb.Choose t = 0 at a point where the waves are in phase atthe place where were listening. (Call that positionx=0.)

The next time the two waves will be in phase will bewhen wave a has gone through exactlyone more cyclethan wave b. The phase difference at this point will haveincreased by 2. Call this time t = Tbeat, the beatperiod.

Let n = number of cycles wave a goes through in thistime; then wave b goes through (n "1) cycles.! Tbeat = nTa! for wave a" Tbeat = (n "1)Tb! for wave b

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Solve one equation for n and plug it into the other, andrearrange. (We want to get rid ifn, but keep Tbeat.)We get:

Since f= 1/T, flip this over:

This started with the definition fa > fb. In general:

! fbeat = |fa "fb|So the beat frequency (the frequency of pulses) is justthe difference in the two original frequencies.

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what constructive interference and destructiveinterference are.

how you can determine what will happen when twowaves interfere by looking at the relative phase.

what happens when two sounds with very similarfrequencies interfere.

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So now you know:


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I think most of you are familiar with the Doppler effect.Think of an engine or a siren going by. (Vreeeee-whooooooom...)

Lets figure out how that works.First, a simulation:http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/doppler.htm

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The Doppler Effect Doppler: Stationary listener

Doppler: Stationary ListenerHow do we determine the frequency of sound thelistener hears, when the source is moving at speed vs?(Note that it will be different in front of or behind the

moving source.)The speed of soundis whatever it is in the medium(e.g. air), regardless of how the source is moving.

The time to generate one cycle of sound is the period(by definition): Ts = 1/fs. This is the time that goes byafter one crest is generated before the next onecomes out.

During this time, the wave move a distance vTs, andthe source moves a distance vsTs.

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The wavelength is the distance between crests in asnapshot. So its the distance to the last crest when thenext crest isgenerated.

Then the wavelength is the distance travelled by the firstcrest plus the distance the source moved beforeproducing the next crest.

In front: = vTs vsTs = (vvs)Ts = (vvs)/fs.

Behind the source: = vTs + vsTs = (v+vs)Ts = (v+vs)/fs.88

ping

v

v vs

ping

v vsvsTsvTs

Time t = 0

Time t = Ts


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To find out what frequencythe listener hears, look atwhat happens to the period:

TL, the period of the waves heard by the listener, is thetime it takes two successive crests to pass the listener(regardless of where they came from). After one crest

goes by, the wave has to travel a distance of for thenext crest to reach the listener.

The waves speed is v. Then TL = /v. That means

fL = v/, and:

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behind the source in front of the source

If the listener is moving in the same direction as thesource, then the speed at which the waves areapproaching the listener is v + vL.

Then the time it takes for the listener to hear twosuccessive crests is going to be TL = /(v + vL).

Plug in lambda from before ((v+vs)/fs or (vvs)/fs) to getthe general Doppler shift equation, for either a movingsource or a moving listener or both:

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Doppler: Moving Listener

The Doppler Effectfor a moving source sand/or a moving listener L.v = speed of sound.

e.g.: Car chases

S L

S L

S L

S L

police youExample: Diana, Duck of Science

Diana, Duck of Science!, fires herassistant Bob the Thrillseeking Catout of a cannon at 30 m/s.

At what frequency does Dianahear Bobs 1000 Hz meow, beforeand after Bob has passed her?

If Diana sends a 10,000 Hz soundpulse at Bob after hes passed her,at what frequency does she hearthe reflection?

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Quark!

!


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Sonic BoomAs we saw in the Doppler Effectsim, the faster an object is movingthe more the waves in front of itbunch up.

It takes an increasing amount of

force to compress the air likethat, the faster the plane goes; thisis the sound barrier.

Once the object is moving fasterthan the speed of sound, itsoutpacing the sound waves; eachwave is generated outside theprevious one.

Waves pile up, and the result is ashock wave.

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Navy Lt. Ron Candiloro's F/A-18 HornetBreaking the Sound Barrier

http://www.defenselink.mil/news/newsarticle.aspx?id=43041

Recall that the wavelength of the sound in front of theobject is given by:

When vs = v, the wave length is zero "representing thewaves piling up on top of each other, as we discussed.

When vs > v, this equation is no longer meaningful!

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The sound created atpoint S1 expands in alldirections by a distancevt in time t. In thattime the object moves

forward by a distancevst.

Then, from the diagram, the angle between theshockwave and the direction of motion is given by:

95shock wave angle

Mach number:or

what the Doppler Effect is, and what causes it.

what the relationship is between the frequency of

sound produced by some source and the frequencydetected by some listener, when one or both ismoving.

what causes a sonic boom.

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So now you know:

chap17 physics unit 1 smh-waves i - waves ii

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