4: Sound Waves (Chapter 16) Phys130, A04 Dr. Robert MacDonald On Sound Sound is any longitudinal (“compression”) wave in a tangible medium, like air, wood, rock, the sun, etc. • At least some aspects of an earthquake are essentially huge sounds. The audible range of sound for humans is about 20–20,000 Hz. Sounds below this range are called infrasonic; above this range they’re called ultrasonic. • Elephants use low frequency (13–35 Hz) sounds to communicate with each other through the earth, over distances of more than 2.5 km. They can find each other, converse, even recognize each other. 2 http://dsc.discovery.com/news/2008/09/30/elephant-communication.html Wavefunction of Sound Sound waves generally travel out in all directions from a source. But for now, we’ll focus on sound waves travelling in a straight line in one direction, which we’ll call the positive x axis (since we can). This is the same wave function we’ve been working with up till now: y(x, t) = A cos(kx – ωt). (The phase constant ϕ 0 isn’t important for now, so we’ll set that to zero to keep things tidy.) • Here I’m using y to represent the displacement of the particles along the direction of wave motion. This is different than on a string, but the principles are the same. 3 Displacement vs Pressure In addition to talking about sound as a wave of particle displacements, 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 it’s how we hear. The eardrum has air on both sides. • Inside your head the eardrum is vented by the Eustachian tube, so it’s always at atmospheric pressure (unless the tube is plugged!). • A sound wave changes the pressure up and down on the outside of the eardrum. The difference moves the eardrum back and forth. 4
Transcript
4: Sound Waves(Chapter 16)
Phys130, A04Dr. Robert MacDonald
On SoundSound is any longitudinal (“compression”) wave in a tangible medium, like air, wood, rock, the sun, etc.
• At least some aspects of an earthquake are essentially huge sounds.
The audible range of sound for humans is about 20–20,000 Hz. Sounds below this range are called infrasonic; above this range they’re called ultrasonic.
• Elephants use low frequency (13–35 Hz) sounds to communicate with each other through the earth, over distances of more than 2.5 km. They can find each other, converse, even recognize each other.
Wavefunction of SoundSound waves generally travel out in all directions from a source. But for now, we’ll focus on sound waves travelling in a straight line in one direction, which we’ll call the positive x axis (since we can).
This is the same wave function we’ve been working with up till now: y(x, t) = A cos(kx – ωt). (The phase constant ϕ0 isn’t important for now, so we’ll set that to zero to keep things tidy.)
• Here I’m using y to represent the displacement of the particles along the direction of wave motion. This is different than on a string, but the principles are the same.
3
Displacement vs PressureIn addition to talking about sound as a wave of particle displacements, 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 it’s how we hear. The eardrum has air on both sides.
• Inside your head the eardrum is vented by the Eustachian tube, so it’s always at atmospheric pressure (unless the tube is plugged!).
• A sound wave changes the pressure up and down on the outside of the eardrum. The difference moves the eardrum back and forth.
4
Since what the wave causes is these variations in pressure, that’s what we’ll use to describe sound — the difference from atmospheric pressure.
Recall how a “high-then-low” longitudinal displacement graph results in particles bunching and spreading. Let’s look at the sinusoidal version of that and figure out how to describe what the pressure’s doing.
When a sound wave passed through, each end of this cylinder will be displaced according to the wavefunction y. The displacement of the left end at time t will be y(x, t); the displacement of the right end will be y(x+Δx, t).
O x x+Δx O x x+Δx
y(x, t) < y(x+Δx, t) y(x, t) > y(x+Δx, t)
>Δx <Δx
8
Displacement → PressureSo the difference in the displacements in either end of our (tiny, imaginary) cylinder results in a change in the length of the cylinder — and a change in the volume of the cylinder, since the area of the ends hasn’t changed.
The change in the volume is the area of the end times the change in the length:! ΔV = S [y(x+Δx, t) - y(x, t)]
Let’s get rid of S by looking at the fractional change in volume:
That last version becomes a derivative if Δx is very small (i.e. the limit Δx→0):
The change in volume results in a change in pressure (you’re compressing or expanding the gas). The amount of 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 = ΔP), then
9
See Section 12.7 formore details about B.
Since we know the wavefunction, y(x, t) = A cos(kx - ωt), we can evaluate ∂y/∂x and determine a formula for p(x,t):
where k is the wavenumber, B is the bulk modulus, A is the displacement amplitude, and ω is the angular frequency.
• More amplitude means more pressure change.
• More bulk modulus means it’s harder to compress the gas, so for a given amplitude you get more pressure change.
• Shorter wavelength means more pressure change (!).10
Difference from atmospheric pressure in a sinusoidal sound wave.
Max difference from atmospheric pressure in a sinusoidal sound wave.
Perception of SoundLoudness depends on the amplitude of the sound wave. But the perceived loudness varies from person to person!
•We don’t have a uniform frequency response — some frequencies sound louder than others, even at the same amplitude. The details vary from person to person.
•We lose sensitivity to sound over time, especially at the higher frequencies.
• Loud sounds damage hearing, too — look at rock musicians and orchestra members.
Pitch is of course mainly dependent on the (fundamental) frequency of the sound wave. But there are a number of “auditory illusions” that make a pitch sound higher or lower. For example, if you hear two tones at the same frequency but different amplitudes, the louder one will tend to sound a little lower.
Timbre (aka, tone colour or quality), the characteristic sound of an instrument, comes from the particular combination of sine waves of various frequencies and amplitudes that make up its sound — that is, its harmonic content. These waves come from the air or string as well as from the instrument body.
From The Physics Hypertextbookhttp://physics.info/music/
15
Noise is a combination of a range of frequencies, not just harmonics.
• “White noise” is a uniform combination of all frequencies of sound (within some range, of course).
• This is why instruments that are out of tune with each other sound so bad: the combination isn’t harmonic but noisy.
We have an expression for the speed of a wave on a string:
What does the speed of a wave (sound) in a fluid depend on?
• We can expect it has something to do with how difficult it is to compress the fluid; this is described by the bulk modulus.
• It’s probably related to how hard it is to get the molecules of gas moving — their mass, or (similar to the string) their mass density.
We can use the wave equation to find the speed.16
Speed of Sound in a Fluid
We’ll start by looking at the speed of sound through a pipe.
• This is a relevant example, since it describes wind instruments and the human voice.
We’ll ignore any effects of the walls of the pipe (like friction); as long as the pipe is reasonably big it shouldn’t have much effect. (Sound through a small tube, it turns out, is slowed down by the walls. This is called the “tube effect.”)
All we care about is that the fluid can’t expand laterally (i.e. can’t squish out the sides of the pipe).
Remember our three ingredients for the wave equation: a restoring force, Newton II (F = ma), and linearization.
17
ρ is the mass density —"the mass per unit volume (usually measured in kg/m3). It’s how “heavy” the air or water etc is.
We can describe the sound by the way it changes the fluid’s pressure as it passes through, like before. The pressure wave p(x,t) gives the amount that the pressure at point x is increased or decreased by the sound wave, at time t.
18
p(x,t) p(x+Δx,t)
Δx A = areaA = areaVolume of pipe segment:! V = AΔxMass of fluid in pipe segment:! m = ρV = ρ A Δx
Greek letter “rho”
x axis
pipe
Remember that pressure is a force distributed over an area (“force per unit area”). For example, a larger parachute 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 the left end (pushing right) is F(x,t) = A p(x,t), and on the right end (pushing left) is F(x+Δx,t) = A p(x+Δx,t).
The net force on the segment of fluid is just the difference of these (remember right = positive):
19
We can put this information into F = ma, as a step towards relating what we know to the Wave Equation.
So we can cancel out A; the speed of sound doesn’t depend on the size of the pipe!
We found previously that which gives us:
20
Net Force massacceleration
That last equation should look vaguely familiar from the derivation of wave speed on a string. Let’s rearrange it a bit:
The left hand side is just the x-derivative of the slope — or the second x-derivative of the position — if Δx is very small. So we put that in:
Huzzah! The Wave Equation! This means that ρ/B=1/v2, or:
21
Speed of sound in a fluid.B is the bulk modulus.ρ is the mass density.
The speed of sound in a solid is more complex than we’re going to deal with here. It’s complicated by the way the surrounding material keeps the wave from squishing the material out sideways.
It depends on the density, the bulk modulus (how hard it is to compress), and the shear modulus (how hard it is to shear the material), and sometimes some other things.
All this adds complexity without adding much more than a few formulas. (If you’re curious, though, it’s in your textbook...) We’ll focus on fluids here.
22
Speed of Sound in a Solid
Speed of Sound in a GasA gas is of course just a type of fluid, but it’s interesting to study specifically. For any fluid the speed of sound is given by:
In a gas, though, the bulk modulus B depends on the pressure of the gas — which can be changed quite easily. So, obviously, does the mass density ρ.
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 (0ºC):! 331 m/s
• Air (20ºC):! 343 m/s
• Helium (0ºC):! 965 m/s
• Water (20ºC):! 1482 m/s
• Lead! 1960 m/s
• Copper! 5010 m/s
• Glass (Pyrex)! 5640 m/s
24
Speeds of Sound
Compare the wavelength of sonar used by bats (in air, say at 20°C) to that used by dolphins. Assume they both use a frequency of 100 kHz = 1.00x105 Hz. Any required constants will be supplied.
• Start by looking for a connection between information we have — frequency —"and information we’re looking for — wavelengths. This is a simple one: v = λf for any wave, so we’ll start with that: λ = v/f.
• Frequency is the same in both situations, but the speed of sound changes.
25
Echolocation: Bat vs Dolphin Dolphins in water:
For water, the density is 1.00 g/cm3 = 1000 kg/m3, and the textbook says B = 2.18x109 Pa. This gives v = 1476 m/s.
Then the wavelength is λ = v/f = (1476m/s)/(1.00x105Hz) or λ = 0.0148 m.
This is independent of temperature, aside from slight changes in the density of water.
26
Bats in air:
The bulk modulus of air at 20°C is about 1.42x105 Pa.The density of air at 20°C is about 1.204 kg/m3.
This gives v = 343 m/s, and λ = v/f = 0.00343 m.
• Dolphins in water: λ = 1.48 cm
• Bats in air: 0.34 cm
Bats can distinguish much smaller objects.(This makes sense — they hunt bugs!)
27
• how the displacement of molecules in a sound wave results in a change in pressure.
• how to describe a sound wave in terms of either displacement or pressure.
• what determines the speed of sound in a liquid or a gas.
28
So now you know:
29
Wave IntensityIn one-dimensional waves (e.g. waves on a string), all of the energy that enters the string at one end makes it to the other end (aside from energy lost to damping).
In three-dimensional waves (e.g. sound, light), the energy spreads out in more than one direction.
The intensity (I) of a (3D) wave is the average rate at which energy is transported across some unit of area. In other words, it’s the average power per unit area.
• Written mathematically, intensity is defined as:! I = P/A.
Consider a spherical wave — a wave which spreads out evenly in all directions. This is the kind of wave you get from a “point” source of light (bulb, candle, star), sound, etc.
At a distance r from the source, the energy of the wave is spread over the surface of a sphere of area 4πr2. So the intensity is given by:
The way intensity drops with the square of distance is called the inverse square law.
Spherical Symmetry
32
Power coming from the source
Surface area of a sphere
Intensity of sound inspherical waves.
Inverse Square LawThe Inverse Square Law describes how intensity changes as a function of distance.
Consider two spheres of radius r1 and r2 centred on the wave source. As long as the medium doesn’t absorb any energy as the wave travels between the spheres (i.e. no damping), the total power P going through each sphere should be the same. Then
Equate these and rearrange and we find:
33
and
Inverse square lawfor intensity of spherical waves.
Example: Solar PanelsConsider a 1 m2 solar panel in use at the equator (so we don’t have to worry about seasons).
The Earth is closest to the Sun in January, at a distance of r1 = 1.47x108 km.
The Earth is furthest from the Sun in July, at a distance of r2 = 1.52x108 km.
If the solar panel generates 200 W of power in January (P1), how much power does it generate in July (P2)?
34
200 W
1.47x108 km
1.52x108 km
36
Intensity and AreaIntensity is defined as power per unit area —"that is, the amount of energy passing through some surface every second (or whatever).
For a spherical wave, at a distance r from the source the original power is distributed over the surface of a sphere of radius r. So the intensity of the wave at this distance is given by I = P ⁄ 4πr2.
If you’re standing at a distance r and hold up a sheet with area A, the amount of power striking that sheet is given by I A — the intensity times the area of the sheet. (Not the area of the imaginary sphere.)
Not all waves are spherical! The formula for intensity depends entirely on the shape of the waves. The intensity of a laser, for example — the amount of energy going into that little dot of light per second — is almost independent of distance r.
So we need to know the shape of the wave in order to figure out the intensity at a distance r from the source.
But if you want to know how much power is striking your photosensor, microphone, etc, it doesn’t matter what the shape of the wave was. All that matters is the intensity of the wave and how big your sensor is. P = I A is always true —"it’s the definition of intensity!
37
Shaping SoundYour voice can be reasonably modelled as a point source radiating in all directions. But if you cup your hands, the sound that would have gone to the sides also gets reflected forward. The result is that the sound doesn't spread out as much, so it decreases more slowly than 1/r2.
Reflections from walls mean that the inverse square law is blown away indoors.
• In fact, theatres (including home theatres!) and concert halls should be carefully designed so you don’t end up with weird dead spots or other effects.
38
Winspear Centre
The Winspear Centre downtown has fantastic acoustics —"there are no bad seats! The whole place was designed around how sound is transported, reflected, focussed, and mixed. No inverse square law!
39
e.g. Intense ConcertDuring a particularly thrilling part of an Edmonton Symphony concert, the sound intensity reaching your eardrum 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 reaching your ear?
40
Examples of Sound IntensityPower Intensity 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 waves and the inverse square law.
Intensity vs Amplitude
42
We should expect the intensity of a sound wave to be related to the displacement amplitude, or equivalently to the pressure amplitude. Let’s figure out what the relationship 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, so we can use this to relate intensity to amplitude.
So if we want the velocity of the air, caused by the change in pressure from the sound wave, look to the wave equation: y(x, t) = A cos(kx – ωt). The derivative∂y/∂t gives the particle velocity we’re looking for.
43
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. We’re almost never interested in this; rather, when we say “intensity” we generally mean the time-averaged intensity.
Remember that the average of sin2 over a period is 1/2.
Then the time average of I(x,t) = BωkA2sin2(kx–ωt) is
44
“rho”
Average intensity of asinusoidal sound wavein a fluid.
The Decibel ScaleHuman sight and hearing are logarithmic. The perceived “brightness” or “loudness” of something goes like the logarithm of the intensity.
So for sound they developed something called the sound intensity level (not to be confused with “sound intensity”!). It’s represented by a β (Greek letter “beta”), and is defined as:
It’s measured in “decibels” or dB. One dB is 1/10 of a “bel”. We pretty much always use decibels, though.
46
sound intensity
reference intensity:10-12 W/m2
base 10 logarithmnot 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 human ear isn’t uniform, so 65 dB will sound louder at some frequencies than at others. There are specialized sound scales and meters that adjust the numbers so that a particular sound level will sound the same at all frequencies.
• The “dBA” scale is one of these. So 65 dBA sounds the same at all frequencies, but high and low frequencies will carry a lot more power.
47
Examples of Sound IntensityIntensity
at your earIntensity 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
It’s often useful to use other reference intensities for various purposes.
Amplifiers and attenuators will sometimes list how much they’re 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 signal from the CD or whatever.
• 0 dB on this scale is an unattenuated signal. (Positive dB means you’re actually amplifying it.)
Decibels and Attenuation
49
Consider an “ideal” bird —"a point source. The bird’s sound will be radiating equally in all directions, in a spherical or hemispherical way, so it follows an inverse square law.
If you go three times farther away from the bird, how much does the sound intensity level of the birdsong change?
How much does the pressure amplitude change?
50
Example: Chirping Bird
Starting with the sound intensity level, first let’s write down the difference:
So it’s basically the usual definition of sound intensity level, but using the original intensity as the reference. (In other words, decibels add!)
Now we’ll apply the inverse square law:
51
So moving three times farther away from the bird reduces the sound intensity level of the song by almost 10"dB.
What does this do to the pressure amplitude?
52
We’re looking for pressure amplitude, and we have the change in sound intensity level. This is related to the sound intensity, which is related to the pressure amplitude by the formula:
Since we’re looking for the relative change in pressure amplitude we don’t actually need to know the density"ρ or the bulk modulus B. (I’ll call the pressure amplitude p here instead of pmax to tidy it up a bit.)
These cancel out;it’s the same air!
A (very large!) electric spark jumps along a straight line of length L = 10 m, and produces a noise with acoustic power of Ps = 1.6x104 W.
The bang travels radially outward from the spark; the wave looks like an expanding cylinder.
• What is the intensity I of the sound at a distance of r"="12"m from the spark?
• At what rate Pd does sound energy reach an acoustic detector of area Ad = 2.0 cm2, aimed at the spark and located 12"m away from it?
54
e.g. Cylindrical sound wave
Consider an imaginary cylinder of radius 12"m and height 10 m, with no end caps.
All of the acoustic energy that leaves the spark must pass through this cylinder, at the same (total!) rate.
• So the total power passing through the cylinder is the same as the total power coming from the source.
55
That energy is uniformly distributed over the whole cylinder — by symmetry.
So the intensity at any point on this cylinder is the total power divided by the cylinder’s area (ignoring the ends!).
Total power passing through the cylinder = Ps
Total area of the cylinder sides = 2π•r•L
Sound intensity at 12 m from the spark:! I = P/A (definition of intensity!)
I = Ps / (2π•r•L)
! = (1.6x104 W) / (2π•(12 m)•(10 m))
! = 21.2 W/m2.
56
So that’s the intensity of sound at the surface of our acoustic detector.
How much acoustic power is the detector receiving?
From the definition of intensity: P = I•A.
Area of the detector = Ad = 2.0 cm2 = 2.0x10–4 m2.
So the detector receives a power of:! Pd = (21.2 W/m2)•(2.0x10–4 m2) ≈ 4.2x10–3 Wor 4.2 mW.
57
• what “intensity” means (definition: I = P/A).
• how the intensity of a wave can change with distance, what the “inverse square law” is, and when it applies.
• how the intensity of a sound wave is related to its displacement amplitude and pressure amplitude.
• what the “decibel” scale is, and how it measures relative intensity.
• That is, the intensity of a wave relative to some reference intensity.
58
So now you know:
The Treachery of Images, René Magritte, c. 1928
When a wave reaches the end of a pipe, whether the end 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 or fixed end.
The incoming and reflected wave interfere (their displacements add), and just as with strings you get a standing wave.
We can describe the standing wave in terms of displacement or pressure, just like with other sound waves.
60
Standing Waves in a Pipe
LongitudinalStanding Wave
Same wavefunction as a transverse standing wave!displacement y of the particle at position x, at time t:! y(x,t) = A sin(kx) sin(ωt)But now the displacement y is along the same direction as the wave motion.Magnutude of pressure change p is largest where displacement y is 0, and p = 0 where |y| is max.
So a good pressure wavefunction is:! p(x,t) = pmax cos(kx) sin(ωt)
Waves and Pipe EndsClosed 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.
62
Wind Instruments
Usually have at least one end open, the other either open or closed.
Generate “noise” by buzzing lips or a reed, or blowing air past a “mouth” (e.g. recorder, pipe organ).
• Noise is a continuous combination of many frequencies!
Frequencies matching the pipe harmonics resonate and are amplified.
63
Davis Organ @ Winspear
Standing Waves in a Pipe
Displacement envelopes shown.
Two Open Ends
66
This is basically the opposite of the “clamped” string, which had displacement nodes at each end. With two open ends we have displacement antinodes and pressure nodes.
The longest wavelength that can produce a standing wave is still λ1=2L, just like with a clamped string.
The situation is basically the same as before, so the formulas turn out the same. (Remember v = λf.)
Allowed standing waves in a pipe with two open ends.
One Closed End (“Stopped”)A pipe’s open end has a displacement antinode. The closed end has a displacement node. This is a little different, obviously.
The distance between a node and its nearest antinode is only λ/4. So the longest wavelength you can get in a “stopped” pipe is λ1=4L — twice as long as before (so twice as low a note!).
The next harmonic will occur when we’ve squeezed in another half cycle, so λ = 4L/3 = λ1/3. After that we’ll get λ = 4L/5, etc...
— only odd harmonics allowed!
67
Allowed standing waves in a pipe with one open end.
Frequency and TemperatureThe wavelengths of standing waves possible in a pipe are determined by the length of the pipe.
Since v = λf (for any wave, remember!) this means the frequency will depend on the speed of sound in the pipe — and that depends on air temperature!
As the temperature changes, the tuning of the instrument changes. (As you may know if you’ve ever played music outdoors.)
This can be a big problem for large pipe organs; often some pipes are warmer than others.
68
Resonance and SoundWe studied forced oscillations previously. It works the same way with sound in a pipe as it does with a simple harmonic oscillator.
If you generate a sound with some frequency near the pipe, the molecules of air in the pipe will oscillate with that frequency (just like the rest of the air). That’s all there is to forced oscillations and sound.
Just like before, if you drive the molecules of air at a frequency they’d like to move at anyway, the oscillations build up, the amplitude increases, and you get resonance.
69
The simple harmonic oscillator has only one resonant frequency; if you let it go it will oscillate with that frequency.
But air in a pipe (or a string!) will be happy to oscillate in any of its harmonics.
So if you apply sound with the same frequency as one (or more!) of the pipe's harmonics the air in the pipe will resonate.
• As the sound reflects back and forth in the pipe, it’s reinforced and added to by the incoming sound.
• You may have already seen this in your labs.
70
e.g.: Sound resonanceYou pick up a cardboard tube of length L = 67.0 cm and hold it near your ear. Random background sounds from the room set up a standing wave in the tube at its fundamental frequency.
(Other standing waves, too, but the fundamental frequency is strongest.)
• Assuming the speed of sound is v = 343 m/s, what fundamental frequency do you hear from the tube?
• If you jam your ear against one end of the tube, what fundamental frequency do you hear from the tube?
71
A 3 m stopped organ pipe (one end open) generates sound in its first harmonic. The speed of sound in the pipe is 350 m/s.
A nearby piece of guitar string is excited into its third harmonic by the sound. The string has linear mass density 5 g/m and tension 50 N.
How long is the guitar string?
72
Example: guitar and pipe
So now you know:• what standing sound waves are possible in a pipe,
and how that depends on whether the pipe ends are open or closed.
• How the wavelength and frequency of a standing sound wave are related to the sound speed and the pipe length (and pipe ends).
• how sound waves can set up standing waves by resonance.
73
Interference of WavesInterference is the term for what happens when two or more waves overlap. Standing waves are a good example.
A different type of interference occurs when you have two or more similar waves travelling in the same direction, or spreading out together in space.
The waves can add constructively, or destructively. Which one you get depends on the relative phase of the two waves at the place you’re looking.
(Remember, the principle of superposition is the statement that when two waves overlap, they just add.)
74
Combinedwaveform
wave 1
wave 2
Waves exactly in phase∆ϕ = (2n)π, n = 0, ±1, ±2, ...
Waves exactly out of phase∆ϕ = (2n+1)π, n = 0, ±1, ±2, ...
At the place where we’re looking, the two waves have different phases (in general). ∆ϕ = difference in phase.
∆ϕ depends on:
• how the waves were created (different ϕ0 values)
• how far they travelled (difference in x/λ (or kx))
• what happened to them on the way (reflections etc)
∆ϕ can have any value! 75
Remember: phase = kx – ωt + ϕ0.
Some DemosSeveral animations showing superposition of 1-D waves, 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
Phase DifferenceThink of a microphone placed near a set of two speakers, producing sound like this:
• the same pure (single frequency) tone.
• produced in phase (simultaneous crests, etc).
The principle of superposition says that the sound wave at the microphone will be the sum of the waves from each source.
• Here we have identical waves, with different phases when they reach the mic.
How they add depends on the difference in phase ∆ϕ at the microphone.
• In this case, ∆ϕ depends only on the difference in path length, ∆L.
77
∆ϕ vs ∆LRemember that phase is kx – ωt + ϕ0.
• We’re told the speakers are emitting in phase, so their wavefunctions have the same ϕ0 values.
• The mic is listening to both speakers at the same time, so their wavefunctions have the same t values.
• The waves travelled different distances to reach the mic, so their wavefunctions have different x values!
• x1 = L1 and x2 = L2.
Use this to relate the phase difference ∆ϕ to the path difference ∆L.
78
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 pirate’s ship and head toward the mast when they’re spotted by a Pirate. They both attack with sound beam guns at the same time. The frequency of the beams is 940 Hz.
Does the Pirate experience the beams as constructive or destructive interference, or somewhere in between?
80
Pirate
Ninja Ninjamast
1.00 m 3.00 m
5.00 msoun
d be
am
sound beam
Speed of sound = 344 m/s.
Beats
81
Beats in sound are one particular case of superposition. When two periodic waves of very close (but not equal) wavelengths overlap, the combined 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 at the place where we’re listening. (Call that position x = 0.)
The next time the two waves will be in phase will be when wave a has gone through exactly one more cycle than wave b. The phase difference at this point will have increased by 2π. Call this time t = Tbeat, the “beat period”.
Let n = number of cycles wave a goes through in this time; then wave b goes through (n –"1) cycles.
! Tbeat = nTa! for wave a" Tbeat = (n –"1)Tb! for wave b
82
Solve one equation for n and plug it into the other, and rearrange. (We want to get rid if n, 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 just the difference in the two original frequencies.
83
• what “constructive interference” and “destructive interference” are.
• how you can determine what will happen when two waves interfere by looking at the relative phase.
• what happens when two sounds with very similar frequencies interfere.
84
So now you know:
I think most of you are familiar with the Doppler effect. Think of an engine or a siren going by. (“Vreeeee-whooooooom...”)
Let’s figure out how that works.
First, a simulation:http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/doppler.htm
85
The Doppler Effect Doppler: Stationary listener
Doppler: Stationary ListenerHow do we determine the frequency of sound the listener 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 sound is 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 by after one crest is generated before the next one comes out.During this time, the wave move a distance vTs, and the source moves a distance vsTs.
87
The wavelength is the distance between crests in a snapshot. So it’s the distance to the last crest when the next crest is generated.
Then the wavelength is the distance travelled by the first crest plus the distance the source moved before producing the next crest.
To find out what frequency the listener hears, look at what happens to the period:
TL, the period of the waves heard by the listener, is the time 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 λ for the next crest to reach the listener.
The wave’s speed is v. Then TL = λ/v. That means fL = v/λ, and:
89
behind the source in front of the source
If the listener is moving in the same direction as the source, then the speed at which the waves are approaching the listener is v + vL.
Then the time it takes for the listener to hear two successive crests is going to be TL = λ/(v + vL).
Plug in lambda from before ((v+vs)/fs or (v–vs)/fs) to get the general Doppler shift equation, for either a moving source or a moving listener or both:
90
Doppler: Moving Listener
The Doppler Effect for 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 her assistant Bob the Thrillseeking Cat out of a cannon at 30 m/s.
At what frequency does Diana hear Bob’s 1000 Hz meow, before and after Bob has passed her?
If Diana sends a 10,000 Hz sound pulse at Bob after he’s passed her, at what frequency does she hear the reflection?
92
Quark!
μ!
Sonic BoomAs we saw in the Doppler Effect sim, the faster an object is moving the more the waves in front of it bunch up.
It takes an increasing amount of force to compress the air like that, the faster the plane goes; this is the “sound barrier”.
Once the object is moving faster than the speed of sound, it’s outpacing the sound waves; each wave is generated outside the previous one.
Waves pile up, and the result is a shock wave.
93
Navy Lt. Ron Candiloro's F/A-18 HornetBreaking the “Sound Barrier”
Recall that the wavelength of the sound in front of the object is given by:
When vs = v, the wave length is zero —"representing the waves piling up on top of each other, as we discussed. When vs > v, this equation is no longer meaningful!
94
The sound created at point S1 expands in all directions by a distance vt in time t. In that time the object moves forward by a distance vst.
Then, from the diagram, the angle α between the shockwave and the direction of motion is given by:
95
shock 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 frequency detected by some listener, when one or both is moving.
