On Sound
4: Sound Waves
(Chapter 16)
Phys130, A04
Dr. Robert MacDonald
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.
http://dsc.discovery.com/news/2008/09/30/elephant-communication.html
2
Wavefunction of Sound
Displacement vs Pressure
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).
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 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.)
This is a useful description since it’s how we hear. The
eardrum has air on both sides.
• 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
•
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.
http://www.animations.physics.unsw.edu.au/jw/sound-pressure-density.htm
5
Surface area S
Undisturbed cylinder
of air:
Δx
O
x
x+Δx
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).
y(x, t) < y(x+Δx, t)
y(x, t) > y(x+Δx, t)
Displacement → Pressure
So 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:
>Δx
O
x
x+Δx
<Δx
O
x
x+Δx
8
That last version becomes a derivative if Δx is very small
(i.e. the limit Δx→0):
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):
Difference from atmospheric
pressure in a sinusoidal sound
wave.
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
See Section 12.7 for
more details about B.
If we define p(x,t) as the difference between the pressure
in our cylinder and atmospheric pressure (i.e. p = ΔP),
then
9
where k is the wavenumber, B is the bulk modulus, A is the
displacement amplitude, and ω is the angular frequency.
Max difference from atmospheric
pressure in a sinusoidal sound wave.
•
•
More amplitude means more pressure change.
•
Shorter wavelength 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.
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Perception of Sound
Loudness 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
http://www.animations.physics.unsw.edu.au/jw/sound-pressure-density.htm
rock musicians and orchestra members.
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From The Physics Hypertextbook
http://physics.info/music/
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.
• The human voice works basically the same way.
http://phet.colorado.edu/simulations/sims.php?sim=Fourier_Making_Waves
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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.
Speed of Sound in a Fluid
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.
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16
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.
A = area
Δx
A = area
pipe
x axis
p(x,t)
p(x+Δx,t)
Volume of pipe segment:
! V = AΔx
Mass of fluid in pipe segment:
! m = ρV = ρ A Δx
Greek letter “rho”
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.
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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.
We can put this information into F = ma, as a step
towards relating what we know to the Wave Equation.
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).
So we can cancel out A; the speed of sound doesn’t
depend on the size of the pipe!
The net force on the segment of fluid is just the
difference of these (remember right = positive):
We found previously that
which gives us:
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mass
Net Force
20
acceleration
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:
Speed of sound in a fluid.
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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.
B is the bulk modulus.
ρ is the mass density.
Speed of Sound in a Gas
A 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.
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Speed of Sound in a Solid
22
Speeds of Sound
Here is the speed of sound in a few different media:
• Air (0ºC):!
• Air (20ºC):!
• Helium (0ºC):!
• Water (20ºC):!
• Lead!
• Copper!
• Glass (Pyrex)!
331 m/s
343 m/s
965 m/s
1482 m/s
1960 m/s
5010 m/s
5640 m/s
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Echolocation: Bat vs Dolphin
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.
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.
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So now you know:
Bats in air:
• how the displacement of molecules in a sound wave
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
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.
Bats can distinguish much smaller objects.
(This makes sense — they hunt bugs!)
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Wave Intensity
Britain’s “Listening Ears”
In 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.
http://www.ajg41.clara.co.uk/mirrors/dungeness.html
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Spherical Symmetry
200 foot
sound mirror
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: Power coming
from the source
Intensity of sound in
spherical waves.
30 foot
sound mirror
Surface area of a sphere
The way intensity drops with the square of distance is
called the inverse square law.
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Inverse Square Law
Example: Solar Panels
The Inverse Square Law describes how intensity
changes as a function of distance.
Consider a 1 m2 solar panel in use at the equator (so
we don’t have to worry about seasons).
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
and
Equate these and rearrange and we find:
The Earth is closest to the Sun in January, at a distance
of r1 = 1.47x108 km.
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Inverse square law
for intensity of
spherical waves.
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)?
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Intensity and Area
Intensity is defined as power per unit area —"that
is, the amount of energy passing through some surface
every second (or whatever).
1.47x108 km
200 W
1.52x108 km
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.)
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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!
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Winspear Centre
Shaping Sound
Your 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.
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e.g. Intense Concert
During 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?
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!
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Examples of Sound Intensity
Power
Threshold of hearing
Intensity at 1 m
away
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
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.
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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.
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
wave speed!
This is the instantaneous intensity.
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“rho”
Average intensity of a ρ = mass density
sinusoidal sound wave B = bulk modulus
ω = angular frequency
in a fluid.
A = displacement amplitude
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Intensity vs Pressure
Remember that the maximum pressure (the “pressure
amplitude”) is given by pmax"="BkA. Then A = pmax/Bk.
Use this in I!=!(1/2)BωkA2:
The Decibel Scale
Human 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:
sound intensity
not a “B”!
p2max
I= √
2 ρB
ρ = Greek letter “rho”
ω = Greek letter “omega”
Average intensity of a pmax = pressure amplitude
sinusoidal sound wave ρ = mass density
B = bulk modulus
in a fluid.
v = wave speed
It’s measured in “decibels” or dB. One dB is 1/10 of a
“bel”. We pretty much always use decibels, though.
45
The threshold of pain (1 W/m2) then corresponds to
(10"dB)log(100/10-12)"="120"dB.
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Examples of Sound Intensity
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.
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reference intensity:
10-12 W/m2
base 10 logarithm
Intensity
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
Decibels and Attenuation
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.
Example: Chirping Bird
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?
• 0 dB on this scale is an unattenuated signal.
(Positive dB means you’re actually amplifying it.)
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50
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!)
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?
Now we’ll apply the inverse square law:
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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!
e.g. Cylindrical sound wave
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 P
d 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
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.
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!).
55
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.
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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 W
or 4.2 mW.
So now you know:
• 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.
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Standing Waves in a Pipe
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.
The Treachery of Images, René Magritte, c. 1928
We can describe the standing wave in terms of
displacement or pressure, just like with other sound
waves.
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Longitudinal
Standing 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.
displacement nodes = pressure antinodes
pressure nodes = displacement antinodes
So a good pressure wavefunction is:
! p(x,t) = pmax cos(kx) sin(ωt)
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.
•
Wind
Instruments
Davis Organ @ Winspear
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).
•
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Noise is a continuous combination
of many frequencies!
Frequencies matching the pipe
harmonics resonate and are amplified.
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Standing Waves in a Pipe
Two Open Ends
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.
Displacement
envelopes shown.
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.
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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!
Frequency and Temperature
The 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.)
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Allowed standing
waves in a pipe
with one open
end.
This can be a big problem for large pipe organs; often
some pipes are warmer than others.
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Resonance and Sound
We 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.
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.
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e.g.: Sound resonance
Example: guitar and pipe
You 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.
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.
(Other standing waves, too, but the fundamental
frequency is strongest.)
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.
• Assuming the speed of sound is v = 343 m/s, what
How long is the guitar string?
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?
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So now you know:
Interference of Waves
• what standing sound waves are possible in a pipe,
Interference is the term for what happens when two
or more waves overlap. Standing waves are a good
example.
• How the wavelength and frequency of a standing
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.
• how sound waves can set up standing waves by
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.
and how that depends on whether the pipe ends are
open or closed.
sound wave are related to the sound speed and the
pipe length (and pipe ends).
resonance.
(Remember, the principle of superposition is the
statement that when two waves overlap, they just add.)
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Some Demos
Combined
waveform
Several animations showing superposition of 1-D
waves, including beats, standing waves, and more:
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.
Remember: phase = kx – ωt + ϕ0.
∆ϕ depends on:
• how the waves were created (different ϕ values)
• how far they travelled (difference in x/λ (or kx))
• what happened to them on the way (reflections etc)
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
0
∆ϕ can have any value!
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∆ϕ vs ∆L
Phase Difference
Think of a microphone placed near a set
of two speakers, producing sound like this:
Remember that phase is kx – ωt + ϕ0.
• We’re told the speakers are emitting in phase, so
•
•
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.
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!
•x
1
= L1 and x2 = L2.
Use this to relate the phase difference ∆ϕ to the
path difference ∆L.
In this case, ∆ϕ depends only on the
difference in path length, ∆L.
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Example: Pirates vs Ninjas
Two 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.
beam
soun
d
1.00 m
Remember, this equation assumes:
the waves were created in phase, and that
nothing happened to them on the way.
•
•
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5.00 m
Ninja
mast
m
bea
Difference in
number of cycles
Radians
per cycle
Does the Pirate experience the
beams as constructive or destructive
interference, or somewhere in
between?
nd
sou
Difference in phase
at the microphone
Pirate
3.00 m
Ninja
Speed of sound = 344 m/s.
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Beats
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”.
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”.
wave 1
wave 2
Let n = number of cycles wave a goes through in this
time; then wave b goes through (n –"1) cycles.
Combined
waveform
!
"
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, and
rearrange. (We want to get rid if n, but keep Tbeat.)
We get:
So now you know:
• what “constructive interference” and “destructive
interference” are.
Since f = 1/T, flip this over:
• 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
This started with the definition fa > fb. In general:
!
frequencies interfere.
fbeat = |fa –"fb|
So the “beat frequency” (the frequency of pulses) is just
the difference in the two original frequencies.
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The Doppler Effect
Doppler: Stationary listener
I think most of you are familiar with the Doppler effect.
Think of an engine or a siren going by. (“Vreeeeewhooooooom...”)
Let’s figure out how that works.
First, a simulation:
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/doppler.htm
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Doppler: Stationary Listener
How 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.
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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.
Time t = 0
v
vs
“ping”
vTs
vsTs
v
v
vs Time t = Ts
“ping”
λ
Then the wavelength is the distance travelled by the first
crest plus the distance the source moved before
producing the next crest.
In front: λ = vTs – vsTs = (v–vs)Ts = (v–vs)/fs.
Behind the source: λ = vTs + vsTs = (v+vs)Ts = (v+vs)/fs.
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To find out what frequency the listener hears, look at
what happens to the period:
Doppler: Moving Listener
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.
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.
The wave’s speed is v. Then TL = λ/v. That means
fL = v/λ, and:
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:
behind the source
Then the time it takes for the listener to hear two
successive crests is going to be TL = λ/(v + vL).
in front of the source
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e.g.: Car chases
police
S
S
you
L
L
S
L
S
L
The Doppler Effect
for a moving source s
and/or a moving listener L.
v = speed of sound.
Example: Diana, Duck of Science
Diana, Duck of Science!, fires her
assistant Bob the Thrillseeking Cat
out of a cannon at 30 m/s.
μ!
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?
Quark!
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Sonic Boom
As we saw in the Doppler Effect
sim, the faster an object is moving
the more the waves in front of it
bunch up.
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!
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.
Navy Lt. Ron Candiloro's F/A-18 Hornet
Breaking the “Sound Barrier”
http://www.defenselink.mil/news/newsarticle.aspx?id=43041
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So now you know:
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.
• 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.
Then, from the diagram, the angle α between the
shockwave and the direction of motion is given by:
or
• what causes a “sonic boom”.
Mach number:
shock wave angle
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