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Sound
What is sound?
How do we make sound?
Why does sound move that fast? What
parameters does the speed of sound depend
on?
How do we work with the pitch and the
volume of sound?
Sound: a form of energy
Sound is a form of energy that moves.
Is this energy carried by particles (that we
call phonons), or is it carried by waves?
[The fact that we can call particles of sound
“phonons” doesn’t necessarily mean that they
exist.]
Since we note that vibrations are involved in
sound, we can try the wave idea.
Sound: what does it go through?
Sound is transmitted through air.
Is sound transmitted through water (and other
liquids)? Can you pipe sound into a
swimming pool? YES! In the navy, they use
sound in sonar to listen for and find things.
Is sound transmitted through solids (like a
knock on a door)? YES! Geologists use this
to “look” for oil!
Is sound transmitted through vacuum? No!
Sound: what is waving?
In waves on a string, the pieces of the string
pulled each other via the tension in the
string.
In sound, the molecules of the gas, liquid, or
solid will pull on each other via the pressure
in the material.
What is waving (or oscillating)? Both the
pressure and the molecules’ positions!
Sound: a travelling wave
We have already considered waves on a string.
We were able to work with Newton’s 2nd law
to get a wave equation for this. Can we do the
same for sound?
YES! We use the fluid equivalent of Newton’s
Second Law to get a wave equation. With this we
have two adjustments: we need a Bulk Modulus (B in
Nt/m2) instead of a Tension (Nt), and we have a
volume density (r in kg/m3) instead of a linear density
(kg/m).
Speed of sound
Newton’s Second law in fluid form gives us
the “wave equation” for sound. From this,
we get for the molecular displacement, y:
y(x,t) = A sin(kx +/- wt)
where v = w/k = (2pf / [2p/l]) = lf
and from the wave equation, v = [B/r]1/2
(this is just like v = [T/m]1/2 for a string).
Volume and pitch
Note that the speed of sound depends on B and
r, and that it relates l and f. Thus, changing the
frequency does NOT change the speed, v; instead it
will change the wavelength. Changing the material
(changing B and/or r) will change the speed.
For sound, then, we see that the pitch is
related to frequency (f, or equivalently, l),
while the volume is related to the amplitude,
A.
Energy, Power and Intensity
In oscillations, we saw that the energy of a
mass (piece of the string) was related to w2A2.
The power (Energy/time) of the wave down
the string was related to w2A2v.
For sound, however, we need the idea of
power/area which we call Intensity.
Intensity
This intensity is also related like power:
I a w2A2v (here A is amplitude).
But as sound spreads out, the area for this
power increases, and so the Intensity falls
off. For a point source, the area for the
power is that of a sphere (4pr2). For a point
source of sound, this takes the form of an
inverse square law for I: I = P / 4pr2 .
Sound in air
For an ideal gas, the bulk modulus, B, is
simply equal to the pressure, P. Thus, the
speed of sound in air is: v = [B/r]1/2 =
[P/r]1/2 . But from the ideal gas law,
PV = nRT, P = nRT/V; by definition,
r = m/V. Thus, v = [P/r]1/2 = [nRT/m]1/2 .
We can replace the m/n (total mass per total
moles) by M, the molar mass):
v = [gRT/M]1/2 , where g = CP/CV = 1.4 for a
diatomic gas like air has to be introduced
due to thermodynamic considerations.
Sound in air
v = [gRT/M]1/2
For air, g = 1.4, R = 8.3 Joules/mole-K, T is
the temperature in Kelvin, and M (a mixture
of N2 and O2) is .029 kg/mole.
Thus at room temperature (75oF=24oC = 297 K),
v = [1.4 * 8.3 J/mole K * 297 K / .029 kg/mole]1/2
= 345 m/s = 770 mph.
At higher altitudes we have lower
temperatures and hence lower speeds.
Human Hearing: Pitch
A standard human ear can hear frequencies from
about 20 Hz to about 20,000 Hz. As you get
older, however, both ends tend to shrink
towards the middle. This will be demonstrated
during class, and you can hear for yourself
what the different frequencies sound like and
what your limits are.
Talking
How do we understand what people say? Does it
have to do with frequency or intensity?
Of course, we can talk loudly or softly, which
means we can talk with high or low intensity.
We can also sing our words at different pitches
(frequencies).
So what goes into talking?
Talking
Along the same lines: both a piano and a guitar
can play the same note, but we can tell whether
a piano or a guitar did play that note. What is
going on?
It turns out that both talking and musical
instruments are based on resonance: standing
waves are set up in the mouths of people and
in the instruments.
Resonance
We can have the same fundamental frequency
set up on a string [ #l/2 = L ] in both a guitar
and a piano. But this indicates that there are
several wavelengths that obey this. These
several wavelengths are called the harmonics,
with the longest wavelength (#=1) being the
fundamental (longest wavelength, shortest
frequency).
Harmonics
Although a guitar and a piano may have the
same fundamental frequency, the higher
harmonics may resonate differently on the
different instruments based on their shape.
In the same way, we can form different words at
the same fundamental frequency by changing
the shape of our mouth.
Fourier Analysis
It turns out that the ear is a great Fourier
Analyzer - that is, it can distinguish many
different frequencies in a sound. (The eye is
not like this at all!)
It is hard to make computers listen to and
understand speech because the computer has
to be taught how to Fourier Analyze the
sounds and interpret that analysis.
Human Hearing: Volume
The volume of sound is related to the intensity
but it is also related to frequency because the
efficiency of the ear is different for different
frequencies.
The ear hears frequencies of about 2,000 Hz
most efficiently, so intensities at this frequency
will sound louder than the same intensity at
much lower or higher frequencies.
Intensity:
2
W/m
The ear is a very sensitive energy receiving
device. It can hear sounds down as low as 1012 W/m2. Considering that the ear’s area is on
the order of 1 cm2 or 10-4 m2, that means the
ear can detect sound energy down to about 1016 Watts!
The ear starts to get damaged at sound levels that
approach 1 W/m2 .
From the lowest to the highest, this is a range of a
trillion (1012)!
Intensity: need for a new unit
Even though we can hear sound down to about
10-12 W/m2, we cannot really tell the difference
between a sound of 10-11 +/- 10-12 W/m2 .
The tremendous range we can hear combined
with the above fact leads us to try to get a more
reasonable intensity measure.
But how do we reduce a factor of 1012 down to
manageable size?
Intensity: decibel (dB)
One way to reduce an exponential is to take
its log: log10(1012) = 12
But this gives just 12 units for the range.
However, if we multiply this by 10, we get
120 units which is a nice range to have.
However, we need to take a log of a
dimensionless number. We solve this
problem by introducing this definition of the
decibel: I(dB) = 10*log10(I/Io) where Io is
the softest sound we can hear (10-12 W/m2) .
Examples
The weakest sound intensity we can hear is what
we define as Io. In decibels this becomes:
I(dB) = 10*log10(10-12 / 10-12) = 0 dB .
The loudest sound without damaging the ear is 1
W/m2, so in decibels this becomes:
I(dB) = 10*log10(1 / 10-12) = 120 dB .
Decibels
It turns out that human ears can tell if one
sound is louder than another only if the
intensity differs by about 1 dB. This does
indeed turn out to be a useful intensity
measure.
Another example: suppose one sound is 1 x 10-6
W/m2, and another sound is twice as intense at
2 x 10-6 W/m2. What is the difference in
decibels?
Decibels
Calculating for each:
I(dB) = 10*log10(1 x 10-6 / 10-12) = 60 dB
I(dB) = 10*log10(2 x 10-6 / 10-12) = 63 dB .
Notice that a sound twice as intense in W/m2
is always 3 dB louder!
This is the result of a property of logs:
If I2 is twice as intense as I1, then in terms of dB:
I2(dB) = 10*log10(2*I1) = 10*[log10(2)+log10(I1)]
= 10*[.3+log10(I1)] = 3dB+I1(dB)
Distance and loudness
For a point source, the intensity decreases as the
inverse square of the distance. Thus if a source
of sound is twice as far away, its intensity
should decrease by a factor of 22 or 4. How
much will its intensity measured in dB
decrease?
I(dB) = 10*log(1 x 10-6 / 10-12) = 60 dB , and
I(dB) = 10*log(1/4 x 10-6 / 10-12) = 54 dB.
(Notice that 4 is two 2’s, so the decrease is two
3dB’s for a total of 6 dB.)
The Doppler Effect
The Doppler Effect is explained nicely in the
Computer Homework program (Vol 4, #5)
entitled Waves and the Doppler Effect.
fR = fS*[(v +/- vR) / (v +/- vS)]
where speeds are relative to the air, not the
ground, and the +/- signs are determined by
directions (use common sense!).
Electromagnetic Waves
For waves on a string and sound waves, we can get a
wave equation from Newton’s Second Law that
predicts the speed of sound.
So far in Physics 251 we’ve talked about electric and
magnetic fields. Can the E & M fields “wave” ?
If the charges or currents oscillate, the fields should then
oscillate. Will this oscillation propagate through
space as a wave? If so, where do we start to try to get
a wave equation for the fields? Will this predict the
speed of these E&M waves?
Electromagnetic Waves
The basic equations for electric and magnetic
fields are the basic four equations we’ve
dealt with in this course:
Gauss’s Law for Electric Fields
Gauss’s Law for Magnetic Fields
Ampere’s Law
Faraday’s Law
Together these four laws are called
Maxwell’s Equations .
Maxwell’s Equations
Gauss’ Law for Electric fields
closed area E  dA = Qenclosed / eo
Gauss’ Law for Magnetic fields
closed area B  dA = 0
Ampere’s Law
closed loop B  dL = moIencircled + moeo d/dt [  E  dA]
Faraday’s Law
closed loop E  dL = -DV = -d/dt [  B  dA ] .
Electric Field Wave Equation
Note that a d/dt of E gives a B in Ampere’s Law, and
that a d/dt of B gives an E in Faraday’s Law.
We’ve written Maxwell’s Equations in integral form, but they
can also be written in differential form using the curl and the
divergence (Calculus III topics). By combining these
equations we get the following wave
equation:
2Ey/x2 = moeo 2Ey/t2
Compare this to the wave equation for a string:
T 2y/x2 = m 2y/t2 .
Electric Field Waves
2Ey/x2 = moeo 2Ey/t2
This has a general solution of E(x,t) = E(x±vt) where
v2 = 1/[moeo], and in particular (with lf = w= v)
it has the solution: Ey = Eo sin(kx  wt + fo)
Note that the phase velocity, v, of this wave depends on
the parameters of the space that the wave is going
through: v = [1/(em)] . Recall that eo = 1/(4pk)
where k = 9 x 109 Nt-m2/Coul2 , and mo = 4p x 10-7 T-m/A .
Thus electric waves should propagate through
vacuum with a speed of (you do the calculation).
Electric and Magnetic Waves
Maxwell’s Equations also predict that whenever
we have a changing Electric Field, we have a
changing Magnetic Field. Thus, we really have
an Electromagnetic Wave rather than just an
isolated Electric Field wave.
Does this E&M wave carry energy? If so, how
does that energy relate to the amplitude of the
field oscillation?
Electric Field and Energy
Recall the energy stored in a capacitor:
Energy = (1/2)CV2 where V = E*d and for a
parallel plate capacitor where the field exists
between the plates: C = KA / 4pkd.
Thus, Energy = (1/2)[KA / 4pkd][Ed]2
= (1/2)eVolE2
where Vol = A*d, and e = K(1/4pk) .
Note that E2 is proportional to the energy per
volume!
Magnetic Field and Energy
Recall that the energy stored in an inductor is:
Energy = (1/2)LI2 where L for a solenoid is
L = m N2 A/Length and B for a solenoid is
B = m(N/Length)I .
Thus, Energy =(1/2)[mN2A/Length][BLength/mN ]2
= (1/2)VolB2/m where Vol = A*Length .
Note that B2 is proportional to the energy
per volume!
Energy and E&M Waves
Energy/Vol = (1/2)eE2 and
Energy/Vol = (1/2)B2/m
Note that Energy density (Energy/Vol) when moving
(m/s) becomes Power/Area, or Intensity.
Also from Maxwell’s equations, E and B are
related, and so we have for E&M waves that:
Eo/Bo = c = [1/(eomo)] .
Putting this together, we have: Intensity = E B / mo.
Intensity and the Poynting Vector
Maxwell’s Equations also predict that electromagnetic
waves will travel in a direction perpendicular to the
directions of both the waving Electric and the waving
Magnetic Fields, and that the direction of the waving
Electric Field must be perpendicular to the direction
of the waving Magnetic Field. This is stated in the
following:
S = (1/mo)E x B
where S is called the Poynting vector and gives the
Intensity of the electromagnetic wave.
This means that E&M waves do not have a longitudinal
polarization, only the two transverse polarizations.
Momentum
From Maxwell’s Equations we also predict that
electromagnetic waves should carry momentum,
where the amount of momentum depends on the
energy per speed:
p = Energy / c .
(In relativity, we will see that electromagnetic energy can be
considered to be carried by photons, where photons have
mass. From relativity, E = mc2, and p = mc, so p = E/c.)
If the light is absorbed the material will receive this
amount of momentum. If the light is reflected, the
material will receive twice this amount of
momentum.
Radiation Pressure
Pressure is Force/Area, and Force is change in
momentum with respect to time. Hence,
electromagnetic radiation should exert a pressure
on objects when it hits them.
Radiation Pressure = F/A =
(dp/dt) / A = [d(Energy/speed)/dt] / A =
[Power/speed] / A = [Power/Area] / speed = S/c.
If the radiation reflects, then the momentum is twice
and so the radiation pressure is also twice.
Computer Homework
Problems dealing with Electromagnetic
Waves are in the Computer Homework
program (Vol 4, #6) entitled
Electromagnetic Waves.
Numerical Example: Sunlight
The intensity of the sun at the earth’s distance
from the sun is about 1,400 W/m2 = S = EB/mo .
E and B are related by E/B = c, so to determine E
we have: S = E(E/c)/mo = E2/(cmo) which gives:
E = [S*c*mo]1/2 = [1,400 W/m2 * 3 x 108 m/s * 4p x 10-7 T*m/A]1/2 =
726 Nt/C. Note that this is the rms average value
of E. The amplitude of the electric field would be
2 * 726 Nt/C  1,000 Nt/C. Since E/B = c,
B = 726 Nt/C / 3x108 m/s = 2.4 x 10-6 T.
Note that these are values above the earth’s atmosphere. Reflection and
absorption by the atmosphere will reduce these at the earth’s surface.
Numerical Example: Sunlight
Radiation pressure (if light is absorbed):
S/c = 1,400 W/m2 / 3x108 m/s = 4.67 x 10-6 Nt/m2.
Recall that if the light reflects, then the
momentum given to the object will be twice
that of light that is absorbed.
“Sailing” in space?
Since light carries both energy and momentum,
could you use sunlight to “sail” in space? If so, would
a white sail or a black sail work best?
Besides emitting light, the very hot and active
surface of the sun also ejects high speed electrons and
protons out into space – called the solar wind. Could
you use this solar wind to “sail” in space?
Sailboats can sail slightly into the wind (called
tacking). Could a “light” sail in space be able to
“tack”?
Sailing in space?
Remember that the earth is orbiting around
the sun, and so are the other planets.
To maintain a circular orbit, there is only
one speed for each radius. If you slow down, you
will go into an elliptical orbit that will get closer
to the central object; if you speed up, you will go
into an elliptical orbit that will get further from
the central object.
Can you use the sail in space to speed up or
slow down? [Consider tilting the sail.]