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## Sound

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### What parameters does the speed of sound depend on? ... Sound is a form of energy that moves. ... that vibrations are involved in sound, we can try the wave ... – PowerPoint PPT presentation

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Title: Sound

1
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?

2
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 doesnt necessarily mean that they
exist.
• Since we note that vibrations are involved in
sound, we can try the wave idea.

3
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!

4
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!

5
Sound a travelling wave
• We have already considered waves on a string. We
were able to work with Newtons 2nd law to get a
wave equation for this. Can we do the same for
sound?
• YES! We use the fluid equivalent of Newtons
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).

6
Speed of sound
• Newtons 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/r1/2
• (this is just like v T/m1/2 for a string).

7
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.

8
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.

9
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 .

10
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/r1/2 P/r1/2 . But
from the ideal gas law,
• PV nRT, P nRT/V by definition,
• r m/V. Thus, v P/r1/2 nRT/m1/2 . We
can replace the m/n (total mass per total moles)
by M, the molar mass)
• v gRT/M1/2 , where g CP/CV 1.4 for a
diatomic gas like air has to be introduced due to
thermodynamic considerations.

11
Sound in air
• v gRT/M1/2
• 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 (75oF24oC 297 K), v
1.4 8.3 J/mole K 297 K / .029 kg/mole1/2
345 m/s 770 mph.
• At higher altitudes we have lower temperatures
and hence lower speeds.

12
Human Hearing Pitch
• A standard human ear can hear frequencies from
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.

13
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?

14
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.

15
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).

16
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.

17
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.

18
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.

19
Intensity W/m2
• The ear is a very sensitive energy receiving
device. It can hear sounds down as low as 10-12
W/m2. Considering that the ears area is on the
order of 1 cm2 or 10-4 m2, that means the ear can
detect sound energy down to about 10-16 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)!

20
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?

21
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)
10log10(I/Io) where Io is the softest sound we
can hear (10-12 W/m2) .

22
Examples
• The weakest sound intensity we can hear is what
we define as Io. In decibels this becomes
• I(dB) 10log10(10-12 / 10-12) 0 dB .
• The loudest sound without damaging the ear is 1
W/m2, so in decibels this becomes
• I(dB) 10log10(1 / 10-12) 120 dB .

23
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?

24
Decibels
• Calculating for each
• I(dB) 10log10(1 x 10-6 / 10-12) 60 dB
• I(dB) 10log10(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) 10log10(2I1)
10log10(2)log10(I1) 10.3log10(I1)
3dBI1(dB)

25
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) 10log(1 x 10-6 / 10-12) 60 dB , and
• I(dB) 10log(1/4 x 10-6 / 10-12) 54 dB.
• (Notice that 4 is two 2s, so the decrease is two
3dBs for a total of 6 dB.)

26
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!).

27
Electromagnetic Waves
• For waves on a string and sound waves, we can get
a wave equation from Newtons Second Law.
• So far in Physics 251 weve talked about electric
and magnetic fields. Can the fields wave ?
• If so, where do we start to try to get a wave
equation for the fields?

28
Electromagnetic Waves
• The basic equations for electric and magnetic
fields are the basic four equations weve dealt
with in this course
• Gausss Law for Electric Fields
• Gausss Law for Magnetic Fields
• Amperes Law
• Together these four laws are called
• Maxwells Equations .

29
Electric Field Wave Equation
• Weve written Maxwells 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 .

30
Electric Field Waves
• ?2Ey/?x2 moeo ?2Ey/?t2
• This has the solution Ey Eo sin(kx ? wt fo)
• and the phase velocity 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).

31
Electric and Magnetic Waves
• Maxwells 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 EM wave carry energy? If so, how does
that energy relate to the amplitude of the field
oscillation?

32
Electric Field and Energy
• Recall the energy stored in a capacitor
• Energy (1/2)CV2 where V Ed and for a
parallel plate capacitor where the field exists
between the plates C KA / 4pkd.
• Thus, Energy (1/2)KA / 4pkdEd2
• (1/2)eVolE2 where Vol Ad, and
• e K(1/4pk) . Note that E2 is proportional to
the energy per volume!

33
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/LengthBLength/mN 2
• (1/2)VolB2/m where Vol ALength . Note
that B2 is proportional to the energy per volume!

34
Energy and EM 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 Maxwells equations, we have for EM
waves that Eo/Bo c ?1/(eomo) .
• Putting this together, we have I E B / mo .

35
Intensity and the Poynting Vector
• Maxwells 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/?o)E x B
• where S is called the Poynting vector and gives
the Intensity of the electromagnetic wave.

36
Momentum
• From Maxwells 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.

37