Soundwaves
•Sound Waves
•The Speed of Sound
•Amplitude & Intensity of Sound Waves
http://www.youtube.com/watch?v=2ieeLVmYFCg
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Sound Waves (~ any longitudinal wave)
A sound wave is similar in
nature to a slinky wave for a
variety of reasons. First, there
is a medium that carries the
disturbance from one location
to another. Typically, this
medium is air; though it could
be any material such as water
or steel.
http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html
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Sound waves are longitudinal. They can be represented
by either variations in pressure (gauge pressure) or by
displacements of an air element.
The middle of a
compression (rarefaction)
corresponds to a pressure
maximum (minimum).
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∆pm = (vρω ) sm
density
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The Speed of Sound Waves
We know the the speed of a mechanical wave: v =
T
µ
The speed of sound in different materials can be determined
as follows:
In fluids
v=
In thin solid rods v =
B
ρ
Y
ρ
B is the bulk modulus of the
fluid and ρ its density.
Y is the Young’s modulus of
the solid and ρ its density.
Materials that have a high restoring force (stiffer) will have a higher sound
speed.
Materials that are denser (more inertia) will have a lower sound speed.
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Hooke’s Law (F∝x) can be written in terms of stress and
strain (stress ∝ strain).
F
∆L
=Y
A
L
YA
The spring constant k is now k =
L
Y is called Young’s modulus and is a measure of an
object’s stiffness. Hooke’s Law holds for an object
to a point called the proportional limit.
The Young'
s modulus (also known as modulus of elasticity, elastic modulus
or tensile modulus) allows the behavior of a material under load to be
calculated. For instance, it can be used to predict the amount a wire will
extend under tension, or to predict the load at which a thin column will buckle
under compression
tensile stress
Y=
tensile strain
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For a volume deformation, Hooke’s Law is (stress∝strain):
∆V
∆P = − B
V
where B is called the bulk modulus. The bulk modulus (B) of a
substance essentially measures the substance'
s resistance to uniform
compression. It is defined as the pressure increase needed to effect a given
relative decrease in volume.
Water: 2.2×109 Pa (value increases at higher pressures)
Air: 1.42×105 Pa
Steel: 1.6×1011 Pa
Solid helium (approximate): 5×107 Pa
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Material
Speed (m/s)
Aluminum
Granite
Steel
Pyrex glass
Copper
Plastic
Fresh water (20
ºC)
Fresh water (0
ºC)
Hydrogen (0 ºC)
6420
6000
5960
5640
5010
2680
1482
Helium (0 ºC)
Air (20 ºC)
Air (0 ºC)
965
343
331
1402
1284
v=
v=
Y
ρ
B
ρ
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In ideal gases
v=
RTemperatureγ
M
v = v0
Temp
TTemp , 0
γ air = 7 / 5 ≈ 1.4 M Air = 29 − 30
γ He = 5 / 3 ≈ 1.66 M He = 4 → vHe ≈ 3v Air
Here v0 is the speed at a temperature T0 (in Kelvin) and v
is the speed at some other temperature T (also in Kelvin).
For air, a useful approximation to the above expression is
v = (331 + 0.606T ) m/s
where T is the air temperature in °C.
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Example: A copper alloy has a Young’s Modulus of 1.1×1011
Pa and a density of 8.92 ×103 kg/m3. What is the speed of
sound in a thin rod made of this alloy?
v=
1.1×1011 Pa
= 3500 m/s
=
3
3
8.9 ×10 kg/m
ρ
Y
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Example: Bats emit ultrasonic sound waves with a frequency
as high as 1.0×105 Hz. What is the wavelength of such a
wave in air of temperature 15.0 °C?
v = (331 + 0.606TC ) m/s
The speed of sound in
air of this temperature
is 340 m/s.
v
340 m/s
−3
λ= =
=
3
.
4
×
10
m
5
f 1.0 ×10 Hz
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Example: A lightning flash is seen in the sky and 8.2
seconds later the boom of thunder is heard. The
temperature of the air is 12.0 °C.
(a) What is the speed of sound in air at that temperature?
The speed of sound in
v = (331 + 0.606TC ) m/s
air of this temperature
is 338 m/s.
(b) How far away is the lightning strike?
d = vt = (338 m/s )(8.2 s ) = 2800 m = 2.8 km
light : t = 2.8km/(3 10 km / s ) = 9 µs
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Find the speed of sound in water
Bwater=2.1 109 N/m2 at 0ºC
3 kg/m3
=
1
10
water
N
2.1⋅10 2
B
m = 1.4km / s ≈ 4 ⋅ v
=
air
ρ
3 kg
1⋅10 3
m
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v=
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Humans can generally hear sounds with frequencies between 20 Hz and 20 kHz
(the audio range) although this range varies significantly with age, occupational
hearing damage, and gender; the majority of people can no longer hear 20,000
Hz by the time they are teenagers, and progressively lose the ability to hear
higher frequencies as they get older. Most human speech communication takes
place between 200 and 8,000 Hz and the human ear is most sensitive to
frequencies around 1000-3,500 Hz. Sound above the hearing range is known as
ultrasound, and that below the hearing range as infrasound.
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Whales, elephants, hippopotamuses, rhinoceros, giraffes, okapi, and alligators are
known to use infrasound to communicate over varying distances of up to many
miles, as in the case of the whale. Infrasound sometimes results naturally from ocean
waves, avalanches, earthquakes, volcanoes, and meteors. Infrasound can also be
generated by man-made processes such as explosions, both chemical and nuclear.
Ultrasound
Medical sonography (ultrasonography) is an ultrasound-based diagnostic medical
imaging technique used to visualize muscles, tendons, and many internal organs, their
size, structure and any pathological lesions.
Ultrasound is also used in Nondestructive testing to find flaws in materials.
A common use of ultrasound is in range finding; this use is also called sonar.
Ultrasonic cleaners
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Bats
Bats use a variety of ultrasonic ranging (echolocation) techniques to detect
their prey.
Dogs
The dog whistle is used to call to a dog. It makes ultrasound at a frequency
in the range of 16 kHz to 22 kHz that dogs can hear.
Dolphins and whales
It is well known that dolphins and some whales can hear ultrasound and
have their own natural sonar system.
Fish
Several types of fish can detect ultrasound. Of the order Clupeiformes,
members of the subfamily Alosinae (shad), have been shown to be able to
detect sounds up to 180 kHz, while the other subfamilies (e.g. herrings) can
hear only up to 4 kHz.
Moths
There is evidence that ultrasound in the range emitted by bats causes flying
moths to make evasive maneuvers, because bats eat moths. Ultrasonic
frequencies trigger a reflex action in the noctuid moth that cause it to drop a
few inches in its flight to evade attack.
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Amplitude & Intensity of Sound
Waves
For sound waves:
I ∝ p02 p0 is the pressure amplitude and
I∝s
2
0
s0 is the displacement amplitude.
The intensity of sound waves also follow an inverse
square law.
p02
I=
2 ρv
∆pm = (vρω ) sm
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Sound Intensity
Notice that sound waves carry
energy. We define the intensity I as
the rate at which energy E flows
through a unit area A perpendicular
to the direction of travel of the wave.
Intensity = Power / Area
I = P / A = E / (At)
For a point source, energy spreads out
in all directions
Area of a sphere A = 4 r2
Intensity with distance from a point
source
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Loudness is not simply sound intensity!
Loudness of a sound is measured by the logarithm of the
intensity.
The threshold of hearing is at an intensity of 10-12 W/m2.
Sound intensity level is defined by
I
β = (10dB) log
I0
dB are decibels
A general "rule of
thumb" for
loudness is that the
power must be
increased by about
a factor of ten to
sound twice as
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loud.
I
β = (10dB )log
I0
I0
β = (10dB )log = 0
I0
10 I 0
β = (10dB )log
= 10dB
I0
100 I 0
β = (10dB )log
= 20dB
I0
1000 I 0
β = (10dB )log
= 30dB
I0
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Example The sound level 25 m from a loudspeaker is 71 dB.
What is the rate at which sound energy is being produced by
the loudspeaker, assuming it to be an isotropic source?
I
Given: β = (10dB) log = 71 dB
I0
Solve for I, the intensity of a sound wave:
I
log = 7.1
I0
I
= 107.1
I0
(
)(
)
I = I 0107.1 = 10 −12 W/m 2 107.1 = 1.3 ×10 −5 W/m 2
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Example continued:
The intensity of an isotropic source is defined by:
P
I=
4πr 2
P = I 4πr 2
= (1.3 ×10 W/m )4π (25 m )
−5
2
2
= 0.10 Watts
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Example: Two sounds have levels of 80 dB and 90 dB.
What is the difference in the sound intensities?
I
β1 = (10dB) log = 80 dB
I0
Subtracting:
I
β 2 = (10dB) log = 90 dB
I0
I1
I2
β 2 − β1 = 10 dB = 10 dB log − log
I0
I0
I2
10 dB = 10 dB log
I1
I2
= 101
I1
I 2 = 10 I1
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Standing Sound Waves
Consider a pipe open at both ends:
The ends of the pipe are open to the atmosphere. The
open ends must be pressure nodes (and displacement
antinodes).
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The distance between two adjacent antinodes is ½λ. Each
pair of antinodes must have a node in between.
The fundamental mode (it has the fewest number of
antinodes) will have a wavelength of 2L.
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The next standing wave pattern to satisfy the conditions at
the ends of the pipe will have one more node and one more
antinode than the previous standing wave. Its wavelength
will be L.
The general result for standing waves in a tube open at
both ends is
2L
λn =
n
v
where n=1, 2, 3,…
nv
fn =
=
= nf1
λn 2 L
f1 is the fundamental frequency.
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