y ( x , t )
=
A sin k ( x
− vt )
Kinetic energy due one mass dm of the medium (string, slinky, etc.
dK
=
1
2 dmv
2 y
=
1
2
μ dx
δ
δ t y
⎞ 2
⇒ dE
=
1
2
μ dx
(
− kvA cos k ( x
− vt )
)
2
K
=
1
2
μ k
2 v
2
A
2
0
λ cos
2 k ( x
− vt ) dx
=
1
2
μ k
2 v
2
A
2
(
λ
2
)
K
=
1
2
μ
(
2
π
λ
)
2
⎛
⎜⎜
F
μ
⎞
⎟⎟
2
A
2
(
λ
2
)
=
A
2
π
λ
2
F
(for one wavelengh t)
I turns out that U = K for a traveling wave and therefore
Total Energy E
=
2
A
2
π
λ
2
F
Power P
=
2
A
2
π
λ
2
F v
λ
5

Total Energy E
=
A
2
π
λ
2
F
Two travelling waves with amplitude ½ A have to have the same energy than a standing wave with A
E
=
2
⎛
2
⎝
A
2
λ
2
π
2
F
=
A
2
π
λ
2
F
6

μ
1 v
1
μ
2 v
2 i: incident wave r: reflection tr: transmitted
A i
+
A r
=
A tr
−
A i k i
+
A r k i
=
( 1 ) _
−
A tr k tr
( A r
−
A i
) k i
= −
A tr k tr from
and y i y r y tr
ω
x
t
Boundary conditions y
1
=y
2
δ
δ y
1 x
=
δ
δ y x
2
=
A i
ω i t
− k i x
x
t
x
t
= v i k i
=
=
=
A r
A tr
ω i t
ω i t
+
− k i k tr x
x
v tr k tr
= const
( A r
−
A i
) v tr
= −
A tr v i
( 2 )
A r
A i
= v v
2
1
− v
1
+ v
2
A tr
A i
= v
1
2 v
2
+ v
2
7

A
A i r
= v v
2
1
− v
1
+ v
2 and
A tr
A i
Wall: μ −> ∞ and v
2
A r
A i
= −
1 and
A
A i tr
=
= 0
0
= v
1
2 v
2
+ v
2 v
=
F
μ
Now
μ
2
< μ
1
A r
A i
= positive
A tr
A i
= positive
μ
2
> μ
1
A r
A i
= negative
A tr
A i
= negative
μ
2
= μ
1
A r
A i
=
0
A tr
A i
=
1
8

•Sound Waves
•The Speed of Sound
•Amplitude & Intensity of Sound Waves http://www.youtube.com/watch?v=2ieeLVmYFCg
9

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
10

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).
11

p m
( v
) s m
12

We know the the speed of a mechanical wave: v
=
T
The speed of sound in different materials can be determined as follows:
In fluids v
=
B
B is the bulk modulus of the fluid and ρ its density.
In thin solid rods v
=
Y
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.
13

Δ
L
L F
F
Apply a force to both ends of a long wire. These forces will stretch the wire from length L to L+ Δ L.
Δ
L
∝
F
Δ
L
∝
L
L
F L
Δ
L
∝
≈
A
14

Hooke’s Law (F ∝ x) can be written in terms of stress and strain (stress ∝ strain).
F
Δ
A
=
Y
L
L
The spring constant k is now k
YA
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
Y
15

An object completely submerged in a fluid will be squeezed on all sides.
volume stress
= pressure
=
The result is a volume strain;
F
A volume strain
=
Δ
V
V
16

For a volume deformation, Hooke’s Law is (stress ∝ strain):
Δ
P
= −
B
Δ
V
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: 2.2×10 9 Pa (value increases at higher pressures)
Air: 1.42×10 5 Pa
Steel: 1.6×10 11 Pa
Solid helium (approximate): 5×10 7 Pa
17

Material Speed (m/s)
Aluminum
Granite
Steel
Pyrex glass
Copper
Plastic
Fresh water (20
ºC)
Fresh water (0
ºC)
Hydrogen (0 ºC)
Helium (0 ºC)
Air (20 ºC)
Air (0 ºC)
6420
6000
5960
5640
5010
2680
1482
1402
1284
965
343
331 v
= v
=
Y
B
http://video.google.com/videoplay?docid=-
7693060040597673377&q=soundwaves 18

In ideal gases v
=
γ air
γ
He
RT emperature
γ v
= v
0
T
M T
0
=
=
7 / 5
≈ 1.4
M
Air
5 / 3
≈ 1.66
M
He
=
=
29
4
−
→
30 v
He
≈
3 v
Air
Here v
0 is the speed at a temperature T
0
(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
=
(
+
T
)
where T is the air temperature in ° C.
19

Example: A copper alloy has a Young’s Modulus of 1.1
× 10 11
Pa and a density of 8.92 × 10 3 kg/m 3 . What is the speed of sound in a thin rod made of this alloy?
v
=
Y
ρ
=
1 .
1
×
10
11
Pa
8 .
9
×
10
3 kg/m
3
=
3500 m/s
20

Example: Bats emit ultrasonic sound waves with a frequency as high as 1.0
× 10 5 Hz. What is the wavelength of such a wave in air of temperature 15.0 ° C?
v
=
λ =
(
v f
+
T
C
)
The speed of sound in air of this temperature is 340 m/s.
=
340 m/s
1.0
×
10
5
Hz
=
3 .
4
×
10
−
3 m
21

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?
v
=
(
331
+
0 .
606 T
C
) m/s
The speed of sound in 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 5 km / s )
=
9
μ s
22

Find the speed of sound in water
B water
ρ water
=2.1 10 9 N/m 2 at 0ºC
= 1 10 3 kg/m 3 v
=
B
=
2 .
1
⋅
10
9
N m
2 kg
1
⋅
10
3 m
3
=
1 .
4 km / s
≈
4
⋅ v air
23

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 . http://video.google.com/videoplay?docid=-
7487415149777549129&q=soundwaves
24

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
25

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. http://www.youtube.com/watch?v=j8rZxmCejD0
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.
26

Two waves are considered coherent if they have the same frequency and maintain a fixed phase relationship.
Two waves are considered incoherent if the phase relationship between them varies randomly.
When waves are in phase, their superposition gives constructive interference .
When waves are one-half a cycle out of phase, their superposition gives destructive interference .
27

For "coherent sources", in which the two waves have the same wave number and frequency
(k
1
= k
2
, w
1
= w
2
): y(x,t) = 2 A cos ( ϕ / 2) sin (k x - w t + ϕ / 2).
Amplitude
28

While testing speakers for a concert, Tomás sets up two speakers to produce sound waves at the same frequency, which is between 100 Hz and 150 Hz . The two speakers vibrate in phase with one another. He notices that when he listens at certain locations, the sound is very soft (a minimum intensity compared to nearby points). One such point is 25.8 m from one speaker and 37.1 m from the other. What are the possible frequencies of the sound waves coming from the speakers? (The speed of sound in air is 343 m/s .)
The intensity minimums imply that the distance
37.1 m
−
25.8 m
=
11.3 m is equal to a whole number of wavelengths m plus one-half wavelength.
m
+
1
2
λ =
11.3 m.
f
= v
=
⎝ m
+
2 ⎠ 11.3 m
=
(30.35 Hz) m
+
15.18 Hz.
f (3)
=
⎛
⎝
3
+
2 ⎠ 11.3 m
=
106 Hz and (4)
=
⎛
4
+
2 ⎠ 11.3 m
=
137 Hz.
106 Hz and 137 Hz.
29

For sound waves:
I
I
∝ p
0
2
∝ s
0
2 p
0 is the pressure amplitude and s
0 is the displacement amplitude.
I
=
2 p
0
2
ρ v
p m
v
s m
The intensity of sound waves also follow an inverse square law.
30

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
Æ Area of a sphere A = 4 π r 2
Intensity with distance from a point source
31

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/m 2 .
Sound intensity level is defined by
=
(
10 dB
) log
I
I
A general " rule of thumb " for loudness is that the power must be increased by about a factor of ten to sound twice as loud.
32

β =
(
10 dB
) log
I
I
0
β =
(
10 dB
) log
I
0
I
0
=
0
β =
(
10 dB
) log
10 I
0
I
0
=
10 dB
β =
(
10 dB
) log
100 I
0
I
0
=
20 dB
β =
(
10 dB
) log
1000 I
0
I
0
=
30 dB
33

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?
Given:
=
(
10 dB
) log
I
=
71 dB
I
0
Solve for I, the intensity of a sound wave:
I log
I
0
I
=
7 .
1
=
I
0
10 7 .
1
=
I
I
0
(
10
=
−
12
10 7 .
1
W/m 2
.
1
=
1 .
3
×
10
−
5 W/m 2
The intensity of an isotropic source is defined by:
I
=
=
P
4
π r 2
P
=
I 4
π r 2
( 1 .
3
×
10
−
5
2
) 4
π (
25
)
2 =
0 .
10
34

Example: Two sounds have levels of 80 dB and 90 dB.
What is the difference in the sound intensities?
β
1
=
(
10 dB
) log
I
I
0
=
80 dB
β
2
=
(
10 dB
) log
I
I
0
=
90 dB
Subtracting:
2
−
1
=
10 dB
=
10 dB
⎛
⎜⎜ log
I
2
I
0
− log
I
I
0
1
⎞
⎟⎟
10 dB
=
10 dB
⎛
⎜⎜ log
I
2
I
1
⎞
⎟⎟
I
2
=
10
1
I
1
I
=
2
10 I
1
35