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Main points of today’s lecture:
Transverse and longitudinal waves
traveling waves vwave = fλ
Wave speed for a string
v=
F
µ
Here F is the tension
and µ is the mass/length.
• 
Intensity of sound I = P A = ΔE ( AΔT )
•  Decibel intensity values
–  Ithreshold=I0 ≈ 10-12 W/m2,
Ipain ≈ 1 W/m2
–  β=10 log10(I/ I0)
–  Spherical waves
I=
P
4πr 2
Types of Waves -- Transverse
•  In a transverse wave, each element that is disturbed moves
perpendicularly to the wave motion
Types of Waves -- Longitudinal
•  In a longitudinal wave, the elements of the medium undergo
displacements parallel to the motion of the wave
•  A longitudinal wave is also called a compression wave
Waves
•  If you tie a spring to a wall, hold one end, and rapidly move your
arm up and down, you will create a pulse that travels down the
spring
–  Transverse – the displacement of the spring is perpendicular to the direction
the pulse moves
•  Rapidly move the spring back and forth, and compressed pulse
travels down the spring
–  Longitudinal – the displacement is parallel to the direction of the pulse
Whatdothesepulses
transport?
•  Ma1er?
•  NO!
Energyistransported!
Snapshot Graphs of an asymmetic wave: space
vs. time
All points at
different times.
Slide 15-14
Constructing a History Graph
one point at all times.
Slide 15-15
Checking Understanding
The graph below shows a snapshot graph of a wave on a string that is
moving to the right. A point on the string is noted. Which of the choices is
the history graph for the subsequent motion of this point?
Slide 15-16
Checking Understanding
The graph below shows a snapshot graph of a wave on a string that is
moving to the left. A point on the string is noted. Which of the choices is
the history graph for the subsequent motion of this point?
WWhat is the motion at this
point?
Slide 15-16
Traveling sinusoidal wave
•  The figure at the right shows a wave
traveling on a string at different times,
measured in units of a period. The round
point shows how wave-front on the
traveling wave moves.
•  At a fixed time and as a function of
distance, the wave reproduces itself after
an x displacement of one wavelength λ.
•  At each point, the wave returns to its
original y displacement after a time T.
•  At time T, the wave front has moved a
distance λ. This makes the wave velocity
v= λ/T= λf. Here we use the relationship
f=1/T between wavelength and
frequency.
λ
Describing a Traveling Wave
λ
t=0
y
each mass point goes up and
down like mass+spring
λ
1 2
E = mvmax
2
x
t=T/4
t=2T/4
t=3T/4
t=T
λ:wavelength=length(m)ofone
oscilla=on.
T:period==meforoneoscilla=on:
frequency(Hz)
Whilethewavehastraveledone
wavelength,eachpointonthewave
hasmadeoneperiodofoscilla=on.
v=Δx/Δt=λ/T=λf
Example
•  A wave traveling in the positive x direction has a frequency of
25.0 Hz as in the figure below. Find the (a) amplitude, (b)
wavelength, (c) period, and (d) speed of the wave.
•  The amplitude is:
–  a) 9 cm
–  b) 18 cm
•  The wavelength is:
–  a) 10 cm
–  b) 20 cm
a) A = 18cm / 2 = 9cm
b) λ = 10cm ⋅ 2 = 20cm
c) T = 1/f =1/25 s = 0.04 s
d) v = λ f = 20cm ⋅ 25Hz = 500cm / s
Wave velocity on string
•  Waves travel faster on strings with higher tension and lower mass/unit
length. The exact relationship is
v=
F
µ
Here F is the string tension and µ is the mass/length.
•  Example: A 0.5 m string is stretched so the tension is 1.7 N. A
transverse wave of frequency 120 Hz and wavelength 0.31 m travels
on the string. What is the mass of the string?
F
m = µL; L = 0.5m; v =
⇒µ= 2
µ
v
= ( 0.31m )(120Hz ) = 37.3m / s
v = λf
1.7N
−3
F
=
=
1.22x10
kg / m
⇒µ= 2
2
(37.3m / s)
v
F
(
)
⇒ m = µL = 1.22x10−3 kg / m (0.5m) = 0.61g
L
0.5 m
F
1.7N
f
120Hz
λ
0.31m
m
?
Example Problem
A particular species of spider spins a web with silk threads of
density 1300 kg/m3 and diameter 3.0 µm. A typical tension in the
radial threads of such a web is 0.007 N. What is the speed of the
waves traveling on the web. If a fly lands in this web, which will
reach the spider first, the sound or the wave on the web silk?
(sound velocity=343 m/s)
a) 172 m/s, no
b) 215 m/s, no
c) 286 m/s. no
d) 430 m/s, yes
3
ρ
=
1300kg
/
m
e) 873 m/s, yes
(
)
µ = ρ i A silk= 1300kg / m 3 i 3.14159 i (3x10−6 / 2m)2 = 9.19x10−9 kg / m
v=
.007kg i m / s 2
=
= 873m / s
−9
µ
9.19x10 kg / m
F
This is faster than the speed of sound in air.
Slide 15-21
Conceptual question
•  A weight is hung over a pulley and attached to a string
composed of two parts, each made of the same material but one
having four times the diameter of the other. The string is plucked
so that a pulse moves along it, moving at speed v1 in the thick
part and at speed v2 in the thin part.What is v1/v2?
a) 1/4
b) 1/2
c) 2
d) 4
v=
⇒ v1 =
F
=
µ
F
F
M/L
F
=
µ
π R12 ρ
M = Vρ = π R 2 Lρ ⇒ µ = M / L = π R 2 ρ
v1
⇒
=
v2
F
/
π/ R 12 ρ/
F/
π/R 22 ρ/
=
1
R 12
1
R 22
⋅
R 12 ⋅ R 22
R ⋅R
2
1
2
2
=
R 22 R 2
=
= 1/ 4
2
R1 R1
Using a Tuning Fork to Produce a Sound Wave
•  A tuning fork will produce a pure musical note of
frequency f
•  As it’s tine swings to the right, it forces the air
molecules near it closer together
•  This produces a high density area in the air
–  This is an area of compression
•  As the tine moves toward the left, the air molecules
to the right of the tine spread out
•  This produces an area of low density
–  This area is called a rarefaction
• As the tuning fork continues to vibrate, a sinusoidal succession
of compressions and rarefactions spread out from the fork
– Crests correspond to
compressions and troughs
to rarefactions
v = λf
λ
Sound Waves
•  Audible range: between 20 Hz to 20,000 Hz
•  Infrasonic waves: f < 20 Hz; Ultrasonic:f > 20,000 Hz
•  Sound velocity : v ≈ elastic _ property
inertial _ property
•  Sound velocity in liquid: v ≈ B ; B is the bulk modulus
ρ
•  Sound velocity in solid:
v ≈ Y ; Y is Young's modulus
ρ
(
•  Sound velocity in air: v ≈ 331m / s
)
T
; where T is in Kelvin or
273K
v ≈ 331m / s + 0.6T ; where T is in Celsius.
Example
•  You are watching a pier being constructed on the far shore of a
saltwater inlet when some blasting occurs. You hear the sound
in the water 4.50 s before it reaches you through the air. How
wide is the inlet? (Assume the air temperature is 20°C and
sound velocity in water is 1530 m/s at its temperature (25°C). )
vs,water = 1530m / s; vs,air = (331m / s ) T 273K = (331m / s) 293
273 = 343m / s
(
t air − t water = d 1/ vs,air − 1/ vs,water
t water = d / vs,water ;
t air = d / vs,air ;
(
⇒ d = ( t air − t water ) / 1/ vs,air − 1/ vs,water
)
)
1
1
⎛
⎞
= 4.5s / ⎜
−
= 1989m
⎟
⎝ 343m / s 1530m / s ⎠
Quiz
•  A dolphin located in sea water at a temperature of 25°C emits a
sound directed toward the bottom of the ocean 150 m below.
How much time passes before it hears an echo? (The sound
velocity in sea water = 1530 m/s at 250C.)
–  a)0.2 s
–  b)0.4 s
–  c)0.6 s
–  d)0.8 s
Δt = 2d / v = 300m / (1530m / s) = 0.196s
–  e) 0.1 s
Sound amplitude and intensity!
•  The amplitude of the sound wave
is proportional to the maximum
velocity of the air as it moves from
the high pressure to the low
pressure domains.
•  The energy and the power of the
sound wave is proportional to the
square of amplitude:
1
2
2
E ∝ ρv 2 ∝ Amplitude 2 ∝ vmax
∝ p 2 ∝ xmax
2
•  More useful than the energy of a sound wave is
the intensity, I, which is the power P that the
sound wave transmits per unit area. I = P A = ΔE ( AΔT )
•  The ear responds logarithmically to the intensity
of sound waves striking the eardrum.
–  Ithreshold=I0 ≈ 10-12 W/m2, Ipain ≈ 1 W/m2
•  This logarithmic behavior motivates the decibel
measure of sound wave intensity.
–  β=10 log10(I/ I0)
v
Fig 14.2, p. 426
Slide 4
Decibel is named after Alexander Graham Bell
How the decibel scale works:
if I=I0, (I is equal to hearing threshold intensity)
β=10log10(I0 /I0)=10log10(1)=0
log10(10n) = n, i.e. log10(1000)=log10(103) = 3
if I=102I0=100I0, then
β=10log10(100I0 /I0)=10log10(100)=20
log10(A/B)=log10(A)-log10(B)
log10(AŸB)=log10(A)+log10(B)
Machine noise
Amachineproducessoundwithalevelof80dB.How
manymachinescanyouaddbeforeexceeding100dB?
1machine
80dB=10log(I/I0)
8=log(I/I0)=log(I/1E-12)
108=I/1E-12
I1=10-4W/m2
Nmachines
100dB=10log(IN/I0)
10=log(IN/I0)=log(IN/1E-12)
1010=IN/1E-12
IN=10-2W/m2
I1/IN=10-4/10-2=1/100
Theintensitymustincreasebyafactorof100;
oneneedstoadd99machines.
Shortcut:x=20sotheincreaseinIis10(x/10)=100.
Example
•  The intensity level of sound A is 5.0 dB greater than that of sound B
and 3.0 dB less than that of sound C. a) Determine the ratio (IC/IB) of
the intensity of sound C to the intensity of sound B.
⎛ I⎞
a) β = 10log10 ⎜ ⎟
⎝ I0 ⎠
⎛
⎜
⎡
⎛ IA ⎞
⎛ IB ⎞ ⎤
⎛ IA ⎞
⎛ IB ⎞
β A − β B = 5 = 10log10 ⎜ ⎟ − 10log10 ⎜ ⎟ = 10 ⎢ log10 ⎜ ⎟ − log10 ⎜ ⎟ ⎥ = 10log10 ⎜
⎝ I0 ⎠
⎝ I 0 ⎠ ⎥⎦
⎝ I0 ⎠
⎝ I0 ⎠
⎜
⎢⎣
⎜⎝
⎛ IA ⎞
⇒ βA − βB = 10log10 ⎜ ⎟ = 5
⎝ IB ⎠
⎛I ⎞
similarly : βC − β A = 10log10 ⎜ C ⎟ = 3
⎝ IA ⎠
⎛I ⎞
⇒ βC − βB = 10log10 ⎜ C ⎟ also, βC − β B = ( βC − β A ) + ( β A − β B ) = 5 + 3 = 8
⎝ IB ⎠
⎛I ⎞
⇒ 10log10 ⎜ C ⎟ = 8
⎝ IB ⎠
⎛ IC ⎞
⇒ log10 ⎜ ⎟ = 0.8
⎝I ⎠
IC
⇒ = 100.8 = 6.3
IB
B
⇒ 10
⎛I ⎞
log10 ⎜ C ⎟
⎝ IB ⎠
= 100.8
IA ⎞
/I0 ⎟⎟
IB ⎟
I/0 ⎟⎠