Sound: Propagation
• Sound Waves: Sound waves are longitudinal
waves (i.e. involve oscillations parallel to the
direction of the wave travel) that propagate
through a medium (e.g. air, water, iron).
Wavefront
Source S
Ray
Medium
oscillations
• Speed of Sound: The speed of any mechanical wave
depends on both the inertial property of the medium
(stores kinetic energy) and the elastic property (stores
potential energy).
elastic (wave speed)
v=
Medium
inertial
Sound
Speed
(m/s)
Sound
Speed
(mi/hr)
Air
(20oC)
343
768
Water
(20oC)
1,482
3,320
Steel
5,941
13,308
• Stretched String (Chapter 11): The speed of
the “transverse” wave along a stretched string is
v=
τ
µ
(string, τ = tension, µ = linear mass density)
• Sound: The speed of the “longitudinal” sound wave is
v=
B
ρ
(sound, B = bulk modulus = -∆
∆P/(∆
∆V/V), ρ = volume mass
density)
R. Field 11/21/2013
University of Florida
PHY 2053
Page 1
Sound: Speed
• Speed of Sound: In Fluids
v=
B
ρ
Sound
Speed
(mi/hr)
Air
(20oC)
343
768
Water
(20oC)
1,482
3,320
Steel
5,941
13,308
(sound, B = bulk modulus = -∆
∆P/(∆
∆V/V), ρ = volume mass
density)
• Speed of Sound: In Air (B proportional to absolute T)
T
T0
v(T ) = v0
Medium
Sound
Speed
(m/s)
where T is measured in degrees Kelvin
with T0 = 273.15 oK and v0 = 331 m/s.
T(in oK) = T(in oC) + 273.15
T (o F ) = (1.8o F / o C )T ( o C ) + 32o F
(degrees Fahrenheit)
• Speed of Sound: In Solids
v=
Y
ρ
(sound, Y = young’s modulus, ρ = volume mass density)
R. Field 11/21/2013
University of Florida
PHY 2053
Page 2
Sound: Traveling Waves
• Traveling Sound Waves in Air: A
traveling sound waves consist of a moving
periodic pattern of expansions and
compressions of the air. As the wave
passes the air elements oscillate
longitudinally in simple harmonic motion.
(1000 Hz sound wave with ∆Pmax at the threshold of pain!)
Longitudinal
displacement
s ( x, t ) = smax cos(kx − ωt )
Pressure
variation
∆P( x, t ) = ∆Pmax sin( kx − ωt )
∆Pmax = (νρω ) smax
(The pressure amplitude is related to the displacement amplitude!)
R. Field 11/21/2013
University of Florida
PHY 2053
Page 3
Sound: Intensity and Level
• Intensity: Traveling sound waves transport energy (kinetic and potential)
from one point to another. The intensity, I, of a sound wave at a surface is
the average energy per unit time per unit area transmitted by the wave to
the surface (i.e. average power per unit area). It is also equal to the average
energy per unit volume in the wave times the speed of propagation of the
wave.
2
(the intensity is proportional to the
∆Pmax
dE
1 dE Ppower
2 2
1
1
I=
square of the amplitude!)
v = 2 ρvω smax = 2
I=
=
dV
ρ
v
Sphere with
A dt
A
2
radius r
Watts/m
Source PS
• Variation with Distance: If sound is emitted isotropically
r
(i.e. equal intensity in all directions) from a point source with
power Ps and if the mechanical energy of the wave is conserved
then
Ps
I=
4πr
(intensity from isotropic point source)
2
• The Decibel Scale: Instead of speaking about the
intensity I of sound, it is more convenient to speak of
the sound level β, where
β = (10dB ) log10 ( I / I 0 )
R. Field 11/21/2013
University of Florida
I = I 0 × 10
(sound level, dB = “decibel”, I0 =
β / 10 dB
10-12
Sound
level
(dB)
Intensity
(W/m2)
Hearing
threshold
0
10-12
Conversation
60
10-6
Pain threshold
120
1
W/m2)
PHY 2053
Page 4
Sound Waves: Example Problem
• At a baseball game, a spectator is 60.0 m away from the batter. How
long does it take the sound of the bat connecting with the ball to travel
to the spectator’s ears? The air temperature is 27.0 oC. Answer: 172.9 ms
The speed of sound in the air depends on the temperature as follows:
v(T ) = v0
T
T0
where T is measured in degrees Kelvin with T0 = 273.15 oK and v0 = 331 m/s.
Also, T(oK) = T(oC) + 273.15. Hence,
v(27o C ) = (331m / s )
300.15
≈ 347.0m / s
273.15
Note that 27oC is about 81oF.
t=
R. Field 11/21/2013
University of Florida
d
60m
=
≈ 0.173s = 172.9ms
o
v(27 C ) 347.0m / s
PHY 2053
Page 5
Sound Waves: Example Problem
• Stan and Ollie are standing next to a train track. Stan puts his ear to
the steel track to hear the train coming. He hears the sound of the train
whistle through the track 2.1 s before Ollie hears it through the air. How
far away is the train? Take the speed of sound in air and steel to be 343
m/s and 5790 m/s, respectively. Answer: 765.7 m
∆t = tair − t steel
d = ∆t
R. Field 11/21/2013
University of Florida
 1
1 
d
d
v −v

 = d steel air
=
−
= d
−
vair vsteel
vair vsteel
 vair vsteel 
vair vsteel
(343m / s )(5790m / s )
= (2.1s )
≈ 765.7 m
vsteel − vair
(5790m / s ) − (343m / s )
PHY 2053
Page 6
Sound Waves: Example Problem
• You drop a stone into a deep well and hear it hit the bottom 3.2 s later.
How deep is the well? Take the speed of sound to be 343 m/s.
Answer: 46.05 m
2d
d
ttot = t drop + t sound =
+
g vsound
d
2d 
=  ttot −
g 
vsound
2

dttot
d2
2
 = ttot − 2
+ 2
vsound vsound

2

2vsound
d −  2ttot vsound +
g

2

2
2
d + vsound
ttot
=0

2


vsound 
2ttot vsound vsound
 ±
+ 2 
d = vsound  ttot +
g
g
g 



343m / s 
2(3.2s )(343m / s ) (343m / s ) 2 
= (343m / s )  3.2s +
±
+
2 
2
2 2 
9
.
8
m
/
s
9
.
8
m
/
s
(
9
.
8
m
/
s
) 


26,159m
= (343m / s )[(38.2s ) ± (38.06573s )] =
46.05m
R. Field 11/21/2013
University of Florida
PHY 2053
Page 7
Sound Waves: Example Problem
• The sound level 25 m from a loudspeaker is 71 dB. What is the rate at
which sound energy is produced by the loudspeaker, assuming it to be
an isotropic source? Answer: 98.9 mW
β = (10dB) log10 ( I / I 0 )
where I0 = 10-12 W/m2. Hence,
I = I 0 × 10 β /10 dB
P = 4πr 2 I = 4πr 2 I 010 β / 10 dB = 4π (25m) 2 (10 −12 W / m 2 )107.1 ≈ 98.9mW
R. Field 11/21/2013
University of Florida
PHY 2053
Page 8