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PHYS2214, Spring 2009
Solutions to Assignment 7
March 23, 2009
1 Energy and power in sound waves
a) As was derived in class, the average intensity of a sound wave is related to the molecular displacement amplitude
s
m
and the
pressure amplitude
p
m
by
I
av
=
1
2
Bkωs
2
m
(1)
=
B
2
v
ω
2
s
2
m
(2)
=
v
2
B
p
2
m
(3)
Therefore for a ﬁxed average intensity
s
m
=
r
2
vI
av
B
1
ω
(4)
∝
1
ω
(5)
and
p
m
=
r
2
BI
av
v
(6)
∝
ω
0
(7)
b) We have 2
ω
= 1 KHz sound waves, one propogating in water (
B
w
= 2
.
2
×
10
9
Pa and
v
w
= 1480 m/s) and one in air
(
B
a
= 1
.
4
×
10
5
Pa and
v
a
= 343 m/s), having equal
I
av
. Using eqn.[4], we see that the ratio of the molecular diplacements is
s
w
m
s
a
m
=
r
v
w
B
a
v
a
B
w
(8)
= 0
.
0166
(9)
Likewise from eqn.(6), we have
p
w
m
p
a
m
=
r
v
a
B
w
v
w
B
a
(10)
= 60
.
35
(11)
1
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View Full Document We see that
p
w
m
p
a
m
=
s
a
m
s
w
m
→
p
w
m
s
w
m
=
p
a
m
s
a
m
. Since also,
I
av
=
1
2
ωp
m
s
m
(12)
and
ω
’s are identical, this
⇒
I
w
av
=
I
a
av
(13)
c) The percieved loudness (
β
) of a soundwave is related to it’s intensity by
β
= 10 log
10
I
I
0
(14)
where
I
0
= 10

12
W
/
m
2
, is the threshold of hearing.
β
is expressed in decibels (dB).
1) We have a 1 W point source of waves, i.e the total energy crossing a sphere of radius
r
about this point, per sec is 1 J. The
intensity, which is power/unit area is hence
I
(
r
) =
1
4
πr
2
W
/
m
2
(15)
The diﬀerence in loudness between a point a distance
r
from the source and one that is twice as far away is
δβ
=
β
(
r
)

β
(2
r
)
(16)
= 10(log
10
I
(
r
)
I
0

log
10
I
(2
r
)
I
0
)
(17)
= 10 log
10
I
(
r
)
I
(2
r
)
(18)
= 10 log
10
4
(19)
= 6
.
02
(20)
The initial ditance from the source does not aﬀect the change in loudness.
2) The intensity as a function of distance is as in eqn.[13]. The sound level as a function of distance is
β
(
r
) = 10 log
10
I
(
r
)
I
0
(21)
= 10 log
10
±
1
4
πr
2
10

12
²
(22)
= 10 log
10
±
10
12
4
πr
2
²
(23)
(24)
We invert the above expression to get the distance at which the intensity is a particular value.
⇒
r
(
β
) =
s
1
4
π
10
12
10
β
(
r
)
10
(25)
r
(120dB) = 0
.
28
m
(26)
r
(65dB) = 158
.
63
m
(27)
r
(20dB) = 28209
m
(28)
(29)
2
120 dB is the threshold of pain. This says that you are at the threshold of pain if you are 28 cm away from a speaker that emits 1 W
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This note was uploaded on 04/01/2009 for the course PHYS 2214 taught by Professor Giambattista,a during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 GIAMBATTISTA,A
 Energy, Power

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