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Unformatted text preview: the same sound power passes through patches 1
and 2, and the intensity at each one is the same, their areas must also be the same, A1 = A2.
The same sound power passes through patch 3, but the intensity at that surface is smaller
than that at patches 1 and 2. Thus, the area A3 of patch 3 is larger than that of surface 1 or 2.
In summary, A3 is the largest area, followed by A1 and A2, which are equal.
SOLUTION
a. The sound intensity at the inner spherical surface is given by Equation 16.9 as IA = P
2.3 W
2
=
= 0.51 W/m
2
2
4 π rA 4 π ( 0.60 m ) This intensity is the same at all points on the inner surface, since all points are equidistant
from the sound source. Therefore, the sound intensity at patches 1 and 2 are equal;
I1 = I2 = 0.51 W/m 2 .
The sound intensity at the outer spherical surface is
IB = P
2.3 W
2
=
= 0.29 W/m
2
2
4 π rB 4 π ( 0.80 m ) 866 WAVES AND SOUND This intensity is the same at all points on outer surface. Therefore, the sound intensity at
patch 3 is I3 = 0.29 W/m 2 .
b. The area of a patch is given by Equation 16.8 as the sound power passing perpendicularly
through that area divided by the sound intensity, A = P/I. The areas of the three patches are:
−3 Patch 1 A1 = P 1.8 × 10 W
=
= 3.5 × 10−3 m 2
I1 0.51 W/m 2 Patch 2 A2 = P 1.8 × 10−3 W
−3
2
=
= 3.5 × 10 m
2
I2
0.51 W/m Patch 3 P 1.8 × 10−3 W
−3
2
A3 = =
= 6.2 × 10 m
2
I 3 0.29 W/m These answers are consistent with our discussion in the REASONING.
______________________________________________________________________________
55. REASONING AND SOLUTION Since the sound spreads out uniformly in all directions,
the intensity is uniform over any sphere centered on the source. From text Equation 16.9,
P
I=
4π r 2
Then,
I1 P /(4π r12 ) r22
=
=
I 2 P /(4π r22 ) r12
Solving for r2, we obtain
I1
2.0 × 10−6 W/m 2
= (120 m )
= 190 m
I2
0.80 ×10−6 W/m 2
______________________________________________________________________________
r2 = r1 56. REASONING AND SOLUTION The intensity of the "direct" sound is given by text
Equation 16.9,
P
I DIRECT =
4π r 2
The total intensity at the point in question is
ITOTAL = IDIRECT + IREFLECTED 1.1 × 10−3 W + 4.4×10−6 W/m 2 = 1.4 ×10 –5 W/m2
=
2 4π (3.0 m) ______________________________________________________________________________ Chapter 16 Problems 57. 867 SSM REASONING AND SOLUTION According to Equation 16.8, the power radiated
by the speaker is P = IA = Iπ r 2 , where r is the radius of the circular opening. Thus, the
radiated power is
P = (17.5 W/m 2 )(π )(0.0950 m) 2 = 0.496 W
As a percentage of the electrical power, this is 0.496 W
× 100 % = 1.98 %
25.0 W
______________________________________________________________________________
The sound intensity I a distance r from a source broadcasting sound
P
(Equation 16.9), where P is the sound
uniformly in all directions is given by I =
4π r 2
power of the source. The total sound intensity Itot that the man hears at either position is the
sum of the intensities I1 and I2 due to the two speakers. 58. REASONING SOLUTION
a. When the man is halfway between the speakers, his distance r from either speaker is half
the 30.0m distance between the speakers: r = 15.0 m. The total sound intensity at that
P
position is, therefore, twice the intensity I =
(Equation 16.9) of either speaker alone:
4π r 2
P
0.500 W
P
I tot = 2 =
=
= 3.54 × 10 −4 W/m 2
2
2
2 4π r 2π r
2π (15.0 m ) b. In part (a), the man is r = 15.0 m from either speaker. After he has walked 4.0 m towards
speaker 1, his distance from that speaker is r1 = 15.0 m − 4.0 m = 11.0 m, and his distance
from speaker 2 is r2 = 15.0 m + 4.0 m = 19.0 m. The total sound intensity at his final
position is, therefore,
I tot = I1 + I 2 = P
P
P1
1 0.500 W 1
1 +
=
+ 2 + 2= 2
2
2
2
r 4π
4π r1 4π r2 4π 1 r2 (19.0 m ) (11.0 m ) = 4.39 × 10−4 W/m 2 868 WAVES AND SOUND 59. REASONING According to Equation 16.8, the intensity I of a sound wave is equal to the
sound power P divided by the area A through which the power passes; I = P/A. The area is
that of a circle, so A = π r2, where r is the radius. The power, on the other hand, is the energy
E per unit time, P = E/t, according to Equation 6.10b. Thus, we have
E
P
I= = t2
A πr All the variables in this equation are known except for the time.
SOLUTION Solving the expression above for the time yields t= E
4800 J
=
2
Iπ r
5.9 × 103 W/m 2 π 1.8 × 10−2 m ( )( ) 2 = 8.0 × 102 s ______________________________________________________________________________
60. REASONING
Because
d
source 1 (the source at the
origin) is more powerful than
x
d−x
source 2, both locations of
equal sound intensity will be
A
B
closer to source 2 than to Source 1
Source 2
source 1. One point, A, is
between the sources, and the other, B, is on the opposite side of source 2 (see the drawing).
Let x be the distance from the source 1 to point A, and d = 123 m be the distance between
the sources. The intensities I1 and I2 of the sound from the s...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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