Unformatted text preview: way from the observer is given by
1
f o = fs v
1+ s
v where fs is the frequency emitted from the source when it is stationary with respect to the
observer, v is the speed of sound, and vs is the speed of the moving source. This expression
can be solved for vs . Chapter 16 Problems 893 SOLUTION We proceed to solve for vs and substitute the data given in the problem
statement. Rearrangement gives
vs fs
–1
=
v
fo Solving for vs and noting that f o / fs = 0.86 yields f 1 vs = v s –1 = (343 m/s) –1 = 56 m/s
f 0.86 o ______________________________________________________________________________
100. REASONING AND SOLUTION The speed of sound in an ideal gas is given by
Equation 16.5, v = γ kT / m . The ratio of the speed of sound v2 in the container (after the
temperature change) to the speed v1 (before the temperature change) is v2
=
v1 T2
T1 Thus, the new speed is
T2 405 K
= 1730 m/s
201 K
T1
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v 2 = v1 = (1220 m/s) 101. SSM REASONING According to Equation 16.10, the sound intensity level β in decibels
(dB) is related to the sound intensity I according to β = (10 dB) log ( I / I 0 ) , where the
quantity I0 is the reference intensity. Since the sound is emitted uniformly in all directions, the intensity, or power per unit area, is given by I = P /(4π r 2 ) . Thus, the sound intensity at
position 1 can be written as I1 = P /(4π r12 ) , while the sound intensity at position 2 can be
written as I 2 = P /(4π r22 ) . We can obtain the sound intensity levels from Equation 16.10
for these two positions by using these expressions for the intensities.
SOLUTION Using Equation 16.10 and the expressions for the intensities at the two
positions, we can write the difference in the sound intensity levels β21 between the two
positions as follows: 894 WAVES AND SOUND I2 I1 − (10 dB ) log I I0 0 β 21 = β 2 − β1 = (10 dB) log I /I I = (10 dB ) log 2 0 = (10 dB ) log 2 I I /I 1 1 0
2 P /(4π r2 ) r2 r
β 21 = (10 dB ) log = (10 dB ) log 12 = (10 dB ) log 1
2
r r P /(4π r1 ) 2
2 r
= ( 20 dB ) log 1
r
2 2 r = ( 20 dB ) log 1 = ( 20 dB ) log (1/ 2 ) = –6.0 dB 2r 1 The negative sign indicates that the sound intensity level decreases.
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I
The intensity level in dB is β = (10 dB)log , I0 Equation 16.10, where β = 14 dB. Therefore, 102. REASONING AND SOLUTION β I
= 1010 dB = 101.4 = 25
I0
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103. REASONING The sound intensity level heard by the gardener increases by 10.0 dB
because distance between him and the radio decreases. The intensity of the sound is greater
when the radio is closer than when it is further away. Since the unit is emitting sound
uniformly (neglecting any reflections), the intensity is inversely proportional to the square
of the distance from the radio, according to Equation 16.9. Combining this equation with
Equation 16.10, which relates the intensity level in decibels to the intensity I, we can use the
10.0dB change in the intensity level to find the final vertical position of the radio. Once
that position is known, we can then use kinematics to determine the fall time.
SOLUTION Let the sound intensity levels at the initial and final positions of the radio be βi and βf , respectively. Using Equation 16.10 for each, we have If I f / I0 I = 10 dB ) log − (10 dB ) log i = (10 dB ) log I I /I ( I0 0 i 0 β f − βi = (10 dB ) log If I
i (1) Chapter 16 Problems 895 Since the radiation is uniform, Equation 16.9 can be used to substitute for the intensities Ii
and If , so that Equation (1) becomes If
β f − βi = (10 dB ) log I
i ( 2 P / 4π h f = (10 dB ) log 2 P / 4π hi ( ) = (10 dB) log h ) 2 h
i
= ( 20 dB ) log i hf h f Since β f – βi = 10.0 dB, Equation (2) gives
h
i hf β f − βi = 10.0 dB = ( 20 dB ) log or hi
(10.0 dB) / ( 20 dB )
0.500
= 10
= 10
= 3.16
hf We can now determine the final position of the radio as follows:
hi
hf = 5.1 m
= 3.16
hf or hf = 5.1 m
= 1.61 m
3.16 It follows, then, that the radio falls through a distance of 5.1 m – 1.61 m = 3.49 m. Taking
upward as the positive direction and noting that the radio falls from rest (v0 = 0 m/s), we can
1
solve Equation 2.8 y = v0t + at 2 from the equations of kinematics for the fall time t: 2 2 (–3.49 m )
2y
=
= 0.84 s
a
–9.80 m/s 2
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t= 104. REASONING AND SOLUTION The maximum acceleration of the dot occurs at the
extreme positions and is
amax = ( 2π f 2
)2 A = 2π ( 4.0 Hz ) ( 5.4 × 10 −3 m ) = 3.4 m/s 2 (10.10) ______________________________________________________________________________
105. SSM WWW REASONING Using the procedures developed in Chapter 4 for using
Newton's second law to analyze the motion of bodies and neglecting the weight of the wire
relative to the tension in the wire lead to the following equations of motion for th...
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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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