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Unformatted text preview: (10 dB ) log = (10 dB) log I I 0 0 I / I0 Thus, 2 β = (10 dB ) log 3 and β = ( 5 dB ) log 3 = 2.39 dB 880 WAVES AND SOUND ______________________________________________________________________________
76. REASONING The observer of the sound (the birdwatcher) is stationary, while the source
(the bird) is moving toward the observer. Therefore, the Dopplershifted observed
frequency is given by Equation 16.11. This expression can be solved to give the ratio of the
bird’s speed to the speed of sound, from which the desired percentage follows directly.
SOLUTION According to Equation 16.11, the observed frequency fo is related to the
frequency fs of the source, and the ratio of the speed of the source vs to the speed of sound v
by
1
fo
fs
v
1
fo = fs =
= 1− s
or
or
1− v / v fs 1 − vs / v
fo
v
s Solving for vs/v gives vs
v = 1− fs
fo = 1− 1250 Hz
= 0.031
1290 Hz This ratio corresponds to 3.1% . 77. SSM REASONING You hear a frequency fo that is 1.0% lower than the frequency fs
emitted by the source. This means that the frequency you observe is 99.0% of the emitted
frequency, so that fo = 0.990 fs. You are an observer who is moving away from a stationary
source of sound. Therefore, the Dopplershifted frequency that you observe is specified by
Equation 16.14, which can be solved for the bicycle speed vo.
SOLUTION Equation 16.14, in which v denotes the speed of sound, states that v
f o = fs 1 − o v Solving for vo and using the fact that fo = 0.990 fs reveal that 0.990 fs f
vo = v 1 − o = ( 343 m/s ) 1 − = 3.4 m/s fs fs ______________________________________________________________________________
78. REASONING The dolphin is the source of the clicks, and emits them at a frequency fs. The
marine biologist measures a lower, Dopplershifted click frequency fo, because the dolphin
is swimming directly away. The difference between the frequencies is the source frequency Chapter 16 Problems 881 1
minus the observed frequency: fs − fo. We will use f o = f s (Equation 16.12), where
vs 1+ v vs is the speed of the dolphin and v is the speed of sound in seawater, to determine the
difference between the frequencies.
1
SOLUTION Solving f o = f s v 1+ s
v fs, we obtain (Equation 16.12) for the unknown source frequency v
fs = f o 1 + s v Therefore, the difference between the source and observed frequencies is v v
fs − f o = fo 1 + s − f o = f o 1 + s − 1
v
v v 8.0 m/s = fo s = ( 2500 Hz ) = 13 Hz
v 1522 m/s 79. REASONING This problem deals with the Doppler effect in a situation where the source of
the sound is moving and the observer is stationary. Thus, the observed frequency is given
by Equation 16.11 when the car is approaching the observer and Equation 16.12 when the
car is moving away from the observer. These equations relate the frequency fo heard by the
observer to the frequency fs emitted by the source, the speed vs of the source, and the speed
v of sound. They can be used directly to calculate the desired ratio of the observed
frequencies. We note that no information is given about the frequency emitted by the
source. We will see, however, that none is needed, since fs will be eliminated algebraically
from the solution.
SOLUTION Equations 16.11 and 16.12 are 1
foApproach = fs 1− v / v s The ratio is (16.11) 1
foRecede = fs 1 + vs / v (16.12) 882 WAVES AND SOUND f oApproach
f oRecede 1
9.00 m/s
fs 1− v / v 1+ v / v 1+ s
=
343 m/s = 1.054
s
=
= 1 − vs / v 1 − 9.00 m/s
1
fs 343 m/s 1+ v / v s As mentioned in the REASONING, the unknown source frequency fs has been eliminated
algebraically from this calculation. 80. REASONING Both the observer (you) and the source (the eagle) are moving toward each
other. According to Equation 16.15, the frequency fo heard by the observer is related to the
frequency fs of the sound emitted by the source by vo 1+ v
f o = fs 1 − vs v where vo and vs are, respectively, the speeds of the observer and source. SOLUTION Substituting in the given data, we find that the frequency heard by the observer
is vo 39 m/s 1+ v 1 + 330 m/s f o = fs = 4.0 ×103 Hz = ( 3400 Hz ) vs 18 m/s 1− 1− 330 m/s v ______________________________________________________________________________
81. SSM REASONING AND SOLUTION The speed vs of the Bungee jumper after she has
fallen a distance y can be obtained from Equation 2.9:
2
2
vs = v0 + 2ay Since she falls from rest, it follows that v0 = 0 m/s. Therefore, taking upward as the positive
direction, we have ( ) vs = 2ay = 2 −9.80 m/s 2 ( −11.0 m ) = 14.7 m/s Chapter 16 Problems 883 Then, from Equation 16.11 1
1 fo = fs = (589 Hz) = 615 Hz . 1− v / v 1 − (14.7 m/s ) / ( 343 m/s ) s ______________________________________________________________________________
82. REASONING If the train were headed directly towards the car, the frequency fo heard by
1 (Equation 16.11), where f is the frequency
the driver would be given by f o = fs s vtrain 1− v of the train’s horn, v is the speed of sound in air, and vtrain is the speed of the train. However,
the driver is 20.0 m south of the crossing, and at the instant when the horn is sounded, the
train is 20.0 m west of the crossing. The train would
vtrain
Train
have to be headed directly southeast (45° south of east)
45°
in order to be moving directly...
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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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