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Unformatted text preview: B ) log I
= 10(115 dB ) /(10 dB) = 1011.5
I0 or Solving for I gives ( ) I = I 01011.5 = 1.00 × 10 –12 W/m 2 1011.5 = 0.316 W/m 2 ______________________________________________________________________________ 890 WAVES AND SOUND 90. REASONING AND SOLUTION
a. The travel time is
t = x/v = (2.70 m)/(343 m/s) = 7.87 × 10 –3 s
b. The wavelength is λ = v/f = (343 m/s)/(523 Hz) = 0.656 m (16.1) The number of wavelengths in 2.70 m is, therefore,
Number of wavelengths = (2.70 m)/(0.656 m) = 4.12
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91. SSM REASONING AND SOLUTION The speed of sound in a liquid is given by Equation 16.6, v = Bad / ρ , where Bad is the adiabatic bulk modulus and ρ is the density of
the liquid. Solving for Bad, we obtain Bad = v ρ . Values for the speed of sound in fresh
water and in ethyl alcohol are given in Table 16.1. The ratio of the adiabatic bulk modulus
of fresh water to that of ethyl alcohol at 20°C is, therefore,
2 2
( Bad )water
vwater ρ water
(1482 m/s) 2 (998 kg/m3 )
=2
=
=
( Bad )ethyl alcohol vethyl alcohol ρethyl alcohol (1162 m/s)2 (789 kg/m3 ) 2.06 ______________________________________________________________________________
92. REASONING The observed frequency changes because of the Doppler effect. As you
drive toward the parked car (a stationary source of sound), the Doppler effect is that given
by Equation 16.13. As you drive away from the parked car, Equation 16.14 applies. SOLUTION Equations 16.13 and 16.14 give the observed frequency fo in each case: fo, toward = fs (1 + vo /v )
1444 24444
4
3
Driving toward parked car and fo, away = f (1 – vo /v )
1444 s
2444
3
Driving away from parked car Subtracting the equation on the right from the one on the left gives the change in the
observed frequency:
fo, toward – fo, away = 2 fs vo /v
Solving for the observer’s speed (which is your speed), we obtain Chapter 16 Problems vo = ( v fo, toward – fo, away 891 ) = (343 m/s )(95 Hz ) = 17 m/s
2 fs
2 ( 960 Hz )
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93. SSM WWW REASONING According to Equation 16.2, the linear density of the string
is given by (m / L ) = F / v 2 , where the speed v of waves on the middle C string is given by
1
Equation 16.1, v = f λ = λ , where T is the period.
T SOLUTION Combining Equations 16.2 and 16.1 and using the given data, we obtain F FT 2 (944 N)(3.82 ×10 –3 s)2
= 2=
= 8.68 × 10 –3 kg/m
v2
λ
(1.26 m) 2
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m/ L = 94. REASONING AND SOLUTION
a. Since 15 boxcars pass by in 12.0 s, the boxcars pass by with a frequency of f= 15
= 1.25 Hz
12.0 s b. Since the length of a boxcar corresponds to the wavelength λ of a wave, we have, from
Equation 16.1, that
v = f λ = (1.25 Hz ) (14.0 m ) = 17.5 m/s
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95. SSM REASONING This is a situation in which the intensities Iman and Iwoman (in watts
per square meter) detected by the man and the woman are compared using the intensity level
β, expressed in decibels. This comparison is based on Equation 16.10, which we rewrite as
follows:
I β = (10 dB ) log man I woman SOLUTION Using Equation 16.10, we have I man I woman β = 7.8 dB = (10 dB ) log Solving for the intensity ratio gives or I
log man
I woman 7.8 dB
= 0.78
= 10 dB 892 WAVES AND SOUND I man = 100.78 = 6.0
I woman
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96. REASONING AND SOLUTION We have I = P/A. Therefore,
P = IA = (3.2 × 10–6 W/m2)(2.1 × 10–3 m2) = 6.7 × 10 –9 W (16.8) ______________________________________________________________________________
97. SSM REASONING When the end of the Slinky is moved up and down continuously, a
transverse wave is produced. The distance between two adjacent crests on the wave, is, by
definition, one wavelength. The wavelength λ is related to the speed and frequency of a
periodic wave by λ = v / f (Equation 16.1). In order to use Equation 16.1, we must first
determine the frequency of the wave. The wave on the Slinky will have the same frequency
as the simple harmonic motion of the hand. According to the data given in the problem
statement, the frequency is f = (2.00 cycles)/(1 s) = 2.00 Hz .
SOLUTION Substituting the values for λ and f, we find that the distance between crests is v 0.50 m/s
=
= 0.25 m
f
2.00 Hz
______________________________________________________________________________ λ= 98. REASONING AND SOLUTION Using Equation 16.1, we find that λ = v/f = (343 m/s)/(4185.6 Hz) = 8.19 × 10 –2 m
(16.1)
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99. SSM REASONING Since you detect a frequency that is smaller than that emitted by the
car when the car is stationary, the car must be moving away from you. Therefore, according
to Equation 16.12, the frequency fo heard by a stationary observer from a source moving
a...
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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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