Unformatted text preview: of 350, so the sound
intensity Iwithout that the crew member would experience without the earplugs is 350 times
larger than the sound intensity Iwith experienced with the earplugs in place:
I without = 350 I with (1) The sound intensity levels βwithout and βwith corresponding to these intensities are found
I
from β = (10 dB ) log (Equation 16.10), where I0 is the reference level sound intensity.
I 0 SOLUTION According to Equation 16.10, the sound intensity level β without is I without I0 (2) 350 I with I0 (3) β without = (10 dB ) log Substituting Equation (1) into Equation (2) yields β without = (10 dB ) log 876 WAVES AND SOUND It is convenient to express the quantity
350 I with
I0 I
= 350 with
I
0 350 I with in Equation (3) as a product: I0 . This permits us to put Equation (3) into a more useful form: I with I with = (10 dB ) log 350 + log I I0 0 β without = (10 dB ) log 350 I
= (10 dB ) log 350 + (10 dB ) log with
I
0 (4) I The term (10 dB ) log with in Equation (4) is the sound intensity level βwith, as we see
I 0
from Equation 16.10. Therefore, the sound intensity level experienced without earplugs is β without = (10 dB ) log 350 + β with = (10 dB ) log 350 + 88 dB = 113 dB 71. SSM REASONING If I1 is the sound intensity produced by a single person, then N I1 is
the sound intensity generated by N people. The sound intensity level generated by N people
is given by Equation 16.10 as NI β N = (10 dB ) log 1 I 0 where I0 is the threshold of hearing. Solving this equation for N yields
βN I N = 0 1010 dB
I 1 (1) We also know that the sound intensity level for one person is
I β1 = (10 dB ) log 1 I 0 β1 or I1 = I 0 1010 dB (2) Equations (1) and (2) are all that we need in order to find the number of people at the
football game. Chapter 16 Problems 877 SOLUTION Substituting the expression for I1 from Equation (2) into Equation (1) gives
the desired result.
βN
10 dB N= I 0 10 β1 = 10 109 dB
10 dB 60.0 dB
10 dB = 79 400 I 0 1010 dB 10
______________________________________________________________________________ 72. REASONING The energy incident on the eardrum is equal to the sound power P (which is
assumed to be constant) times the time interval t (see Equation 6.10b): Energy = Pt (1) According to Equation 16.8, the sound power is equal to the intensity I of the wave times the
area A of the eardrum through which the power passes, or P = I A . Substituting this relation
into Equation (1) gives
(2)
Energy = P t = ( I A ) t
We can obtain an expression for the sound intensity by employing the definition of the
sound intensity level β (Equation 16.10)
I I0 β = (10 dB ) log where I0 is the threshold of hearing. Solving this expression for I gives
β I = I 010 10 dB Substituting this relation for I into Equation (2), we have that
β I 1010 dB At
Energy = ( I A ) t = 0 SOLUTION Noting that I0 = 1.00 × 10−12 W/m2 and that t = 9.0 h = 3.24 × 104 s, the
energy incident on the eardrum is
β 10 dB I 10
Energy = 0 At
90.0 dB −12
2)
−4
2
4
−3
10 dB (
(1.00 × 10
=
W/m 10 2.0 × 10 m )( 3.24 × 10 s ) = 6.5 × 10 J
______________________________________________________________________________ 878 WAVES AND SOUND 73. REASONING Suppose we knew the sound intensity ratio
knew the sound intensity ratio
ratio IC
IB IC
IA IA
. In addition, suppose we
IB . Then, all that would be necessary to obtain the desired would be to multiply the two known ratios: IA I
B I C I A IC
= IB This is exactly the procedure we will follow, except that first we will obtain the intensity
ratios from the given intensity levels (expressed in decibels). The intensity level β in
decibels is given by Equation 16.10.
SOLUTION Applying Equation 16.10 to persons A and B gives IA IB β A/B = 1.5 dB = (10 dB) log I 1.5 dB
log A = I 10 dB = 0.15 B IA
= 100.15
IB or In a similar fashion for persons C and A we obtain IC IA β C/A = 2.7 dB = (10 dB ) log I 2.7 dB
log C = I 10 dB = 0.27 A IC or IA = 100.27 Multiplying the two intensity ratios reveals that I I = A C = 100.15 100.27 = 100.42 = 2.6 I B I B I A IC ( )( ) Chapter 16 Problems 879 74. REASONING AND SOLUTION The intensity level at each point is given by
I= P
4π r 2 Therefore, r = 2 I 2 r1 I1 2 Since the two intensity levels differ by 2.00 dB, the intensity ratio is I1
= 100.200 = 1.58
I2
Thus,
2 r2 = 1.58
r 1
We also know that r2 – r1 = 1.00 m. We can then solve the two equations simultaneously by
substituting, i.e., r2 = r1 1.58 gives
r1 1.58 – r1 = 1.00 m so that
r1 = (1.00 m)/[ 1.58 – 1] = 3.9 m
and
r2 = 1.00 m + r1 = 4.9 m
______________________________________________________________________________ 75. SSM REASONING AND SOLUTION The sound intensity level β in decibels (dB) is related to the sound intensity I according to Equation 16.10, β = (10 dB) log ( I / I 0 ) , where
the quantity I0 is the reference intensity. According to the problem statement, when the
sound intensity level triples, the sound intensity also triples; therefore, 3I
3β = (10 dB ) log I
0 Then, 3I I 3I / I 0 3β – β = (10 dB ) log –...
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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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