Physics Solution Manual for 1100 and 2101

The changing velocity is related to the acceleration

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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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