Physics Solution Manual for 1100 and 2101

# Therefore the expression for the doppler shifted

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Unformatted text preview: I I /I ( 1 I0 0 1 0 β 2 − β1 = (10 dB) log Therefore, I2 I1 = 10( β 2 – β1 ) / (10 dB) When the difference in sound intensity levels is β 2 – β1 = 1.0 dB , the ratio of the sound intensities is I2 = 10(1 dB) /(10 dB) = 1.3 I1 ______________________________________________________________________________ 872 WAVES AND SOUND 64. REASONING a. The intensity I1 of the sound from the motor when she is a distance d away is given by P (Equation 16.9), where P is the total sound power emitted by the motor. As she I1 = 4π d 2 moves away from the motor, its total sound power P does not change. We will use this fact, together with Equation 16.9, to determine the sound intensity I2 when she is a distance 2d from the motor. b. Once the sound intensity I2 at the woman’s final position has been determined, the sound I intensity level β2 at that position will be found from β = (10 dB ) log (Equation 16.10), I0 where I0 is the reference sound intensity, in this case the threshold of hearing. SOLUTION a. Applying Equation 16.9 to the woman’s second position, a distance 2d from the motor, we find that P P (1) I2 = = 2 16π d 2 4π ( 2 d ) Solving I1 = P (Equation 16.9) for the total sound power P of the motor, we find that 4π d 2 P = 4π d 2 I1 (2) Substituting Equation (2) into Equation (1) yields I2 = 4 π d 2 I1 I1 3.2 ×10−3 W/m 2 P = == = 8.0 ×10−4 W/m 2 4 4 16 π d 2 16 π d 2 b. From Equation 16.10, the sound intensity level at a position twice as far from the motor is I 8.0 × 10 −4 W/m 2 β 2 = (10 dB ) log 2 = (10 dB ) log = 89 dB 1.0 × 10 −12 W/m 2 I0 where we have used I0 = 1.0×10−12 W/m2 for the threshold of hearing. Chapter 16 Problems 65. 873 SSM REASONING AND SOLUTION The intensity level β in decibels (dB) is related to the sound intensity I according to Equation 16.10: I I0 β = (10 dB ) log where the quantity I0 is the reference intensity. Therefore, we have I2 I I 2 / I0 I − (10 dB ) log 1 = (10 dB ) log = (10 dB ) log 2 I 1 I0 I0 I1 / I 0 β 2 − β1 = (10 dB ) log Solving for the ratio I2 / I1 , we find I 30.0 dB = (10 dB ) log 2 I 1 or I2 = 10 3.0 = 1000 I1 Thus, we conclude that the sound intensity increases by a factor of 1000 . ______________________________________________________________________________ 66. REASONING The sound intensity level outside the room is I outside I0 β outside = (10 dB ) log (16.10) where I0 is the threshold of hearing. Solving for the intensity Ioutside gives βoutside I outside = I 010 10 dB These two relations will allow us to find the sound intensity outside the room. SOLUTION From the problem statement we know that βoutside = βinside + 44.0 dB. We can evaluate β inside by applying Equation 16.10 to the inside of the room: I inside 1.20 × 10−10 W/m2 = (10 dB ) log −12 2 = 20.8 dB 1.00 × 10 W/m I0 βinside = (10 dB ) log Thus, the sound intensity level outside the room is β outside = 20.8 dΒ + 44.0 dB = 64.8 dB. The sound intensity outside the room is 874 WAVES AND SOUND βoutside I outside = I 010 10 dB ( = 1.00 × 10 −12 2 ) W/m 10 64.8 dB 10 dB = 3.02 × 10−6 W/m 2 ______________________________________________________________________________ 67. REASONING The intensity level β is related to the sound intensity I according to Equation 16.10: I β = (10 dB ) log I 0 where I0 is the reference level sound intensity. Solving for I gives β I = I 010 10 dB SOLUTION Taking the ratio of the sound intensity Irock at the rock concert to the intensity Ijazz at the jazz fest gives β rock I rock I jazz = I 0 10 I 0 10 115 dB 10 dB 10 dB β jazz 10 dB = 10 10 95 dB 10 dB = 1.0 ×102 ____________________________________________________________________________________________ 68. REASONING A threshold of hearing of –8.00 dB means that person 1 can hear a sound whose intensity is less than I0 = 1.00 × 10−12 W/m2, which is the intensity of the reference level. Person 2, with a threshold of hearing of +12.0 dB, can only hear sounds that have intensities greater than I0 = 1.00 × 10−12 W/m2. Thus, person 1 has the better hearing, because he can hear sounds with intensities less than 1.00 × 10−12 W/m2. We expect, then that the ratio I1/I2 is less than one. SOLUTION The relation between the sound intensity level β and the sound intensity I is given by Equation 16.10: β I β = (10 dB ) log or I = I 0 1010 dB I 0 The threshold-of-hearing intensities for the two people are Chapter 16 Problems β1 875 β2 I1 = I 0 1010 dB and I 2 = I 0 1010 dB Taking the ratio I1/ I2 gives β1 10 dB −8.00 dB 10 dB I 10 I1 10 −2 =0 = = 1.00 ×10 β2 +12.0 dB I2 10 10 dB I 0 1010 dB This answer is less than one, as expected. ______________________________________________________________________________ 69. REASONING AND SOLUTION a. PA PB β = (10 dB) log 250 W = 7.4 dB = (10 dB) log 45 W b. No , A will not be twice as loud as B, since it requires an increase of 10 dB to double the loudness. ______________________________________________________________________________ 70. REASONING The earplugs reduce the sound intensity by a factor...
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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.

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