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RrgPJo3rTB5w_1210016326_jwk572

# RrgPJo3rTB5w_1210016326_jwk572 - homework 11 KIM JI Due...

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homework 11 – KIM, JI – Due: Nov 14 2007, 4:00 am 1 Question 1, chap 17, sect 2. part 1 of 1 10 points An explosive charge is detonated at a height of several kilometers in the atmosphere. At a distance of 488 m from the explosion the acoustic pressure reaches a maximum of 14 . 1 N / m 2 . The density of the air is 1 . 4 kg / m 3 and the speed of sound is 343 m / s and sound waves in air are absorbed at a rate of approximately 9 . 52 dB / km. What will be the sound level at a distance 2 . 53 km from the explosion? Consider the atmosphere to be homogeneous over the dis- tances considered. Correct answer: 79 . 426 dB (tolerance ± 1 %). Explanation: Let : λ β = 9 . 52 dB / km , L 1 = 488 m , P m = 14 . 1 N / m 2 , ρ = 1 . 4 kg / m 3 , v = 343 m / s , and L 2 = 2 . 53 km . The intensity at L 1 is I 1 = P 2 m 2 ρ v = (14 . 1 N / m 2 ) 2 2 (1 . 4 kg / m 3 ) (343 m / s) = 0 . 207007 W / m 2 . Since the intensity is inversely proportional to the distance squared, I 2 I 1 = L 2 1 L 2 2 I 2 = I 1 L 2 1 L 2 2 = (0 . 207007 W / m 2 ) (488 m) 2 (2 . 53 km) 2 = 0 . 00770167 W / m 2 . The sound level at L 2 is β 2 = 10 log parenleftbigg I 2 I 0 parenrightbigg = 10 log parenleftbigg 0 . 00770167 W / m 2 1 × 10 12 W / m 2 parenrightbigg = 98 . 8658 dB , so allowing for absorption of the wave over the distance traveled, β 2 = β 2 - λ β ( L 2 - L 1 ) = 98 . 8658 dB - (9 . 52 dB / km) (2 . 53 km - 0 . 488 km) = 79 . 426 dB . Question 2, chap 17, sect 2. part 1 of 1 10 points A rock group is playing in a bar. Sound emerging from the door spreads uniformly in all directions. The intensity level of the music is 102 dB at a distance of 7 . 22 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing? Disregard absorption. Correct answer: 908944 m (tolerance ± 1 %). Explanation: Let : r = 7 . 22 m β = 102 dB , and I 0 = 1 × 10 12 dB . I 1 r 2 , so I I 0 = R 2 0 r 2 . The decibel level is β = 10 log parenleftbigg I I 0 parenrightbigg = 10 log parenleftbigg R 2 0 r 2 parenrightbigg = 20 log parenleftbigg R 0 r parenrightbigg , so R 0 r = 10 β/ 20 R 0 = r 10 β/ 20 = (7 . 22 m) 10 (102 dB) / (20 dB) = 908944 m .

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homework 11 – KIM, JI – Due: Nov 14 2007, 4:00 am 2 Question 3, chap 17, sect 2. part 1 of 1 10 points A source of sound (1000 Hz) emits uni- formly in all directions. An observer 4 . 18 m from the source measures a sound level of 62 . 3 dB. Calculate the average power output of the source. Correct answer: 372 . 874 μ W (tolerance ± 1 %). Explanation: Let : β = 62 . 3 dB , I 0 = 1 × 10 12 W / m 2 , and L = 4 . 18 m . The intensity of the sound is I = I 0 10 β/ 10 = (1 × 10 12 W / m 2 ) 10 (62 . 3 dB) / (10 dB) = 1 . 69824 × 10 6 W / m 2 so the average power output is P = 4 π L 2 I = 4 π (4 . 18 m) 2 (1 . 69824 × 10 6 W / m 2 ) = 0 . 000372874 W = 372 . 874 μ W . Question 4, chap 17, sect 2. part 1 of 1 10 points Two sources have sound levels of 65 . 8 dB and 114 dB. What is their combined intensity? Correct answer: 0 . 251192 W / m 2 (tolerance ± 1 %). Explanation: Let : β 1 = 65 . 8 dB and β 2 = 114 dB . The intensities of the sources are I 1 = I 0 10 β 1 / 10 = 1 × 10 12 W / m 2 10 (65 . 8 dB) / (10 dB) = 3 . 80189 × 10 6 W / m 2 I 2 = I 0 10 β 2 / 10 = 1 × 10 12 W / m 2 10 (114 dB) / (10 dB) = 0 . 251189 W / m 2 .
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RrgPJo3rTB5w_1210016326_jwk572 - homework 11 KIM JI Due...

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