HW13 - homework 13 GAUTHIER, JOSEPH Due: Dec 4 2007, 1:00...

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Unformatted text preview: homework 13 GAUTHIER, JOSEPH Due: Dec 4 2007, 1:00 am 1 Question 1, chap 17, sect 2. part 1 of 1 10 points A rifle is fired in a valley with parallel ver- tical walls. The echo from one wall is heard 1 . 37 s after the rifle was fired. The echo from the other wall is heard 3 . 64 s after the first echo. The velocity of sound is v = 343 m / s . How wide is the valley? Correct answer: 1094 . 17 m (tolerance 1 %). Explanation: Let : t 1 = 1 . 37 s , t 2 = 3 . 64 s , and v = 343 m / s . Let the distance from the closer wall be d 1 and the distance from the farther wall be d 2 . The first echo is heard after traveling a distance of 2 d 1 = v t 1 . The second echo is heard after traveling a distance of 2 d 2 = v ( t 1 + t 2 ). Each time the sound traveled from your position to the wall and back to you. 2 d = v t 1 + v ( t 1 + t 2 ) = v (2 t 1 + t 2 ) Thus the distance d between these walls is d = v bracketleftBig 2 t 1 + t 2 bracketrightBig 2 = (343 m / s) bracketleftBig 2 (1 . 37 s) + (3 . 64 s) bracketrightBig 2 = 1094 . 17 m . Question 2, chap 17, sect 2. part 1 of 1 10 points A rock band plays at a 72 dB sound level. How many times greater is the sound pressure from another rock band playing at 127 dB? Correct answer: 562 . 341 (tolerance 1 %). Explanation: Let : 1 = 72 dB and 2 = 127 dB . For a given sound level, every increase of 20 dB corresponds to a pressure amplitude 10 times larger than the original pressure ampli- tude, so n = 10 ( 2 1 ) / 20 = 10 (127 dB 72 dB) / (20 dB) = 562 . 341 . Question 3, chap 17, sect 2. part 1 of 1 10 points A stereo speaker is placed between two ob- servers who are 121 m apart, along the line connecting them. If one observer records an intensity level of 24 . 6 dB, and the other records an intensity level of 29 . 6 dB, how far is the speaker from the closer observer? Correct answer: 43 . 5521 m (tolerance 1 %). Explanation: Let : 1 = 24 . 6 dB , 2 = 29 . 6 dB , and L = 121 m . I r 2 , so I 2 I 1 = r 2 1 r 2 2 . 1 = 10 log parenleftbigg I 1 I parenrightbigg and 2 = 10 log parenleftbigg I 2 I parenrightbigg , so 2 1 = 10 log parenleftbigg I 2 I 1 parenrightbigg = 10 log parenleftbigg r 1 r 2 parenrightbigg 2 = 20 log parenleftbigg r 1 r 2 parenrightbigg 2 1 20 = log parenleftbigg r 1 r 2 parenrightbigg r 1 r 2 = 10 ( 2 1 ) / 20 r 1 = 10 ( 2 1 ) / 20 r 2 homework 13 GAUTHIER, JOSEPH Due: Dec 4 2007, 1:00 am 2 L = r 2 + r 1 , = r 2 + 10 ( 2 1 ) / 20 r 2 , so r 2 = L 1 + 10 ( 2 1 ) / 20 = 121 m 1 + 10 (29 . 6 dB 24 . 6 dB) / (20 dB) = 43 . 5521 m . Question 4, chap 17, sect 2. part 1 of 1 10 points A vacuum cleaner has a measured sound level of 70 . 5 dB....
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HW13 - homework 13 GAUTHIER, JOSEPH Due: Dec 4 2007, 1:00...

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