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Unformatted text preview: homework 13 – VENNES, ROSS – Due: Nov 30 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 . 04 s after the rifle was fired. The echo from the other wall is heard 3 . 33 s after the first echo. The velocity of sound is v = 343 m / s . How wide is the valley? Correct answer: 927 . 815 m (tolerance ± 1 %). Explanation: Let : t 1 = 1 . 04 s , t 2 = 3 . 33 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 . 04 s) + (3 . 33 s) bracketrightBig 2 = 927 . 815 m . Question 2, chap 17, sect 2. part 1 of 1 10 points A rock band plays at a 61 dB sound level. How many times greater is the sound pressure from another rock band playing at 123 dB? Correct answer: 1258 . 93 (tolerance ± 1 %). Explanation: Let : β 1 = 61 dB and β 2 = 123 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 ( β 2 − β 1 ) / 20 = 10 (123 dB − 61 dB) / (20 dB) = 1258 . 93 . Question 3, chap 17, sect 2. part 1 of 1 10 points A stereo speaker is placed between two ob servers who are 48 . 1 m apart, along the line connecting them. If one observer records an intensity level of 63 . 9 dB, and the other records an intensity level of 66 . 9 dB, how far is the speaker from the closer observer? Correct answer: 19 . 9375 m (tolerance ± 1 %). Explanation: Let : β 1 = 63 . 9 dB , β 2 = 66 . 9 dB , and L = 48 . 1 m . I ∝ r − 2 , so I 2 I 1 = r 2 1 r 2 2 . β 1 = 10 log parenleftbigg I 1 I parenrightbigg and β 2 = 10 log parenleftbigg I 2 I parenrightbigg , so β 2 − β 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 β 2 − β 1 20 = log parenleftbigg r 1 r 2 parenrightbigg r 1 r 2 = 10 ( β 2 − β 1 ) / 20 r 1 = 10 ( β 2 − β 1 ) / 20 r 2 homework 13 – VENNES, ROSS – Due: Nov 30 2007, 1:00 am 2 L = r 2 + r 1 , = r 2 + 10 ( β 2 − β 1 ) / 20 r 2 , so r 2 = L 1 + 10 ( β 2 − β 1 ) / 20 = 48 . 1 m 1 + 10 (66 . 9 dB − 63 . 9 dB) / (20 dB) = 19 . 9375 m . Question 4, chap 17, sect 2. part 1 of 1 10 points A vacuum cleaner has a measured sound level of 65 . 9 dB....
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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