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Unformatted text preview: oldhomewk 36 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 17, sect 3. part 1 of 1 10 points Given: The speed of sound in air is 344 m / s. A student uses an audio oscillator of ad justable frequency to measure the depth of a water well. Two successive resonant frequen cies are heard at 156 Hz and 180 Hz. What is the depth of the well? Correct answer: 7 . 16667 m (tolerance ± 1 %). Explanation: Basic Concept: For a tube closed at one end, resonances occur at λ = 4 L 2 n 1 , where n = 1 , 2 , 3 , ··· . Solution: Call L the depth of the well and v s the speed of sound. Then for some integer n L = (2 n 1) λ 1 4 = (2 n 1) v s 4 f 1 , (1) then n = 2 f 1 v s parenleftbigg L + v s 4 f 1 parenrightbigg . (2) Presuming that f 2 > f 1 , let n = n + 1 for the next resonant frequency f 2 L = [2 ( n + 1) 1] λ 2 4 = (2 n + 1) v s 4 f 2 , (3) then n = 2 f 2 v s parenleftbigg L v s 4 f 2 parenrightbigg . (4) Eliminating n from Eqs. (2) and (4), we obtain f 1 parenleftbigg L + v s 4 f 1 parenrightbigg = f 2 parenleftbigg L v s 4 f 2 parenrightbigg ( f 2 f 1 ) L = v s 2 from where we can obtain L L = v s 2 [ f 2 f 1 ] (5) = (344 m / s) 2 [(180 Hz) (156 Hz)] = 7 . 16667 m . From Eq. (4), we have n = 2 f 2 L v s 1 2 = 2 (180 Hz) (7 . 16667 m) 344 m / s 1 2 = 7 . To check this result, using Eq. (1) and (3), we have L = [2 n 1] v s 4 f 1 = [2 (7) 1] (344 m / s) 4 (156 Hz) = 7 . 16667 m , and L = [2 n + 1] v s 4 f 2 = [2 (7) + 1] (344 m / s) 4 (180 Hz) = 7 . 16667 m . The wavelengths are λ 1 = v s f 1 = 2 . 20513 m λ 2 = v s f 2 = 1 . 91111 m . Question 2, chap 17, sect 2. part 1 of 1 10 points A copper rod is given a sharp compressional blow at one end. The sound of the blow, traveling through air at 4 . 74 ◦ C, reaches the opposite end of the rod 2 . 83 ms later than the sound transmitted through the rod. The speed of sound in copper is 3690 m / s and the speed of sound in air at 4 . 74 ◦ C is 331 m / s. What is the length of the rod? Correct answer: 1 . 02904 m (tolerance ± 1 %). Explanation: oldhomewk 36 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 2 The difference in the times of travel of sound through the copper rod and the air is t = L parenleftbigg 1 v air 1 v Cu parenrightbigg , where L is the distance traveled ( i.e. , the length of the rod). Then L = t bracketleftbigg v air v Cu v Cu v air bracketrightbigg = (0 . 00283 s) × bracketleftbigg (331 m / s) (3690 m / s) (3690 m / s) (331 m / s) bracketrightbigg = 1 . 02904 m . Question 3, chap 17, sect 2. part 1 of 3 10 points Compare sound intensities in terms of the difference in decibels for two sound waves. Denote the angular frequency, the amplitude and the intensity of the first sound wave to be ω 1 , A 1 , and I 1 , respectively, and those of the second to be ω 2 , A 2 , and I 2 , respectively....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.
 Spring '09
 KLEINMAN
 Physics

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