OLDHW36 - oldhomewk 36 PAPAGEORGE, MATT Due: Apr 28 2008,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = 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) 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] 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 1 = v s f 1 = 2 . 20513 m 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 1 , A 1 , and I 1 , respectively, and those of the second to be 2 , A 2 , and I 2 , respectively....
View Full Document

Page1 / 8

OLDHW36 - oldhomewk 36 PAPAGEORGE, MATT Due: Apr 28 2008,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online