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OLF FINAL 08

# OLF FINAL 08 - oldfinal 01 VALENCIA DANIEL Due May 6 2008...

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oldfinal 01 – VALENCIA, DANIEL – Due: May 6 2008, 4:00 am 1 Question 1, chap 16, sect 3. part 1 of 1 10 points While waiting for Stan Speedy to arrive on a late passenger train, Kathy Kool notices beats occurring as a result of two trains blow- ing their whistles simultaneously. One train is at rest and the other is approaching her at a speed of 8 . 16 km / h. Assume that both whistles have the same frequency and that the speed of sound is 344 m / s. If Kathy hears beat frequency at 9 . 26 Hz, what is the frequency of the whistles? Correct answer: 1396 . 08 Hz (tolerance ± 1 %). Explanation: The frequency of sound waves produced by the moving whistle at Kathy’s location is given, according to Doppler Effect, by f = parenleftbigg v sound v sound v train parenrightbigg f = parenleftbigg 344 m / s 344 m / s 2 . 26667 m / s parenrightbigg f = (1 . 00663) f . The beat frequency equals the difference in frequency between the two sources. f b = | f f | = (1 . 00663 1) f . Therefore, f = f b (1 . 00663 1) = 9 . 26 Hz (1 . 00663 1) = 1396 . 08 Hz . Question 2, chap 16, sect 4. part 1 of 1 10 points An elastic string of mass 4 . 23 g is stretched to a length of 2 . 649 m by the tension force 58 N. The string is fixed at both ends and has fundamental frequency f 1 . When the tension force increases to 3120 . 4 N the string stretches to a length 8 . 92713 m and its fundamental frequency becomes f 2 . Calculate the ratio f 2 f 1 . Correct answer: 3 . 99555 (tolerance ± 1 %). Explanation: Let : L 1 = 2 . 649 m , F 1 = 58 N , L 2 = 8 . 92713 m , and F 2 = 3120 . 4 N . The fundamental mode of a string has wavelength λ = 2 L and therefore frequency f = v 2 L , where v = radicalBigg F μ = radicaltp radicalvertex radicalvertex radicalbt F m L = radicalbigg F L m is the speed of waves on the string. Alto- gether, f = 1 2 L radicalbigg F L m = 1 2 radicalbigg F m L . Thus the ratio is f 2 f 1 = 1 2 radicalbigg F 2 m L 2 1 2 radicalbigg F 1 m L 1 = radicalbigg F 2 L 1 F 1 L 2 = radicalBigg (3120 . 4 N) (2 . 649 m) (58 N) (8 . 92713 m) = 3 . 99555 . Note: You don’t need the value of the string’s mass since it cancels out of the frequency ratio. Question 3, chap 17, sect 3. part 1 of 1 10 points A variable-length air column is placed just below a vibrating wire that is fixed at the both ends. The length of air column open at

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oldfinal 01 – VALENCIA, DANIEL – Due: May 6 2008, 4:00 am 2 one end is gradually increased from zero until the first position of resonance is observed at 20 . 1 cm. The wire is 143 . 6 cm long and is vibrating in its third harmonic. Find the speed of transverse waves in the wire if the speed of sound in air is 340 m / s. Correct answer: 404 . 842 m / s (tolerance ± 1 %). Explanation: Given : L = 20 . 1 cm , L wire = 143 . 6 cm , and v sound = 340 m / s . The length of the air column when the first resonance is heard is L = λ air 4 , where λ air is the wavelength of the sound in air. The frequency of sound wave, and hence the vibrating wire producing sound, is f = v sound λ air = v sound 4 L .
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