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final_oldfinal1 - oldnal 01 PAPAGEORGE MATT Due May 6 2008...

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oldfinal 01 – PAPAGEORGE, MATT – 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 13 . 9 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 7 . 39 Hz, what is the frequency of the whistles? Correct answer: 651 . 011 Hz (tolerance ± 1 %). Explanation: The frequency of sound waves produced by the moving whistle at Kathy’s location is given, according to Doppler E ff ect, by f = v sound v sound - v train f = 344 m / s 344 m / s - 3 . 86111 m / s f = (1 . 01135) f . The beat frequency equals the di ff erence in frequency between the two sources. f b = | f - f | = (1 . 01135 - 1) f . Therefore, f = f b (1 . 01135 - 1) = 7 . 39 Hz (1 . 01135 - 1) = 651 . 011 Hz . Question 2, chap 16, sect 4. part 1 of 1 10 points An elastic string of mass 5 . 2 g is stretched to a length of 1 . 791 m by the tension force 44 N. The string is fixed at both ends and has fundamental frequency f 1 . When the tension force increases to 624 . 8 N the string stretches to a length 8 . 68635 m and its fundamental frequency becomes f 2 . Calculate the ratio f 2 f 1 . Correct answer: 1 . 71109 (tolerance ± 1 %). Explanation: Let : L 1 = 1 . 791 m , F 1 = 44 N , L 2 = 8 . 68635 m , and F 2 = 624 . 8 N . The fundamental mode of a string has wavelength λ = 2 L and therefore frequency f = v 2 L , where v = F μ = F m L = F L m is the speed of waves on the string. Alto- gether, f = 1 2 L F L m = 1 2 F m L . Thus the ratio is f 2 f 1 = 1 2 F 2 m L 2 1 2 F 1 m L 1 = F 2 L 1 F 1 L 2 = (624 . 8 N) (1 . 791 m) (44 N) (8 . 68635 m) = 1 . 71109 . 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 – PAPAGEORGE, MATT – Due: May 6 2008, 4:00 am 2 one end is gradually increased from zero until the first position of resonance is observed at 24 . 9 cm. The wire is 127 . 2 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: 289 . 478 m / s (tolerance ± 1 %). Explanation: Given : L = 24 . 9 cm , L wire = 127 . 2 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 . When the wire vibrates in its third harmonic, its wavelength is related to its length as fol- lows; L wire = 3 2 λ wire , where λ wire is the wavelength of waves trav- eling in the wire. Therefore, λ wire = 2 3 L wire , and the speed of transverse waves in the wire is v wire = λ wire f = 2 3 L wire v sound 4 L = L wire v sound 6 L = (127 . 2 cm) (340 m / s) 6 (24 . 9 cm) = 289 . 478 m / s .
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