final_oldfinal1 - oldnal 01 PAPAGEORGE, MATT Due: May 6...

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oldfnal 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±ect, by f ± = ± v sound v sound - v train ² f = ± 344 m / s 344 m / s - 3 . 86111 m / s ² f = (1 . 01135) f. The beat Frequency equals the di±erence in Frequency between the two sources. f b = | f ± - f | = (1 . 01135 - 1) 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 fxed 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 . 8N . The Fundamental mode oF a string has wavelength λ =2 L and thereFore Frequency f = v 2 L , where v = ³ F μ = ´ µ µ F m L = · FL m is the speed oF waves on the string. Alto- gether, f = 1 2 L · m = 1 2 · F mL . Thus the ratio is f 2 f 1 = 1 2 · F 2 2 1 2 · F 1 1 = · F 2 L 1 F 1 L 2 = ³ (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 fxed at the both ends. The length oF air column open at
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oldfnal 01 – PAPAGEORGE, MATT – Due: May 6 2008, 4:00 am 2 one end is gradually increased From zero until the frst position oF resonance is observed at 24 . 9 cm. The wire is 127 . 2 cm long and is vibrating in its third harmonic. ±ind 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 frst 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 ² ³ 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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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.

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final_oldfinal1 - oldnal 01 PAPAGEORGE, MATT Due: May 6...

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