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Unformatted text preview: hinojosa (jlh3938) – oldmidterm5 01 – Turner – (58185) 1 This printout should have 33 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points While waiting for Stan Speedy to arrive on a late passenger train, Kathy Kool notices beats occurring as a result of two trains blowing their whistles simultaneously. One train is at rest and the other is approaching her at a speed of 23 . 8 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 6 . 82 Hz, what is the frequency of the whistles? Correct answer: 348 . 049 Hz. Explanation: Let : v sound = 344 m / s , v train = 23 . 8 km / h = 6 . 61111 m / s , and f b = 6 . 82 Hz . 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 − 6 . 61111 m / s parenrightbigg f = 1 . 01959 f . The beat frequency equals the difference in frequency between the two sources. f b =  f ′ − f  = (1 . 01959 − 1) f . Therefore, f = f b 1 . 01959 − 1 = 6 . 82 Hz 1 . 01959 − 1 = 348 . 049 Hz . 002 10.0 points An elastic string of mass 8 . 5 g is stretched to length L 1 by the tension force 47 . 9 N. The string is fixed at both ends and has funda mental frequency F 1 . When the tension force increases to 3788 . 89 N the string stretches to length 6 . 3825 m and its fundamental frequency be comes F 2 . Calculate the ratio F 2 F 1 . Correct answer: 3 . 70898. Explanation: Let : L 1 = 1 . 11 m , F 1 = 47 . 9 N , L 2 = 6 . 3825 m , and F 2 = 3788 . 89 N . The fundamental mode of a string has wavelength λ = 2 L and therefore frequency f = v 2 L where v is the speed of waves on the string, v = radicalBigg F μ = radicalBigg F m/L . Altogether, f = 1 2 L radicalbigg F L m = 1 2 radicalbigg F mL . At this point, we may evaluate f 1 and f 2 using the above formula twice, then calculate the ratio, but it’s simpler to do the algebra before the arithmetic: f 2 f 1 = 1 2 radicalbigg F 2 mL 2 1 2 radicalbigg F 1 mL 1 = radicalbigg F 2 F 1 radicalbigg L 1 L 2 = radicalBigg (3788 . 89 N) (47 . 9 N) (1 . 11 m) (6 . 3825 m) = 3 . 70898 . hinojosa (jlh3938) – oldmidterm5 01 – Turner – (58185) 2 Note: You don’t need the value of the string’s mass as it cancels out of the frequency ratio. 003 10.0 points A variablelength air column is placed just below a vibrating wire that is fixed at the both ends. The length of air column open at one end is gradually increased from zero until the first position of resonance is observed at 35 . 7 cm. The wire is 119 . 4 cm long and is vibrating in its third harmonic....
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This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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