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# 1-2 - Create assignment 00224 Homework 2 Jan 21 at 5:20 pm...

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Create assignment, 00224, Homework 2, Jan 21 at 5:20 pm 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Temperature and Sound Waves 17:02, trigonometry, numeric, > 1 min, nor- mal. 001 A sound wave has a frequency of 700 Hz in air and a wavelength of 0 . 5 m. What is the temperature of the air? As- sume the velocity of sound at 0 C is 331 m / s. Correct answer: 32 . 2409 C. Explanation: Let : f = 700 Hz , λ = 0 . 5 m , and v 0 = 331 m / s . The velocity of sound is v s B ρ . Since the density decreases as the tempera- ture increases, the speed will increase with the square root of the temperature. Assuming a linear dependence, we have ρ ρ 0 = 1 - T T 0 v v 0 = v u u u t B ρ 0 B ρ 0 1 - T T 0 r 1 + T T 0 , or T T 0 = v v 0 2 - 1 . Since the speed of the sound wave is v = λ f = (0 . 5 m) (700 Hz) = 350 m / s , wehave T = T 0 " v v 0 2 - 1 # = (273 C) " 350 m / s 331 m / s 2 - 1 # = 32 . 2409 C . keywords: Car Passing Stationary Siren 17:05, trigonometry, numeric, > 1 min, nor- mal. 002 A car, moving at 80 mi / hr, passes a station- ary police car whose siren has a frequency of 500 Hz. The velocity of sound in air is 343 m / s . What is the frequency change heard by an observer in the moving car as he passes the police car? (1 mi = 1.609 km) Correct answer: 104 . 244 Hz. Explanation: Let : v o = 80 mi / hr = 35 . 7556 m / s , and f = 500 Hz . When the observer is approaching the po- lice car, the frequency he hears is f b = f 1 + v o v sound . While he is moving away from the police car, he hears the following frequency f a = f 1 - v o v sound . So the change of the frequency he hears is Δ f = | f a - f b | = 2 f v o v sound = 2 (500 Hz) (80 mi / hr) (343 m / s) = 104 . 244 s - 1 .

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Create assignment, 00224, Homework 2, Jan 21 at 5:20 pm 2 keywords: Guitar String Comparison 18:04, trigonometry, numeric, > 1 min, nor- mal. 003 Two guitar strings, of equal length and lin- ear density, are tuned such that the second harmonic of the first string has the same fre- quency as the third harmonic of the second string. The tension of the first string is 100 N.
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