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Unformatted text preview: Overton, Mays – Homework 35 – Due: May 2 2005, 4:00 am – Inst: Turner 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points An ambulance is traveling north at 58 m / s and is moving towards a car that is traveling south at 33 . 9 m / s. The ambulance driver hears his siren at a frequency of 750 Hz. Velocity of sound v sound = 330 m / s . Ambulance 58 m / s 33 . 9 m / s Police What wavelength does the driver of the car detect from the ambulance’s siren? Correct answer: 0 . 362667 m. Explanation: Due to the motion of the ambulance, the wavelength detected by the driver of the car is compressed. λ c = λ Δ λ = v s v a f = 0 . 362667 m 002 (part 2 of 2) 10 points At what frequency does the driver of the car hear the ambulance’s siren? Correct answer: 1003 . 4 Hz. Explanation: f c = v s + v c v s v a f = 1003 . 4 Hz 003 (part 1 of 1) 10 points A train is moving parallel and adjacent to a highway with a constant speed of 34 m / s. A car is traveling in the same direction as the train at 68 m / s. The train’s whistle sounds at 360 Hz. The speed of sound is 343 m / s. When the car is behind the train what frequency does an occupant of the car observe for the train whistle? Correct answer: 392 . 467 Hz. Explanation: Basic Concepts: Doppler effect f = f µ v a ± v o v a ∓ v s ¶ , where v a is the velocity of sound in air, v o is the velocity of the observer, and v s is the ve locity of the source. Use the upper sign (+) in the numerator when the observer is mov ing towards the source and the lower sign ( ) in the numerator when the observer is moving away from the source. Use the upper sign ( ) in the denominator when the source is moving towards the observer and the lower sign (+) in the denominator when the source is mov ing away from the observer. Solution: In this case the observer is moving toward the source with v c = 68 m / s and the source is moving away from the observer with v t = 34 m / s. We have f 1 = f t µ v a + v c v a + v t ¶ , where f t = 360 Hz is the frequency of the train whistle and v a = 343 m / s is the speed of sound in air. Thus, f 1 = 360 Hz µ 343 m / s + 68 m / s 343 m / s + 34 m / s ¶ = 392 . 467 Hz . 004 (part 1 of 1) 10 points You are approaching a police car at 58 . 2 mph and the police car is approaching you at 95 . 7 mph. The wind is blowing from be hind you at 43 . 2 mph toward the police car....
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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