Turner Final2 Spring 2005

Turner Final2 Spring 2005 - Trevino, Denise Fnmakeup 1 Due:...

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Trevino, Denise – Fnmakeup 1 – Due: Jun 4 2005, 11:00 am – Inst: Turner 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 2) 10 points A police car is traveling at a speed, v c , to the le±t. A truck is traveling at a speed, v t , to the right. The ±requency o± the siren on the police car is f c . The speed o± sound in air is v a . Let v t be the speed o± the observer in the truck, and v c be the speed o± the source , the police car. Police v c v t Truck What is the ±requency, f t , heard by an observer in the moving truck? 1. f t = v a - v t v a + v c f c correct 2. f t = v a - v t v a - v c f c 3. f t = v a + v t v a + v c f c 4. f t = v a + v t v a - v c f c Explanation: Basic Concepts: The Doppler shi±ted ±re- quency, f 0 , heard in the truck is f 0 = v a ± v 0 v a v s f , (1) where v a is the speed o± sound in air, v o is the speel o± the observer, and v s is the speed o± the source, The upper sign is used when the relative velocities are toward one-another, and vice versa . Solution: The relative velocity o± the ob- server is away ±rom the source so the lower sign is used in the numerator ( ± → - ), and the relative velocity o± the source is away ±rom the observer so the lower sign is used in the denominator ( ∓ → +). There±ore Eq. 2 be- comes f t = v a - v t v a + v c f c . This is version two o± ±our versions. 002 (part 2 o± 2) 10 points A police car is traveling at a speed, v c , to the le±t. A truck is traveling at a speed, v t , to the le±t. A wind is blowing in the same direction as that o± the truck with a speed, v w , to the le±t. The ±requency o± the siren on the police car is f c . The speed o± sound in air is v a . Police v c v t Truck v w wind What is the ±requency, f t , heard by an observer in the moving truck? 1. f t = v a - v t + v w v a + v c - v w f c 2. f t = v a + v t + v w v a + v c - v w f c 3. f t = v a + v t - v w v a - v c + v w f c 4. f t = v a - v t - v w v a + v c - v w f c 5. f t = v a - v t - v w v a - v c + v w f c 6. f t = v a - v t + v w v a - v c + v w f c 7. f t = v a + v t - v w v a + v c - v w f c correct 8. f t = v a + v t + v w v a - v c + v w f c Explanation: The problem must be worked in the ±rame o± re±erence relative to the air. “ v t - v w ” is the relative velocity o± the truck (observer), v o . “ v c - v w ” is the relative velocity o± the car (source), v s , there±ore f t = v a ± ( v t - v w ) v a ( v c - v w ) f c . (2)
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Trevino, Denise – Fnmakeup 1 – Due: Jun 4 2005, 11:00 am – Inst: Turner 2 The relative velocity of the observer is to- wards the source so the upper sign is used in the numerator ( ± → +), and the relative ve- locity of the source is away from the observer so the lower sign is used in the denominator ( ∓ → +). Therefore Eq. 2 becomes f t = v a + ( v t - v w ) v a + ( v c - v w ) f c , so = v a + v t - v w v a + v c - v w f c . This is version one of eight versions.
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Turner Final2 Spring 2005 - Trevino, Denise Fnmakeup 1 Due:...

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