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Unformatted text preview: homework 14 – ALIBHAI, ZAHID – Due: May 2 2007, 4:00 am 1 Question 1 part 1 of 2 0 points An ambulance is traveling north at 49 m / s and is moving towards a car that is traveling south at 26 . 5 m / s. The ambulance driver hears his siren at a frequency of 690 Hz. The velocity of sound is 330 m / s . Ambulance 49 m / s 26 . 5 m / s Police What wavelength does the driver of the car detect from the ambulance’s siren? Correct answer: 0 . 376974 m (tolerance ± 1 %). Explanation: Let : v a = 49 m / s , v c = 26 . 5 m / s , f a = 690 Hz , and v s = 330 m / s . Due to the motion of the ambulance, the wavelength detected by the driver of the car is compressed. λ c = λ a parenleftbigg v s − v a v s + v c parenrightbigg = v s f a parenleftbigg v s − v a v s + v c parenrightbigg = 330 m / s 690 Hz parenleftbigg 330 m / s − 49 m / s 330 m / s + 26 . 5 m / s parenrightbigg = . 376974 m . Question 2 part 2 of 2 10 points At what frequency does the driver of the car hear the ambulance’s siren? Correct answer: 875 . 391 Hz (tolerance ± 1 %). Explanation: f c = v s + v c v s − v a f a = 330 m / s + 26 . 5 m / s 330 m / s − 49 m / s (690 Hz) = 875 . 391 Hz . Question 3 part 1 of 2 10 points The equation for the Doppler shift of a sound wave of speed v , reaching a moving detector, is f ′ = f parenleftbigg v ± v d v ∓ v s parenrightbigg , where v d is the speed of the detector, v s is the speed of the source, f is the frequency of the source, and f ′ is the frequency at the detec tor. In the numerator the + / − indicates when the detector is moving toward/away from the source. In the denominator the − / + indi cates when the source is moving toward/away from the detector. A train moving toward a detector at 25 m / s blows a 350 Hz horn. The velocity of sound is v sound = 343 m / s . What frequency is detected by a stationary train? Correct answer: 377 . 516 Hz (tolerance ± 1 %). Explanation: Let : v = 343 m / s , v d = 0 m / s , v s = 25 m / s , and f = 350 Hz . The source is moving toward the detector, (which is stationary), so v s > 0 and since v d = 0 , we have f ′ = f parenleftbigg v v − v s parenrightbigg = (350 Hz) parenleftbigg 343 m / s 343 m / s − 25 m / s parenrightbigg = 377 . 516 Hz . homework 14 – ALIBHAI, ZAHID – Due: May 2 2007, 4:00 am 2 Question 4 part 2 of 2 10 points What frequency is detected by a train moving toward the first train at a speed of 15 . 5 m / s? Correct answer: 394 . 575 Hz (tolerance ± 1 %). Explanation: Let : v d = 15 . 5 m / s . f ′ = f parenleftbigg v + v d v − v s parenrightbigg = (350 Hz) parenleftbigg 343 m / s + 15 . 5 m / s 343 m / s − 25 m / s parenrightbigg = 394 . 575 Hz ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
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