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Homework 37

Homework 37 - homework 37 BAUTISTA ALDO Due May 1 2006 4:00...

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homework 37 – BAUTISTA, ALDO – Due: May 1 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Standing at a crosswalk, you hear a fre- quency of 571 Hz from the siren on an ap- proaching police car. After the police car passes, the observed frequency of the siren is 491 Hz. Assume the speed of sound in air to be 343 m / s. Determine the car’s speed from these ob- servations. Correct answer: 25 . 838 m / s (tolerance ± 1 %). Explanation: Basic Concepts: The Doppler Effect. When the police car is approaching, the ob- served frequency is f 0 = f (1 - v s /v ) After the police car passes, the observed fre- quency is f 00 = f (1 + v s /v ) Thus, f 0 f 00 = 1 + v s v 1 - v s v Solving for v s , v s = f 0 f 00 - 1 f 0 f 00 + 1 v = ( 571 Hz 491 Hz - 1 ) ( 571 Hz 491 Hz + 1 ) 343 m / s = 25 . 838 m / s Question 2 Part 1 of 2. 10 points. The velocity of sound in air is 343 m / s . An ambulance is traveling east at 49 . 7 m / s . Behind it there is a car traveling along the same direction at 34 . 9 m / s . The ambulance driver hears his siren with a wavelength of 0 . 42 m . 34 . 9 m / s Car 49 . 7 m / s Ambulance What is the measured wavelength of the sound of the ambulance’s siren when you are holding your measuring device behind the am- bulance? Correct answer: 0 . 480857 m (tolerance ± 1 %). Explanation: Let : v sound = 343 m / s , v car = 34 . 9 m / s , v ambulance = 49 . 7 m / s , λ siren = 0 . 42 m , Basic Concept: v wave f wave λ wave (1) f 0 = v sound ± v observer v sound v source f (2) Solution: The frequency of the sound emitted as heard by the ambulance driver is f siren v sound λ siren = (343 m / s) (0 . 42 m) = 816 . 667 Hz . In terms of the original frequency, f 0 = v sound + v person v sound + v amb f = (343 m / s) + (0 m / s] (343 m / s) + (49 . 7 m / s) × (816 . 667 Hz) = 713 . 31 Hz , so λ 0 v sound f 0 = (343 m / s) (713 . 31 Hz) = 0 . 480857 m .

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homework 37 – BAUTISTA, ALDO – Due: May 1 2006, 4:00 am 2 The positive sign arises because the ambu- lance driver is traveling in the opposite direc- tion as the sound waves. Here the wave form is stretched out, which results in a longer wavelength. Question 3 Part 2 of 2. 10 points. What is the measured wavelength of the sound of the ambulance’s siren when your measuring device is on the car’s hood? Correct answer: 0 . 436449 m (tolerance ± 1 %). Explanation: In terms of the original frequency, f 0 = v sound + v car v sound + v amb f = (343 m / s) + (34 . 9 m / s] (343 m / s) + (49 . 7 m / s) × (816 . 667 Hz) = 785 . 888 Hz , so λ 00 v sound f 0 = (343 m / s) (785 . 888 Hz) = 0 . 436449 m .
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