PHY 303K - Homework 26

# PHY 303K - Homework 26 - homework 26 FIERRO JEFFREY Due...

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homework 26 – FIERRO, JEFFREY – Due: Apr 29 2008, 11:00 pm 1 Question 1, chap 17, sect 4. part 1 of 2 10 points An ambulance is traveling north at 48 m / s and is moving towards a police car that is traveling south at 28 m / s. The ambulance driver hears his siren at a frequency of 770 Hz. The velocity of sound is 330 m / s . Ambulance 48 m / s 28 m / s Police What wavelength does a stationary person at the position of the police car detect from the ambulance’s siren? Correct answer: 0 . 366234 m (tolerance ± 1 %). Explanation: Let : v a = 48 m / s , v c = 28 m / s , f a = 770 Hz , and v s = 330 m / s . Due to the motion of the ambulance, the wavelength detected by a stationary person at the position of the police car is compressed. λ c = λ a bracketleftbigg v s v a v s bracketrightbigg = v s f a bracketleftbigg v s v a v s bracketrightbigg = 330 m / s 770 Hz bracketleftbigg (330 m / s) (48 m / s) (330 m / s) bracketrightbigg = 0 . 366234 m . Question 2, chap 17, sect 4. part 2 of 2 0 points At what frequency does the driver of the police car hear the ambulance’s siren? Correct answer: 770 Hz (tolerance ± 1 %). Explanation: f c = v s + v c v s v a f a = 330 m / s + 28 m / s 330 m / s 48 m / s (770 Hz) = 770 Hz . Question 3, chap 17, sect 4. part 1 of 2 10 points The velocity of sound in air is 343 m / s . An ambulance is traveling east at 50 . 1 m / s. Behind it there is a car traveling along the same direction at 32 . 1 m / s. The ambulance driver hears his siren at a frequency of 717 Hz. 32 . 1 m / s Car 50 . 1 m / s Ambulance What is the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car? Correct answer: 0 . 548257 m (tolerance ± 1 %). Explanation: Let : v car = 32 . 1 m / s , v amb = 50 . 1 m / s , v sound = 343 m / s , and f 0 = 717 Hz . The wavelength of the sound emitted be- hind the ambulance is λ = v sound v amb f 0 = v sound + v amb f 0 = 343 m / s + 50 . 1 m / s 717 Hz = 0 . 548257 m . The positive sign arises because the ambu- lance driver is traveling in the opposite direc- tion as the sound waves. Here the wave form

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homework 26 – FIERRO, JEFFREY – Due: Apr 29 2008, 11:00 pm 2 is stretched out, which results in a longer wavelength. Question 4, chap 17, sect 4. part 2 of 2 10 points At what frequency does the driver of the car hear the ambulance’s siren? Correct answer: 684 . 169 Hz (tolerance ± 1 %). Explanation: The observed frequency is f = v λ . The car driver sees the sound wave overtaking him with a velocity of v = v sound + v car , so the car driver hears a sound with a frequency f = v sound + v car λ . In terms of the original frequency, f = parenleftbigg v sound ± v car v sound v amb parenrightbigg f 0 = parenleftbigg v sound + v car v sound + v amb parenrightbigg f 0 = parenleftbigg 343 m / s + 32 . 1 m / s 343 m / s + 50 . 1 m / s parenrightbigg (717 Hz) = 684 . 169 Hz .
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