Homework 14 Solutions

# Homework 14 Solutions - homework 14 GADHIA TEJAS Due May 2...

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homework 14 – GADHIA, TEJAS – Due: May 2 2007, 4:00 am 1 Question 1 part 1 of 1 0 points A submarine traveling at 14 m / s toward an aircraft carrier emits a 2490 Hz sonar pulse. The reflected pulse returns with a frequency of 2547 Hz. What is the speed of the aircraft carrier? (Positive/negative means the carrier is mov- ing toward/away from the submarine.) The speed of sound in water is 1500 m/s. Correct answer: 2 . 94111 m / s (tolerance ± 1 %). Explanation: Let : f = 2490 Hz , f r = 2547 Hz , v = 1500 m / s , and v s = 14 m / s . The image travels with a speed of 14 m / s + 2 v , so 1500 m / s + 14 m / s 1500 m / s (14 m / s + 2 v ) (2490 Hz) = 2547 Hz 1514 m / s = 2547 Hz 2490 Hz (1486 m / s 2 v ) v = 1 2 bracketleftbigg (1514 m / s)(2490 Hz) 2547 Hz 1486 m / s bracketrightbigg = 2 . 94111 m / s Question 2 part 1 of 2 10 points An ambulance is traveling east at 54 . 5 m / s . Behind it there is a car traveling along the same direction at 33 . 8 m / s . The ambulance driver hears his siren with a wavelength of 0 . 46 m . 33 . 8 m / s Car 54 . 5 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 ambulance? The velocity of sound in air is 343 m / s . Correct answer: 0 . 53309 m (tolerance ± 1 %). Explanation: Let : v sound = 343 m / s , v car = 33 . 8 m / s , v ambulance = 54 . 5 m / s , and λ siren = 0 . 46 m . v wave f wave λ wave (1) f = bracketleftbigg v sound ± v observer v sound v source bracketrightbigg f (2) The frequency of the sound emitted as heard by the ambulance driver is f siren v sound λ siren = 343 m / s 0 . 46 m = 745 . 652 Hz . In terms of the original frequency, f = parenleftbigg v sound + v person v sound + v amb parenrightbigg f = parenleftbigg 343 m / s + 0 m / s 343 m / s + 54 . 5 m / s parenrightbigg × (745 . 652 Hz) = 643 . 418 Hz , so λ v sound f = 343 m / s 643 . 418 Hz = 0 . 53309 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 14 – GADHIA, TEJAS – Due: May 2 2007, 4:00 am 2 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 . 485271 m (tolerance ± 1 %). Explanation: In terms of the original frequency, f = parenleftbigg v sound + v car v sound + v amb parenrightbigg f = parenleftbigg 343 m / s + 33 . 8 m / s 343 m / s + 54 . 5 m / s parenrightbigg × (745 . 652 Hz) = 706 . 822 Hz , so λ ′′ v sound f = 343 m / s 706 . 822 Hz = 0 . 485271 m . The car is a moving reference frame with re- spect to the fixed reference frame of the air (Earth). Question 4 part 1 of 2 0 points The “red shift” of radiation from a distant galaxy consists of the light known to have a wavelength of 433 nm when observed in the laboratory, appearing to have a wavelength of 460 nm. What is the speed of galaxy in the line of sight relative to the Earth? The speed of light is 2 . 99792 × 10 8 m / s.
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