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Unformatted text preview: practice 14 – ALIBHAI, ZAHID – Due: Apr 29 2007, 4:00 am 1 Question 1 part 1 of 2 10 points An ambulance is traveling east at 58 . 8 m / s . Behind it there is a car traveling along the same direction at 33 . 1 m / s . The ambulance driver hears his siren with a wavelength of . 42 m . 33 . 1 m / s Car 58 . 8 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 . 492 m (tolerance ± 1 %). Explanation: Let : v sound = 343 m / s , v car = 33 . 1 m / s , v ambulance = 58 . 8 m / s , and λ siren = 0 . 42 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 . 42 m = 816 . 667 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 + 58 . 8 m / s parenrightbigg × (816 . 667 Hz) = 697 . 154 Hz , so λ ′ ≡ v sound f ′ = 343 m / s 697 . 154 Hz = . 492 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 is stretched out, which results in a longer wavelength. Question 2 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 . 4487 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 . 1 m / s 343 m / s + 58 . 8 m / s parenrightbigg × (816 . 667 Hz) = 764 . 431 Hz , so λ ′′ ≡ v sound f ′ = 343 m / s 764 . 431 Hz = . 4487 m . The car is a moving reference frame with re- spect to the fixed reference frame of the air (Earth). Question 3 part 1 of 2 10 points In order to be able to determine her speed, a skydiver carries a tone generator. A friend on the ground at the landing site has equip- ment for receiving and analyzing sound waves. practice 14 – ALIBHAI, ZAHID – Due: Apr 29 2007, 4:00 am 2 While the skydiver is falling at terminal speed, her tone generator emits a steady tone of 1250 Hz. Assume the air is calm and that the sound speed is 343 m / s, independent of altitude. If her friend on the ground (directly be- neath the skydiver) receives waves of fre- quency 2390 Hz, what is the skydiver’s speed of descent? Correct answer: 163 . 607 m / s (tolerance ± 1 %). Explanation: Let : v = 343 m / s , f e = 1250 Hz , and f g = 2390 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.
- Spring '08