# HW 34 - Morby Grant – Homework 34 – Due noon – Inst...

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Unformatted text preview: Morby, Grant – Homework 34 – Due: Apr 28 2006, noon – Inst: Drummond 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The four figures below represent sound waves emitted by a moving source. Which picture represents a source moving at a speed bigger than zero but less than the speed of sound? 1. correct 2. 3. 4. Explanation: In the following figure the wavefront cre- ated when the source was at the center of cir- cle 1 has traveled the distance between lines 1 and 3. 1 2 3 Similarly, the wavefront created when the source was at the center of circle 2 has trav- eled the distance between lines 2 and 3. This means that the source had to move at exactly the speed of the sound waves - else the wave emitted at 2 would not catch up with that emitted at 1. The result of this motion at the speed of sound is a piling up of sound waves in front of the source at 3. In figure all the wavesfronts are concentric, indicating that the source is stationary. Figure shows a moving source of sound, but the dis- placement of the wavefronts is less than in the first figure so the source is moving at a speed slower than the speed of sound. So this figure is our answer. In figure Morby, Grant – Homework 34 – Due: Apr 28 2006, noon – Inst: Drummond 2 the wavefronts emitted later have moves be- yond those emitted earlier, indicating that the source moves faster than the speed of sound. 002 (part 1 of 1) 10 points An ambulance is traveling east at 45 . 8 m / s. Behind it there is a car traveling along the same direction at 33 . 4 m / s. The ambulance driver hears his siren at a frequency of 870 Hz. The speed of sound is 343 m/s. 33 . 4 m / s Car 45 . 8 m / s Ambulance At what frequency does the driver of the car hear the ambulance’s siren? (in units of Hz) Correct answer: 842 . 253 Hz. Explanation: The wavelength of the sound emitted in the back of the ambulance is λ = ( v sound + v amb ) /f . The positive sign arises because the ambulance driver is traveling in the opposite direction as the sound waves. Here the wave form is stretched out, which results in a longer wavelength. The frequency observed by the car driver is: f = v λ . The car driver sees the sound wave overtaking him with a velocity of v = v sound + v car . Therefore the car driver hears a sound with a frequency f = ( v sound + v car ) /λ . In terms of the original frequency, f = µ v sound + v car v sound + v amb ¶ f = µ 343 m / s + 33 . 4 m / s 343 m / s + 45 . 8 m / s ¶ (870 Hz) = 842 . 253 Hz . 003 (part 1 of 1) 10 points A driver in a Car is traveling North at a velocity represented by the vertical vector....
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## This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.

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HW 34 - Morby Grant – Homework 34 – Due noon – Inst...

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