# panda 5 - pokharel(pp7242 Sound walther(16180 This...

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pokharel (pp7242) – Sound – walther – (16180) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Good luck, this is your home work on Ch. 12 C± Walther 001 10.0 points Consider the Following: A) the speed oF the source B) the speed oF the observer C) the loudness oF the sound In the Doppler e²ect For sound waves, which Factors a²ect the Frequency that the observer hears? 1. A and B only correct 2. C only 3. B and C only 4. None oF these. 5. A, B, and C 6. A only 7. A and C only 8. B only Explanation: In the Doppler e²ect oF sound waves, both the speed oF the source and the speed oF the observer a²ect the Frequency that the observer hears. (The directions oF the movement are Factors, too.) On the other hand, the loudness oF the sound (the intensity oF the sound wave) is irrelevant. 002 10.0 points A small vibrating object on the surFace oF a ripple tank is the source oF waves oF Frequency 20 Hz and speed 60 cm/s. The source S is moving to the right with speed 20 cm/s, as shown. S C A B D At which oF the labeled points will the Fre- quency measured by a stationary observer be greatest? 1. It will be the same For all Four points. 2. B 3. C 4. A correct 5. D Explanation: The Doppler e²ect For stationary observers is f = v sound v sound v source f 0 . Since the source is moving directly toward point A , the measured Frequency at point A by a stationary observer will be greatest. 003 (part 1 oF 2) 10.0 points An ambulance is traveling south at 62 . 9 m / s, away From a car that is traveling north at 30 . 5 m / s. The ambulance driver hears his siren at a Frequency oF 535 Hz. Ambulance 62 . 9 m / s 30 . 5 m / s Car What wavelength does a person who is standing between the car and the ambulance detect From the sound oF the ambulance’s siren? The velocity oF sound in air is 343 m / s. Correct answer: 0 . 758692 m.

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pokharel (pp7242) – Sound – walther – (16180) 2 Explanation: Let : v amb = 62 . 9 m / s , f 0 = 535 Hz , and v = 343 m / s . From the Doppler shift, f = v sound ± v observer v sound v source f . The ambulance moves away from the car, so f = v v + v amb f 0 and v = f λ λ = v f = v + v amb f 0 = 343 m / s + 62 . 9 m / s 535 Hz = 0 . 758692 m . The wave form is stretched out, which re- sults in a longer wavelength. 004 (part 2 of 2) 10.0 points At what frequency does the driver of the car hear the ambulance’s siren? Correct answer: 411 . 893 Hz. Explanation: Let : v car = 30 . 5 m / s . The car moves away from the ambulance and the ambulance moves away from the car, so f = v v car v + v amb f 0 = p 343 m / s 30 . 5 m / s 343 m / s + 62 . 9 m / s P (535 Hz) = 411 . 893 Hz .
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panda 5 - pokharel(pp7242 Sound walther(16180 This...

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