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Midterm 04

# Midterm 04 - Bautista Aldo Midterm 4 Due 10:00 pm Inst...

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Bautista, Aldo – Midterm 4 – Due: Dec 12 2005, 10:00 pm – Inst: Maxim Tsoi 1 This print-out should have 17 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 An ambulance is traveling North at 47 . 2 m / s, approaching a car that is also travel- ing North at 31 . 5 m / s. The ambulance driver hears his siren at a frequency of 848 cycles / s. The velocity of sound is 343 m / s. Ambulance 47 . 2 m / s 31 . 5 m / s Car What is the wavelength at the car driver’s position for the sound from the ambulance’s siren? Correct answer: 0 . 348821 m. Explanation: Basic Concepts: The Doppler Effect: Wavelength of Sound Created by Source with Rest Frequency f and Speed v source is λ = v sound ± v source f . Wave Speed Relative to Moving Observer is v 0 = v sound ± v observer . Observed Frequency is f 0 = v 0 λ . Note: The wavelength is specified in the reference frame of the medium of propagation. Solution: Sound waves always travel at a given speed with respect to their medium of propagation. Note: Both the ambulance and car drivers, as well as any observers at rest (on the side of the road, for example), will measure the same wavelength for the sound from the siren, because length measurements will not depend on the velocity of the measurer. However, as sources/observers move through the medium at different velocities, they see the sound waves move past them at different velocities. As a result, the number of wavefronts passing them in a given time in- terval; i.e. , the frequency of the sound, must change. The relationship between observed fre- quency and observed wavelength is always given by λ 0 f = v rel , where v rel is the relative speed of the sound wave and the observer/source. Therefore, the wavelength of the sound emitted in front of the ambulance is λ 0 = v sound - v amb f = (343 m / s) - (47 . 2 m / s) (848 cycles / s) = 0 . 348821 m . The negative sign arises because the ambu- lance driver is traveling in the same direction as these sound waves and therefore perceives them as being slower than sound waves emit- ted when the ambulance is at rest. This re- sults in a smaller wavelength; intuitively, the wavefronts are compressed together by the motion of the siren. 002 (part 1 of 1) 10 points A skyrocket explodes 114 m above the ground. Three observers are spaced 81 m apart, with observer A directly under the point of the explosion. A B C 114 m 81 m 81 m Find the ratio of the sound intensity heard by observer A to that heard by observer B. Caution : Obviously the observer-size is not drawn to scale . For practical purpose, you

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Bautista, Aldo – Midterm 4 – Due: Dec 12 2005, 10:00 pm – Inst: Maxim Tsoi 2 may treat each observer as a point on the ground. Correct answer: 1 . 50485 . Explanation: Given : h = 114 m and d = 81 m .
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