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Standing at a crosswalk, you hear a frequency of 505 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 428 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)

Standing at a crosswalk, you hear a frequency of 505 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 428 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Standing at a crosswalk, you hear a frequency of 505 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 428 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)
Transcribed Image Text:Standing at a crosswalk, you hear a frequency of 505 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 428 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)
S Since the ambulance is the source of the sound, let v represent the magnitude of the velocity of the ambulance. Approaching the ambulance, you hear the frequency V f ( √ ² √₂ ) ² = (₁ - √ ₁ /V)² f V- (1 - and departing the ambulance, you hear the frequency f' = f (1 − (− Vs/v))* The negative sign appears because the source is moving toward the observer. The opposite sign describes the ambulance moving away. Since f' = 505 Hz and f" = 428 Hz, we have the following. f" = 505 which gives S The correct answer is not zero. Hz = 0 Solving for v from the equation above, we have = Hz)(1-2) - (428 V = 0 X The correct answer is not zero. V 0 X = The correct answer is not zero. ( 28.3 m/s. Hz H₂) (1 + ~/3), S V 0 X The correct answer is not zero. Hz. 343 m/s)
Transcribed Image Text:S Since the ambulance is the source of the sound, let v represent the magnitude of the velocity of the ambulance. Approaching the ambulance, you hear the frequency V f ( √ ² √₂ ) ² = (₁ - √ ₁ /V)² f V- (1 - and departing the ambulance, you hear the frequency f' = f (1 − (− Vs/v))* The negative sign appears because the source is moving toward the observer. The opposite sign describes the ambulance moving away. Since f' = 505 Hz and f" = 428 Hz, we have the following. f" = 505 which gives S The correct answer is not zero. Hz = 0 Solving for v from the equation above, we have = Hz)(1-2) - (428 V = 0 X The correct answer is not zero. V 0 X = The correct answer is not zero. ( 28.3 m/s. Hz H₂) (1 + ~/3), S V 0 X The correct answer is not zero. Hz. 343 m/s)
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