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 f' P² = ( √ ~ √₂ ) ² = (₁ - √ ₁ / 1 ) ² f= V-V. (1 - V and departing the ambulance, you hear the frequency f (1 - (- Vs/v))* The negative sign appears because the source is moving toward the observer. The opposite sign describes the Hz we have the ambulance moving away Since f¹ - 505 Hz and f" - 128 f" =

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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-
f =
V-V
S
(1 - Vs/v)'
-
and departing the ambulance, you hear the frequency
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
f
0
=
The correct answer is not zero. Hz
Hz) V
Hz)(1)-(428
=
V
= 28.3
Solving for v from the equation above, we have
=
0
X
The correct answer is not zero. V
0
X
The correct answer is not zero. 343
m/s.
Hz) (1 + Y3),
0
The correct answer is not zero. Hz.
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- f = V-V S (1 - Vs/v)' - and departing the ambulance, you hear the frequency 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 f 0 = The correct answer is not zero. Hz Hz) V Hz)(1)-(428 = V = 28.3 Solving for v from the equation above, we have = 0 X The correct answer is not zero. V 0 X The correct answer is not zero. 343 m/s. Hz) (1 + Y3), 0 The correct answer is not zero. Hz. 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.)
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.)
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