Asked by AlphaPlus
A source passing a stationary observer is emitting a frequency of 560 Hz. If the speed of sound is 345 m/s, what must speed of the source be if the frequency speed is 480 hz?
Here's what I've tried...
Fo=480 Hz
FObs = 480 + 560
VSource = ?(x)
Vsound = 345 m/s
I try to rationalize the denominator by moving the x to the right side. That being said, I always get a number slightly off from the answer.
(Answer: 56 m/s)
Here's what I've tried...
Fo=480 Hz
FObs = 480 + 560
VSource = ?(x)
Vsound = 345 m/s
I try to rationalize the denominator by moving the x to the right side. That being said, I always get a number slightly off from the answer.
(Answer: 56 m/s)
Answers
Answered by
Henry
Fo = ((Vs-Vo)/(Vs+Ve)) * Fe = 480 Hz.
(345-0)/(345+Ve) * 560 = 480,
345/(345+Ve) * 560 = 480,
193,200/(345+Ve) = 480,
165,600 + 480Ve = 193,200,
480Ve = 27600, Ve = 57.5 m/s. = Velocity of the emitter or source.
v
(345-0)/(345+Ve) * 560 = 480,
345/(345+Ve) * 560 = 480,
193,200/(345+Ve) = 480,
165,600 + 480Ve = 193,200,
480Ve = 27600, Ve = 57.5 m/s. = Velocity of the emitter or source.
v
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