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4hv.org :: Forums :: General Science and Electronics
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Acoustic doppler effect problem

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Steve Conner
Sat Mar 28 2009, 09:50PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
I don't get this. You posted the equation for stationary observer and receding source. Are you now saying that this equation doesn't apply, because the problem is not a case of stationary observer and receding source? Maybe the observer is moving too, but that wasn't clear from your original post.
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likewhat
Sat Mar 28 2009, 10:35PM
likewhat Account deactivated by user request on 6/11/2009.
Registered Member #1071 Joined: Fri Oct 19 2007, 02:13AM
Location:
Posts: 44
I was just including the time it takes for the frequency emitted by the MOVING emitter to reach the STATIONARY receiver. The distance between the receiver and the emitter is 1/2*a*t^2. Since the waves travel with a speed c the time it takes for the waves to travel from the emitter to the receiver is distance/speed so call that ttravel = 1/2*a*t^2/c. That means that the waves you receive at time t where actually emitted from the emitter earlier at time t-ttravel.

My first post was just including time in the equation for the doppler shifted frequency, this correction seems to be what the question is asking for.
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Marko
Thu Apr 02 2009, 02:35PM
Marko Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Hi guys,

Getting closer, but no match yet.
I guess enough time has passed, so I can post the answer:

f (t) = f/sqrt(1+(2a*t/c)

And yeah, it's just as unintuitive for me... Now I challenge you to match your math to this solution.

Marko
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