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Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Hi guys,
At my last physics competition I've ran into a problem that caused much of confusion due to highly unintuitive solution. I have the solution but no proper explanation at all and hence my understanding is very limited; so I'd appreciate to see whether anyone here falls into the same trap as I did:
A sound transmitter are at first at the same spot in the time 0; simultaneously, the transmitter starts accelerating away with acceleration a and transmitting a sound of frequency f. Speed of sound in air is c. The problem asks for a relationship of the time elapsed and frequency read on the receiver.
The acoustic doppler effect equation for stationary observer and receeding source is
fnew = fsource/(1+(v/c)) where v is the velocity of the source.
Account deactivated by user request on 6/11/2009. Registered Member #1071
Joined: Fri Oct 19 2007, 02:13AM
Location:
Posts: 44
you have a constant acceleration a so the velocity at time t is just
v=a*t
you can see this because acceleration has units of m/s^2 and time is in units of s so m/s^2 * s = m/s , which is in units of velocity. you can put this velocity which is now in terms of a and t into your equation for frequency and then you get frequency as a function of time.
Registered Member #191
Joined: Fri Feb 17 2006, 02:01AM
Location: Esbjerg Denmark
Posts: 720
i got pretty much the same equation for constant velocity, except i got 1-V/C. its probably just a different convention for V.
However, if you just replace V with A*T, it will give you the instantaneous frequency at the transmitter. Which means you are kinda traveling at with the transmitter but somehow being stationary at the same time. <--- lol
Anyways, the Fnew is going down for sure, distance between the transmitter and the receiver is increasing as well while the wave is getting slower. I'll have to work this out, but I'm going to guess that Fnew will never really reach zero.
Account deactivated by user request on 6/11/2009. Registered Member #1071
Joined: Fri Oct 19 2007, 02:13AM
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Anyways, the Fnew is going down for sure, distance between the transmitter and the receiver is increasing as well while the wave is getting slower. I'll have to work this out, but I'm going to guess that Fnew will never really reach zero.
What about when the velocity of the transmitter exceeds the speed of sound?
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
likewhat wrote ...
you have a constant acceleration a so the velocity at time t is just
v=a*t
you can see this because acceleration has units of m/s^2 and time is in units of s so m/s^2 * s = m/s , which is in units of velocity. you can put this velocity which is now in terms of a and t into your equation for frequency and then you get frequency as a function of time.
This is what I and my professor arrived at firstly - and it is a trap indeed. If something looks too simple to be right, in case of this competition it is! I think Electroholic is on the right track - I'd really love to see how would you guys solve this before I post the solution.
Electroholic wrote ... i got pretty much the same equation for constant velocity, except i got 1-V/C. its probably just a different convention for V.
However, if you just replace V with A*T, it will give you the instantaneous frequency at the transmitter. Which means you are kinda traveling at with the transmitter but somehow being stationary at the same time. <--- lol
Anyways, the Fnew is going down for sure, distance between the transmitter and the receiver is increasing as well while the wave is getting slower. I'll have to work this out, but I'm going to guess that Fnew will never really reach zero.
The used equation is in concern of static observer and receding transmitter; this lowers the received frequency, and hence the f0 divisor is supposed to be larger than 1.
You seem to be on the right track; the actual solution is somewhat more complex though.
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
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Is there some additional term in the equation to allow for the time it takes sound from the transmitter to actually reach the receiver (which is getting further and further away with the passage of every second) ?
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Acoustic doppler effect problem
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