I got this right on accident. Either way, my thought process was in order to have the v vector perpendicular to the road, the 2 additive vectors must include one going towards the right and one going downward, which means the velocity decreases.... But this has nothing to do with the problem after I read the explanation. Can you please shine some light on my thought process and how this is irrelevant?
Also, what's the significance of the v vector being perpendicular to the road? and how do we know that the observer and source are not moving closer together or further apart?
Doppler: the v vector being perpendicular to the road
Doppler: the v vector being perpendicular to the road
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Re: Doppler: the v vector being perpendicular to the road
Hey, this one almost got lost.
I've attached a little drawing of the scenario. Let me know if that helps!
Essentially, for the instant that the ambulance is perpendicular to the observer, it is neither approaching nor departing. Therefore, for just that one instant, no Doppler shift is observed.
Edit: The image disappeared for a while, it is up now.
I've attached a little drawing of the scenario. Let me know if that helps!
Essentially, for the instant that the ambulance is perpendicular to the observer, it is neither approaching nor departing. Therefore, for just that one instant, no Doppler shift is observed.
Edit: The image disappeared for a while, it is up now.
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 ambulancedoppler.pdf
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Re: Doppler: the v vector being perpendicular to the road
I knew it has something to do with the 90 degree angle..How do you know that it's cosine. What equation is this? and what's your thought process to this?

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Re: Doppler: the v vector being perpendicular to the road
That is just a decomposition of the approach vector using trigonometry. Since the ambulance is not approaching you directly, but driving past, it's velocity vector has a component approaching you and a component perpendicular to you. Only the component approaching you contributes to the Doppler effect.
In the instant the question is asking about, the ambulance is moving PURELY perpendicular to the observer. Therefore no Doppler shift should occur (for that one instant) and the true frequency should be observed.
Consider the same road, and yourself standing on the side. You understand that when the ambulance is first approaching, most of its motion is directly at you. When it is passing you, is entire motion is perpendicular to you. Then the next instant, it starts departing. That is all that is going on here.
In the instant the question is asking about, the ambulance is moving PURELY perpendicular to the observer. Therefore no Doppler shift should occur (for that one instant) and the true frequency should be observed.
Consider the same road, and yourself standing on the side. You understand that when the ambulance is first approaching, most of its motion is directly at you. When it is passing you, is entire motion is perpendicular to you. Then the next instant, it starts departing. That is all that is going on here.
Re: Doppler: the v vector being perpendicular to the road
This is a very silly question cause it's not obvious to me: how do you know the object's entire motion is perpendicular to me when it's away from me?

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Re: Doppler: the v vector being perpendicular to the road
I hope this addresses the hangup. Note that in order to observe a Doppler shift, the emitter must be approaching or departing. If we can prove that the emitter is doing neither, there must be no shift at that instant.
So that is why we spend all this time puzzling out whether the ambulance itself is approaching or departing that very instant.
So that is why we spend all this time puzzling out whether the ambulance itself is approaching or departing that very instant.
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