Question that made my brain hurt :)

[fishslayer]

(firstly not sure if this goes here but if it doesnt then please move it)

 

I was set this question by somebody and i have not been able to answer it. I have a rough idea what the answer is but ive not a clue how i can prove it. Does anybody have any help?

 

Here is the question:

 

A person stands on the sidwalk when he/she hears the siren of an ambulance which approaches with velocity V. The ambulance passes in front of him/her and continues its way, getting further and further with the same velocity V. Let f1 and f2 be the frquencies of the siren heard by this person when the ambulance is approaching and when the ambulance is getting further respectively. If f0 is the frequency of the siren as heard by the ambulance driver, which statement below is true?

 

A - f1 < f2 and f0 is slightly larger than (f1+f2)/2

B - f1 < f2 and f0 is slightly smaller than (f1+f2)/2

C - f1 > f2 and f0 is slightly larger than (f1+f2)/2

D - f1 > f2 and f0 is slightly smaller than (f1+f2)/2

 

P.S. I narrowed it down myslef to C or D and i am confident it's C but i do not know if thats true or how to prove im right/wrong.

[ydoaPs]

I'm not sure why this is in Relativity. Anyway, look up Doppler Effect.

[RyanJ]

Try putting the information here. That should help you understand it

[fishslayer]

RyanJ - i dont think that will help as it doesnt have values (and if you put values in i think it screws up)

 

ydoaPs - yeah i wasnt sure where to put it so i put it here

[RyanJ]

RyanJ - i dont think that will help as it doesnt have values (and if you put values in i think it screws up)

 

Maybe not the exact one I listed there but there are other options for different equation sets there too. One of those should have what you want.

[fishslayer]

ok but does anybody have an answer for this exact question?

[swansont]

ok but does anybody have an answer for this exact question?

 

The goal is to provide you with the information to answer it yourself, rather than just answer it for you. Why do you think C is correct?

[fishslayer]

actually personally i would jsut like the answer and the proof, ive spent at least a week on it and im getting nowhere. Everybody ive gone to for help doesnt seem to know the answer .

 

Why i think its C. I worked out it had to be D or C and due to sum weird thinking i came up with C.

[swansont]

It shouldn't require weird thinking. A/B vs B/C can be narrowed by whether the frequency shifts up or down as the siren approaches. Within each pair, the location of the center frequency, as compared to the average, depends on the frequency shift — if it were a symmetric shift (i.e. the same amount up and down), they would be equal. Is the up-shift larger or smaller than the down-shift?

[fishslayer]

all the information i gave in the question i wrote out is all i have to answer it.

[swansont]

all the information i gave in the question i wrote out is all i have to answer it.

 

And you can answer it if you know the Doppler shift equation, which was given in the link provided by RyanJ.

[fishslayer]

ok, so i have relooked over the question, done some calcualtions etc and still havent got that far. i personally dont think i have learnt enough of physics to understand some of it but could somebody please tell me what is the right answer and why it is. (i have tried to use the equations on the doppler effect etc and havent had much sucsess)

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oh lol managed to answer is it

 

you can elimanate (sp?) A and B beacuse of the doppler shift effect and beacuse E=hf energy is proportional to frequency so it must be d.

[dalemiller]

The number of oscillations involved with the sound should be unchanged. An approaching sound source delivers them all in less time than an equivalent stationary source, hence total oscillations delivered in less time yields higher frequency to stationary observer. A departing sound source should increase delivery time by the same value. Since fixed time delay has less lowering effect upon the lower frequency f2 and more pitch raising effect upon the higher frequency f1, then condition D should result.

 

Fixed time variation takes a smaller bite out of longer wave length.

[Sorcerer]

The number of oscillations involved with the sound should be unchanged. An approaching sound source delivers them all in less time than an equivalent stationary source, hence total oscillations delivered in less time yields higher frequency to stationary observer. A departing sound source should increase delivery time by the same value. Since fixed time delay has less lowering effect upon the lower frequency f2 and more pitch raising effect upon the higher frequency f1, then condition D should result.

 

Fixed time variation takes a smaller bite out of longer wave length.

 

He said "D"... is he right?

[fishslayer]

Yeah as i D was right. It seems simple now. As f and E are connected (E=hf) you can narrow choices down. It cant be A or B as they dont make sense. In C in order for it to be a higher f it must have gained energy from something. There is nothnig for it to gain energy from so its D.