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Doppler Effect and car alarm

  • Thread starter exi
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  • #1
exi
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Homework Statement



A parked car's alarm goes off, producing a sound at 961 Hz.

As you drive toward, pass, and drive away from this parked car, you observe a frequency change of 98 Hz. At what speed are you driving?

(Speed of sound is 343 m/s.)

Homework Equations



[tex]f_o = f_s\left( 1-\frac{v_o}{v} \right)[/tex] (Observer moving away from source)

And possibly

[tex]f_o = f_s\left( 1+\frac{v_o}{v} \right)[/tex] (Observer moving towards source)

The Attempt at a Solution



Interesting problem. Using the above formula for Doppler shift when an observer is moving away from a source, I was coming up with

[tex]f_o = 863\left( 1-\frac{v_o}{343 m/s} \right)[/tex]

which solves to 34.9781 m/s, an incorrect answer.

Any hints on what I'm doing wrong? I was wondering if two-stepping this and using both equations would get anywhere, but without knowing for sure what the observer's frequency would be on that half of the trip, I'm a bit unsure of things.

Thanks.
 
Last edited:

Answers and Replies

  • #2
bel
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First you derive the expression for the change of frequency as you move towards the parked car, then you add it to the frequency change as you move away. What you have now is the total frequency change, and you have a number to which your expression could be equated, and you have only one unknown since your velocity was assumed to be constant.
 
  • #3
exi
85
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First you derive the expression for the change of frequency as you move towards the parked car, then you add it to the frequency change as you move away. What you have now is the total frequency change, and you have a number to which your expression could be equated, and you have only one unknown since your velocity was assumed to be constant.
Hmm... so:

[tex]f_o = f_s\left( 1+\frac{v_o}{v} \right) + f_s\left( 1-\frac{v_o}{v} \right)[/tex]

Distributing the frequency of the source and solving for the observer's velocity results in Vo terms that cancel out. Seems I'm doing something stupid, but... what?
 
  • #4
Dick
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bel suggested you add the CHANGES in frequency. You can also do this by taking the difference instead of the sum. You want the change of frequency, which is a DIFFERENCE.
 
  • #5
exi
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bel suggested you add the CHANGES in frequency. You can also do this by taking the difference instead of the sum. You want the change of frequency, which is a DIFFERENCE.
Don't I feel silly. I'm coming up with

[tex]f_o = 2*f_s*v_o[/tex]

I've tried several different numbers, but I seem to keep getting stupidly small (no greater than 4 m/s) velocities. The frequency change suggests to me that it should be higher, so I'm thinking I'm making a mistake as to just which number is equated to the above equation (in place of f_o).
 
  • #6
Dick
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What happened to the division by v?? And what are you substituting for the various variables?
 
  • #7
exi
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What happened to the division by v?? And what are you substituting for the various variables?
I think I just caught another algebra / calculation mistake.

Well, correcting the above

[tex]f_o = f_s\left( 1+\frac{v_o}{v} \right) - f_s\left( 1-\frac{v_o}{v} \right)[/tex]

seems to allow f_s distribution to arrive at

[tex]f_o = f_s+f_s\frac{v_o}{v} - f_s + f_s\frac{v_o}{v}[/tex]

and multiplying terms by v to remove that fraction gives

[tex]f_o(v) = f_s(v) + f_s(v_o) - f_s(v) + f_s(v_o)[/tex]

the f_s(v) terms cancel, and that boils down to, if I'm correctly understanding Bel,

[tex]98(343) = 2(961)(v_o)[/tex], or [tex]1922(v_o) = 33614[/tex]

And 17.4891 m/s.

Ouais?
 
  • #8
Dick
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Ouais, certainment. (One of your intermediate eqns is missing some stuff, but let's pass on that).
 
  • #9
exi
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Any other fundamental errors I'm making here? I'd hate to miss an almost-right question. o:)
 
  • #10
Dick
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Do you believe it? Does it feel right? Can you think of anyway to check it? I'm not going to be here all the time to check.
 
  • #11
exi
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Do you believe it? Does it feel right? Can you think of anyway to check it? I'm not going to be here all the time to check.
Not sure, yes, and not offhand (although I did stumble upon a quadratic form of an equation to do this problem, which gave me a number within .04 of my own answer).

Et c'était une réponse correcte aussi; merci encore.
 
  • #12
Dick
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Glad to hear the 'yes'. I believe it. Bon chance! Here's what I've checked. 98Hz is about 10% of 961Hz. That means 5% shift approaching and 5% retreating. 5% of the speed of sound is 15m/sec. And that means you are right. Try doing problems in without a calculator sometimes. Just roughly.
 
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