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Doppler effect Indianapolis 500 problem

  • #1

Homework Statement



At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise between approaching and receding cars. Suppose the sound of a certain car drops by a factor of 1.50 as it goes by on the straightaway. How fast is it going? (Take the speed of sound to be 343 m/s.)

answer given in km/h

Homework Equations




v= (1- (f/(f prime/1.5))v

where the second v is the speed of sound.

Sorry it isnt a cleaner equation, I couldn't figure out how to get my mathtype equation from microsoft word here.




The Attempt at a Solution


The equation above is my attempt at a solution, since the sound of the cars frequency drops by 1.5 thats what I divided my f prime by. I just dont know how to figure out the f values. Maybe I am using the wrong equation, I just can't figure out any other way to solve for the speed of the car. Any hints would be greatly appreciated and I thank you for your time.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
1.5 IS f/f'. You don't need to know the values of f.
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
618
And be careful. When the car is coming toward you it's doppler shifted up in pitch and when it leaves you it is shifted down. As I read the question there is a factor of 1.5 between the ratios of those two pitches.
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,599
203

Homework Statement



At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise between approaching and receding cars. Suppose the sound of a certain car drops by a factor of 1.50 as it goes by on the straightaway. How fast is it going? (Take the speed of sound to be 343 m/s.)

answer given in km/h

Homework Equations




v= (1- (f/(f prime/1.5))v

where the second v is the speed of sound.

Sorry it isnt a cleaner equation, I couldn't figure out how to get my mathtype equation from microsoft word here.




The Attempt at a Solution


The equation above is my attempt at a solution, since the sound of the cars frequency drops by 1.5 thats what I divided my f prime by. I just dont know how to figure out the f values. Maybe I am using the wrong equation, I just can't figure out any other way to solve for the speed of the car. Any hints would be greatly appreciated and I thank you for your time.
Welcome to the forums!

I don't quite see how you got your equation but I have a feeling that you used incorrectly the factor of 1.5.

Set up an equation for the frequency of the sound when the car is going toward you. Call this frequency [itex] f^' [/itex]. It will contain the frequency emitted by the source that I will call [itex] f_{source} [/itex] to be clear. Now find an expression for the frequency when the car is moving away from you. Let's call this f double prime = [itex] f^{''}[/itex].

What they are telling you is that [itex] \frac{f^'}{f^{''}} = 1.5 [/itex]. So take the ratio of your two equations and you will see that [itex] f_{source}[/itex] will drop out, leaving you with an equation for the speed of the source (the car) in terms of the speed of sound.
 

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