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Doppler Effect physics homework

  • Thread starter thatnewkid
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  • #1
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So a factory whistle emits a sound at 875Hz. The temperature of the air is 25.9oC. What frequency will be heard by an observer in a car traveling at 27m/s away from the source

i know this is a two step problem but i have no idea where to start! anyone lead me in the right direction?
 

Answers and Replies

  • #2
lewando
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Have you considered the Doppler Effect?
 
  • #3
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i haven't herd of that but im guessing that's possible...how does that work?
 
  • #4
Redbelly98
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i haven't herd of that but im guessing that's possible...how does that work?
Find the discussion of Doppler effect in your physics textbook, with the equation you can use. Then post back with your attempt at solving the problem.
 
  • #5
lewando
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  • #6
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ok ok hold up im trying it
 
  • #7
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is it 0.36518857142857? i understand what the source velocity is and the initial wavelength frequency is but i think i may have messed up where temperature plays a role
 
  • #8
lewando
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Let's use some common sense here. The source frequency is 875Hz. Your result should be in this neighborhood. Higher? Lower? What is your experience watching NASCAR? As the car approaches you what does it sound like? As it goes away what does it sound like? Also: way too many digits--use sig figs.
 
  • #9
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alright so i have f1= 875(v-27/v+vs)

how do i find the speed of the source with the temperature?

v=speed of wave
vo=speed of observer
vs=speed of source
f=frequency at source
f'=frequency observed by observer.
 
  • #10
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no nm i think i got it is it 811.75Hz?
 
  • #11
lewando
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You need to show your work. How did you arrive at the speed of sound at 25.9C?
 
  • #12
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V of sound = 331 + .6(temp)=331 + .6(25.9)=346.54m/s

f = actual f (v - v of observer/ v+ v of source)
f= 875(346.54/346.54 +27)
f= 811.75Hz
 
  • #13
lewando
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Your speed of sound looks right. Your result does not. Your v of source should be zero. Your v of observer is 27m/s
 
  • #14
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ok i just got so confused, help me out with what your trying to get at? (sorry i may seem dumb but i haven't done this in a long time and im trying to learn it again to prepare for my final)
 
  • #15
lewando
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You are so close, hang in there!

You correctly stated:
f = actual f (v - v of observer/ v+ v of source)
but this was wrong:
f= 875(346.54/346.54 +27)
Just evaluate the equation with "v of observer" = 27m/s and "v of source" with 0m/s
 
  • #16
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so...

875(346.54-27/346.54 + 0)=

806.826Hz?
 
  • #17
lewando
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Good work by you. Remember to watch your signs based on how the source/observer is moving. I am not a significant digit maniac, but I think you should go with 807Hz.
 

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