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Doppler effect and some doubts

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Revered members,
    I have attached the image of Doppler effect explanation.

    2. Relevant equations

    I have the following doubts
    1)t1 = L/v
    2) t2 = T0 +( L + vsT0)/v
    But why T0 comes here, instead of ( L + vsT0)/v

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 25, 2011 #2
    A stationary source emits waves of frequency F and period 1/F. The period is the time it takes successive wave fronts to reach you. How does the period change when the source moves away from you? First you need to know how far the source moves in the time 1/F, that is simply 1/F times the source velocity v_s. Now how much is the period lengthened? By the extra time it takes wave to go the extra distance v_s/F which is delta_t =v_s/Fv where v is the wave velocity. So the new period between wave fronts of a moving source is

    1/F' = 1/F + v_s/Fv = 1/F(1+ v_s/v)

    Makes more sense to me thinking of the period then the frequency?

    If you are troubled by what is written derive the formula yourself.
     
  4. Sep 27, 2011 #3
    A stationary source emits waves of frequency F and period 1/F. The period is the time it takes successive wave fronts to reach you. How does the period change when the source moves away from you? First you need to know how far the source moves in the time 1/F, that is simply 1/F times the source velocity v_s.
    Thanks for the reply Spinnor. I don't understand the line given in quote.
     
  5. Sep 28, 2011 #4
    Lets say you are in a car and beep your horn once a second. Everyone who is at rest relative to the car will also hear the horn beep once a second. Now suppose you drive away from me at some large velocity, say 1/6 the speed of sound all the while beeping your horn once a second. With this information you should be able to compute the amount of time I hear between beeps. Yes? It will be longer then one second because the beep has longer to travel. You only need three numbers, the period of the beeps, the velocity of the car, and the velocity of sound in air.
     
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