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Dopple Effect (Frequency vs Time)

  1. Mar 12, 2010 #1
    Doppler Effect (Frequency vs Time)

    1. The problem statement, all variables and given/known data
    A jet airplane is flying above an observer parallel to the ground with speed vj. Given the observed frequency vs time graph, would you be able to calculate the altitude of the jet airplane? Explain, also, how to determine observed frequency as a function of x, where x is the initial horizontal distance of the jet airplane to the vertical directly above an observer.

    2. Relevant equations
    v = [tex]\lambda[/tex]f
    x = vt

    3. The attempt at a solution
    I am really clueless on how to start this - I've been staring at this question for over an hour.
    Is it true that if I find two values of time so that the inverse of the difference is a value of frequency on the graph, then that frequency is the frequency of the jet airplane? (The only problem is that if the jet airplane is initially moving away, this value will not be on the graph.)
    Can someone give me a lead please?
     
    Last edited: Mar 12, 2010
  2. jcsd
  3. Mar 12, 2010 #2

    rl.bhat

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    Homework Helper

    Resolve vj into two components. vj*sinθ towards the observer.And another component vj*cosθ perpendicular to the line joining jet and the observer. θ is the angle between vertical and line joining the jet and the observer.
    vj*sinθ changes the frequency where as vj*cosθ does not affect the frequency.
    So f' = [v/(v-vj*sinθ)]*f for approaching jet and f' = [v/(v + vj*sinθ)]*f for receding jet.
    Sinθ = x/sqrt(x^2 + h^2) and x = vj*t.
     
  4. Mar 13, 2010 #3
    Ohh I get it now. Thanks a lot!
     
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