Doppler Effect

Discussion in 'Physics' started by logearav, Sep 26, 2011.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
    248
    0
    Revered members,
    I have attached the image of Doppler effect explanation
    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
     
  2. BillO

    Well-Known Member

    Nov 24, 2008
    985
    136
    Because the next wave peak does not get emitted until To (one period of the wave) has passed. Then to that we need to add the time of propagation, ( L + vsT0)/v, so that t2 (the next time you detect the peak) = T0 + ( L + vsT0)/v.
     
  3. logearav

    Thread Starter Member

    Aug 19, 2011
    248
    0
    Sir, I can't understand.
     
  4. BillO

    Well-Known Member

    Nov 24, 2008
    985
    136
    Read 15.8.1 very carefully. It is well explained there.

    T0 is the period of time between peaks emitted by the moving source, so some multiple of T0 must be a part of the time the observer detects between each successive peak after t1, as well as the time of propagation from the point of emission to the observer.

    So, we have the following sequence;

    <br />
t_{1}\ =\ L/v<br />
t_{2}\ =\ T_{0}\ +\ (L\ +\ v_{s}T_{0})/v<br />
t_{3}\ =\ 2T_{0}\ +\ (L\ +\ 2v_{s}T_{0})/v<br />
t_{4}\ =\ 3T_{0}\ +\ (L\ +\ 3v_{s}T_{0})/v<br />
.<br />
.<br />
.<br />
t_{n+1}\ =\ (n)T_{0}\ +\ (L\ +\ (n)v_{s}T_{0})/v<br />
.<br />
.<br />
.<br />
     
    logearav likes this.
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