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Physics: doppler effect (observer & source in motion)

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    A train is moving parallel and adjacent to a highway with a constant speed of 25 m/s. A car, at time t0, is 42 m behind the train, traveling in the same direction as the train at 49 m/s, and accelerating at 4 m/s^2. The train's whistle blows at a frequency of 460Hz. The speed of sound in air is 343 m/s. At the moment t0, the car's driver hears a whistle from the train.

    What frequency does the car's driver hear?

    Part B. @ a time t, after the car has passed the train and is a distance 32 m ahead of the train, the whistle blows again. What frequency does the driver hear?


    2. Relevant equations

    f' =f [ (1+ v(observer) / v) / (1- v(source) / v)] <---equation 1

    x(final) = x(initial) + v t +.5 a t2 <---equation 2

    f'= f [1- (v(observer)/v)] <---equation 3

    3. The attempt at a solution

    Part A:

    460[(1+(49/343)) / (1-(25/343)) = 567.044 = WRONG ANSWER


    Part B:


    I treated the train as stationary to find the time,I determined the car to be traveling 24 m/s
    So using equation 2
    74 = 0 + 24 t + ( .5 * 4 ) t2
    v(final)=49 + ( 2.54 * 4 ) = 59.16 m/s



    then using equation 3 I did

    f'=460(1- (59.16/343) ) = 380.66=WRONG
     
  2. jcsd
  3. Mar 12, 2008 #2
    Is this what you get on simplifying the above equation?

    [tex] f'=f(\frac{v+v_{obs}}{v-v_{s}} )[/tex]

    Try not to apply the equation directly. Since, the frequency is received by the car at that moment [tex]t_{0}[/tex] surely the sound wave of that frequency which we have to find out must have left some time [tex]t_{b}[/tex] ago for it to reach at that instant. Apply this, it might help.

    As for computing the frequency during acceleration, you cannot simply apply the standard equations. For every time period T that the wave moves forward a length [tex]\lambda[/tex] the car moves to cover a length of [tex] \frac{1}{2}aT^2[/tex]. The effective wavelength should now become: [tex] \lambda' = \lambda - \frac{1}{2}aT^2 [/tex].
     
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