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Doppler Shift/ Doppler effect HELP

  1. Jun 5, 2013 #1
    Doppler Shift/ Doppler effect!! HELP

    1. A police car is traveling at a constant speed of 110. km/h with a radar gun which generates and detects radiowaves. The police car and the ambulance drive directly towards each other. The radar gun produces radiowaves with a wavelength of 0.0440 m.
    An ambulance is traveling at 64.0 km/h towards an intersection
    The ambulance has a siren which produces sound at a frequency of 4.30 kHz. Assume the speed of sound in air is 340. m/s.

    (i) What is the relative speed between the police car and the ambulance?

    (ii) What doppler shift (Δf = f′–f) is detected by the radar gun after the radiowave has been reflected by the ambulance?




    2. Relevant equations
    relative speed = v1 + v2

    f obs = fs ((vw + v obs) / (vw + vs))
    Vw = 340 m/s (speed of sound...)
    v obs = 64km/h (17.778 m/s)
    vs = 110 km/h (30.556 m/s)
    fs = 4.3 kHz

    (Δf = f′–f)
    f = 4.3 kHz

    v = fλ (I got λ = 0. 044 so what am I using it for??)
    3. The attempt at a solution
    A) Easy 110 +64 = 174 km/h

    B) The problem child.
    I plug in with the equation above, with + because the ambulance and the police are moving towards each other, but I don't get the right answer (2,200 Hz)
    I know I need to find Δf, but how do I go about to find it? Why is this equation not working, and which equation am I meant to use? Whats the radar wavelength for? How do I implement it?

    Any help would be much appreciated!!
     
  2. jcsd
  3. Jun 5, 2013 #2
    you have taken a wrong value of Vs. Check it.
     
  4. Jun 5, 2013 #3
    What do you mean?
    I don't understand.
    Vs is the source, and the ambulance is the source of the siren sound.
    It is travelling towards the observer which is the police car (with the radar detector of λ wavelength)....

    If the ambulance isn't the source, then what is? The police car?
    Then the ambulance is the observer?
    I'm so confused....
     
  5. Jun 5, 2013 #4
    The sound wave is changing frequency twice. One the frequency observed by the observor in ambulance. Another one is the that observed b ht observor in the police car(that is radar detector). So the equation

    f obs = fs ((vw + v obs) / (vw + vs))

    will be used twice.

    NOTE: Use the values of Vs and Vobs carefully as the observor and the source is changing and keep in mind when the value will be taken negative/positive
     
  6. Jun 5, 2013 #5
    Okay,
    so I'm meant to use the equation twice.

    In that case,
    f obs (police) = fs (( vw + vobs) / (vw + vs ) )

    I keep fs = 4.3kHz and as both sources are still going towards each other (+)
    vw = 340 m/s
    V obs (detector) = v = fλ = 4,300 * 0.044 = 189.2 km/h (???) (52.56 m/s)
    Vs = 64 km/ h (17.778 m/s)

    f (obs police) = 4.718 kHz

    f obs (ambulance) = fs (( vw + vobs) / (vw + vs ) )

    As above keep vw and (+)
    v/λ = fs = 110 km/h * 0.044 = 1.34 kHz (???)
    Swap Vobs = 64 km/h (17.778 m/s)
    vs (police) = 110 km/h (30.558 m/s)

    f (obs ambulance) = 1.293 kHz

    Δf = 4.718 - 1.293 = 3.42 kHz ??? (not the right answer of 2,200 Hz)...

    Mind pointing out what I'm getting wrong?
     
  7. Jun 6, 2013 #6
    I give up.
    Don't get it.
    Good luck to the rest!
     
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