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Doppler shift with microwaves i think

  1. Mar 12, 2010 #1
    Doppler shift with microwaves....i think....

    1. The problem statement, all variables and given/known data
    Microwaves, which travel at the speed of light, are reflected from a distant airplane approaching the microwave source. It is found that when the reflected waves are beat against the waves radiating from the source, the beat frequency is given by 969 Hz. If the source microwaves are 150 mm in wavelength, what is the approach speed (in km/h) of the airplane?


    2. Relevant equations
    I think that the 3 relevant equations are [tex]C=\lambda[/tex] [tex]\nu[/tex]
    and dopplers equation

    [tex]F= \left(v+v_{r}\right)/\left(v+v_{s}\right)*F_{0}[/tex]

    [tex]F_{beat} = \left|F_1-F_2\right|[/tex]

    3. The attempt at a solution
    My attempt at the solution is to take the speed of light, divide by .150 m, to get the frequency of my microwave. This value gives me [tex]\nu =1.998*10^9[/tex].
    I then put it into the beat formula and solve for [tex]F_1= 1.997*10^9[/tex].
    I would then sub into my dopplers formula, except i don't know if the source is moving towards it or if it is moving away. I am also not sure if i did any of this right because the microwave frequency is so high, that the beats have virtually no affect. Any insight would be appreciated, thank you in advance.
     
    Last edited: Mar 12, 2010
  2. jcsd
  3. Mar 13, 2010 #2

    rl.bhat

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    Re: Doppler shift with microwaves....i think....

    Hi Dorbo, welcome to PH.
    In the case of light, the Doppler effect is given by
    Δf/f = v*cosθ/C.
     
  4. Mar 13, 2010 #3
    Re: Doppler shift with microwaves....i think....

    I am sorry, what is [tex]\Delta f[/tex] in your equation? And how do beats fit into it? From what i understand [tex]\theta = \pi[/tex] because the waves are going in opposite directions. I do understand what C is though :p.
     
  5. Mar 13, 2010 #4

    rl.bhat

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    Re: Doppler shift with microwaves....i think....

    Δf is the beat frequency.
    The airplane is approaching. So the source appears to be coming towards the observer. Hence θ = 0.
     
  6. Mar 13, 2010 #5
    Re: Doppler shift with microwaves....i think....

    Ok i put in [tex]\frac{969*C}{2*10^9}[/tex], i then got a value of 145.35 m/s, i then converted it to 523.26 km/h. I then submitted this answer and got it wrong. Any ideas, and thank you for the help so far. I got the 2*10^9 by [tex] c= \lambda \nu[/tex]
     
  7. Mar 13, 2010 #6

    rl.bhat

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    Re: Doppler shift with microwaves....i think....

    Δf/f = v*cosθ/C.
    This formula is for a source approaching the stationary observer.
    In this case, when the reflector moves x meter per second with respect to the source, the image of the source in the reflector approaches the source with 2x meter per second. So try the answer 2*523.26 km/h.
     
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