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Doppler Effect of a siren

  1. Sep 13, 2013 #1

    dwn

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    1. The problem statement, all variables and given/known data

    A state trooper chases a speeder along a straight stretch of road; both vehicles move
    at 160 km/h. The siren on the trooper's vehicle produces sound at a frequency of 500 Hz.
    What is the Doppler shift in the frequency heard by the speeder?


    2. Relevant equations

    This is what I need the answer to.

    3. The attempt at a solution

    I understand why the answer is 500 Hz. They are both traveling at the velocity with respect to one another, so there is no doppler effect taking place---it is cancelled out by the equal velocity and direction of the vehicles. What I do not understand is, how do I prove it?


    Thanks!
     
  2. jcsd
  3. Sep 13, 2013 #2

    cepheid

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    Do you have any equations pertaining to the Doppler shift in your book or notes? That would be the place to start. There may even be a derivation.

    It's the relative velocity between the source and the observer that matters for the Doppler effect (as you will see when you find the equation). In this case, that's zero. (The source is stationary *relative* to the observer). All motion is relative: velocity is only meaningful if you specify the frame of reference i.e. what it's measured relative to. In the frame of reference of the road, the vehicles may be moving at 160 km/h, but in the frame of reference of the police car, the speeder is stationary (0 km/h) and vice versa.
     
  4. Sep 13, 2013 #3

    dwn

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    Good to go on that question....dumb-dumb didn't check the book first.

    Maybe you could help me get this one started instead. I'm having trouble understanding the beats equation and breaking it down into it's components.

    [tex] D(x,t)=Dcos(2∏(Δf/2)t)sin(2∏ft)[/tex]

    Two automobiles are equipped with the same single frequency horn. When one is at
    rest and the other is moving toward an observer at 15 m/s, a beat frequency of 5.5 Hz is
    heard. What is the frequency the horns emit? Temp. 20°C

    Any help appreciated!
     
  5. Sep 13, 2013 #4

    cepheid

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    When you superpose two sine waves that are very similar in frequency, but not exactly the same, you get beating: the beats are another low frequency component whose frequency Δf is equal to the difference between the frequencies of the two sine waves. One of my favourite examples of this from everyday life: windshield wipers (I think I first saw this on a city bus). Say the left and right wipers on the windshield are just slightly out of sync. It takes the left one just a teeny bit longer to swipe back and forth than the right one. So its swipe cycle frequency is slightly lower. When the two start out they are like this: //. They both swipe towards the other side of the windshield, ending up more or less like this: \\. However, the left one took slightly longer to get there: it lags behind. So, eventually they get out of sync. If you wait for enough swipe periods, they will end up like doing this /\ --> \/, i.e. being completely out of phase. If you wait long enough, they'll get back in phase again, and end up looking like this // -->\\ again. The beat frequency is the frequency with which the wipers go from in phase to out of phase (from \\ // to \/ /\). This beating is a very low frequency (long timescale) oscillation, compared to the high frequency oscillation of the individual swipes.

    If we make a plot of your equation above, it looks something like the attachment below. The high frequency oscillation there is the sine wave (2πf = 50 in this case) and the long slow modulation that it gets multiplied by (the "envelope") is the cosine wave, in this case 2π(Δf/2) = 1. This results from two sine waves of very similar frequency being added together. You should be able to derive the beat frequency equation above by just adding two sine functions together, one of frequency f, and the other of frequency f + Δf. It will require some trig identities to get it into its final form though.


    Side note: when using LaTeX, try using \pi to get ##\pi##
     

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