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Doppler effect

  1. Aug 8, 2009 #1
    When an ambulance approaches, its siren sounds shriller than when I relatively motionless to it, and vice versa. Such is an example of doppler effect, isn't it?

    However, there one thing I am not sure: does the siren gradually becomes more shriller when the ambulance approaches, or not?

    (Assuming constant speed for the ambulance)
     
  2. jcsd
  3. Aug 8, 2009 #2
    annatar,0

    This is the equation for doppler effect of sound.

    Example:

    A sound source of 4000 hz is moving towards a stationary person at a speed of 80 m/s.

    f = frequency at observer, c = speed of sound, V= velocity of source, f1= frequency of sound source.

    f= f1 x c/ (c-v)

    = 4000 x 340/ (340-80) = 5230 Hz

    remember this will continue to vary.
     
  4. Aug 8, 2009 #3

    Doc Al

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    If the ambulance comes directly at you, no, the observed frequency will be steady. But if it passes by you, then the observed frequency will vary from its highest to its lowest as it passes you. (What matters is the radial velocity of the siren with respect to you.)
     
  5. Aug 8, 2009 #4
    Doc,

    I'm confused, i've put up a good example of one scenerio, are you saying that on a US battle ship that the sailors will hear a steady frequency of Kamikazi pilots ? LOL!
     
    Last edited by a moderator: Aug 10, 2009
  6. Aug 8, 2009 #5

    Doc Al

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    Read exactly what I wrote. If you are stationary in the water, and the plane is headed straight towards you, then yes, the observed frequency is steady.

    Edit: It seems I misinterpreted your scenario with the planes. In the ambulance example, the speed of the ambulance was constant. (I thought you agreed; After all, you used a single speed in your calculation.) So I assumed you meant for the planes to have a constant speed as well, since you were comparing that example to the ambulance example. Of course, if the approaching sound source is accelerating, then the observed frequency will increase as the speed increases. (And as others have pointed out, it's likely that the planes are accelerating.) Sorry if I confused you.
     
    Last edited: Aug 10, 2009
  7. Aug 8, 2009 #6
    You use the word steady? do you mean a frequency of 3000 HZ will remain 3000 Hz or steady means the frequency will build up constantly?
     
  8. Aug 8, 2009 #7

    Doc Al

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    By "steady" I mean unchanging. (Not steadily increasing!)
     
  9. Aug 8, 2009 #8
    Hmm,so when the ambulance is coming/leaving, the sound heard will have a higher/lower frequency than the wave source, but remains steady at that frequency?

    If the ambulance is accelerating directly towards you...I guess the frequency will rise then?
     
  10. Aug 8, 2009 #9
    f = frequency at observer, c = speed of sound, V= velocity of source, f1= frequency of sound source.


    Example 2:

    A Jap plane heads towards a US battle ship:

    f1 = 1500 Hz

    V= 100 m/s

    f= f1 x c/ (c-v)

    f = 1500 x 340/ (340-100) = steady 1500 Hz :rofl:
     
  11. Aug 8, 2009 #10

    Doc Al

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    Yes. (If it's heading directly towards you at constant speed.)

    Sure. The speed is increasing.
     
  12. Aug 8, 2009 #11
    Thanks guys...I think I got it :smile:
     
  13. Aug 8, 2009 #12

    russ_watters

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    Note that for an ambulance heading almost directly toward you, the velocity toward you is always decreasing and so the doppler shift is always changing (frequency is decreasing).
     
  14. Aug 9, 2009 #13
    http://www.school-for-champions.com/science/sound_doppler_equations.htm" [Broken]

    Frequency

    The equation for the observed frequency of sound when the source is traveling toward you is:

    fo = fv/(v − vt)

    where

    * fo is the observed frequency
    * f is the emitted frequency
    * v is the velocity of sound
    * vt is the velocity of the source toward you
    * v > vt (vt is less than v)

    Note that this equation does not work if the speed of the source is equal to the speed of sound. In such as case, you would be dividing by 0, which is impossible.

    This matches my previous equations.
     
    Last edited by a moderator: May 4, 2017
  15. Aug 9, 2009 #14
    so 1500 x 340/240 = 1500.....
    thats new.....
     
  16. Aug 9, 2009 #15

    Vanadium 50

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    If an object is moving towards you at the speed of sound, you can't hear it until it's right on top of you. The equation is telling you something.

    By the way "Jap" is considered an ethnic slur by many.
     
  17. Aug 9, 2009 #16
    Vanadium,

    You or others will not convince me untill you put forward some calculations.

    I've put mine forward with also a back up from that web site i posted the link.

    BTW i'm from the UK we aren't so frivolous here regarding "alleged slurs"
     
  18. Aug 9, 2009 #17

    Doc Al

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    Convince you of what? Is there something you don't understand about the Doppler effect?
     
  19. Aug 9, 2009 #18

    turin

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    Turv, Kamikazes surely accelerate, since they are on a downward trajectory (I don't believe that they apply their brakes as they head for their target). So the pitch will increase due to their increased speed as they approach.
     
  20. Aug 9, 2009 #19

    russ_watters

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    The ironic thing is that the calculation you posted doesn't imply what you concluded at the ende of your post! So if you don't want to believe your own math you'll have to go find some more math!
    Now you know. Please incorporate this newfound undestanding into your future contributions to this site.
     
  21. Aug 9, 2009 #20

    russ_watters

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    Good point!

    It does include a couple of assumptions, that may or may not always be true:
    1. It assumes that they are always on a downward trajectory (they probably usually are).
    2. It assumes they haven't reached their terminal velocity.
     
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