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Doppler effect of an ambulance

  1. Jul 22, 2009 #1
    1. The problem statement, all variables and given/known data
    An ambulance is running on an expressway at a speed of 60 km/h from east to west (from A to B) with a siren of 880 Hz. Let the frequency of the siren sound detected by an observer located at a point O be Va and Vb when the vehicle just passes point A and point B respectively. Point C is just north of O and OC = AC = BC = 100 m. Here the sound velocity in calm air is 340 m/s.

    a. In the case of calm weather (no wind), what is the approximate value of (Va / Vb) - 1
    b. In the case of an east wind of 5 m/s, what is the approximate value of (Va / Vb) - 1
    c. In the case of an north wind of 5 m/s, what is the approximate value of (Va / Vb) - 1

    pic-1.jpg

    2. Relevant equations

    [tex]f ' = \frac{V sound \pm V observer}{V sound \mp Vsource} f [/tex]

    3. The attempt at a solution

    Angle CAO = 45 degree
    V source = 60 km/h cos 45 = 50/3 cos 45 m/s
    V observer = 0

    a. [tex]Va = \frac{340}{340 - 50/3 cos 45} 880 [/tex]

    [tex]Vb = \frac{340}{340 + 50/3 cos 45} 880 [/tex]

    (Va / Vb) - 1 = 0.07

    b. Because east wind is from A to B as well, so V source = (50/3 + 5) cos 45 ???

    c. Taking OA as x-axis, I break the velocity of north wind to other components and get the velocity along x-axis
    V source = (50/3 - 5) cos 45 ????

    thx
     
  2. jcsd
  3. Jul 22, 2009 #2
    Hi songoku. :smile:

    There are a couple of questions regaring your problem statement. First, where did you get the angle of 45 degrees? It never specifically said so in the problem statement, so you may want to check that out. Assuming this is correct, however, let's take a look at parts b and c. It may be helpful to transform your reference frame so that the wind is at rest; that way you can use your formula. In the new reference frame for part b, the ambulance and the observer get a "boost" of 5 m/s in the negative x direction. Let me know if this works for you.
     
  4. Jul 22, 2009 #3
    Hi Sam_Goldberg :)

    About 45 degrees, I think that COA is an isosceles triangle because OC = AC. With angle OCA = 90 degrees, angle CAO = (180 - 90) / 2 = 45

    To make the wind at rest, all the system is moving with velocity 5 m/s to the left ?

    thx
     
  5. Jul 23, 2009 #4

    turin

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    Sounds correct to me.
     
  6. Jul 23, 2009 #5
    For b, V source = (50/3 + 5) cos 45 and V observer = 5 cos 45 ???

    Thx
     
    Last edited: Jul 23, 2009
  7. Jul 23, 2009 #6

    turin

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    I believe that those are the projections of the velocities onto AO (in m/s) in the frame of reference at rest w.r.t. the air, assuming that "east wind" blows toward the East (Usually, the wind direction is labeled as the direction where the wind is coming from).
     
  8. Jul 23, 2009 #7
    Yes, those are the projections of the velocities onto AO. But i think i get those values with assumption that east wind blows toward west, i.e. comes from east and moves to west, which is the wind moves from A to B.

    I assume that all the system moves to the left with velocity 5m/s. Because the wind moves in the same direction as the ambulance, i think the velocity will increase and V source becomes (50/3 + 5) cos 45.

    Then :
    [tex]Va = \frac{340 + 5 cos 45}{340 - (50/3 + 5) cos 45} 880 [/tex]


    [tex]Vb = \frac{340 - 5 cos 45}{340 + (50/3 + 5) cos 45} 880 [/tex]

    (Va / Vb) - 1 = 0.117

    I know this is wrong because the answer is 0.07, but i don't know where my mistake is...

    thx
     
  9. Jul 24, 2009 #8

    turin

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    You got it backward. If the wind moves in the same diretcion as the ambulance, then the speed of the ambulance in the air's rest frame is less. For instance, imagine that the ambulance is moving at the exact same speed as the wind, in the same direction. Then, what would be the speed of the ambulance in the air's rest frame? Would it be twice as much as the wind? (No.)
     
  10. Jul 25, 2009 #9
    Oh now i get it.

    And for (c), how to determine the effect of the wind?
    I don't know because the wind is perpendicular to the direction of the ambulance..

    thx
     
  11. Jul 26, 2009 #10

    rl.bhat

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    Wind changes the velocity of the sound.
    If wind moves in the direction of the sound, it adds up.
    If it is in the opposite direction, it will decreases.
    In this problem, the velocity of sound will increase with north wind, i.e wind blowing from north towards the observer.
    When the wind is perpendicular to the sound velocity, it does not change
     
  12. Jul 26, 2009 #11
    So, for (b) the velocity of the sound will be 345 ?
     
  13. Jul 26, 2009 #12

    rl.bhat

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    No. Due to east wind, the velocity of sound along AO will be 340 + 5cos45 and along BO will be 340 - 5cos45.
     
    Last edited: Jul 26, 2009
  14. Jul 26, 2009 #13
    and due to east wind, the speed of the ambulance will be (50/3 - 5) cos 45 and the speed of the observer will be 5 cos 45 along AO?

    thx
     
  15. Jul 26, 2009 #14

    rl.bhat

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    Let observer remain stationary and speed of ambulance remain unaffected by the wind.
    You may get the equation as in your post #7
     
  16. Jul 26, 2009 #15
    Sorry i don't get why the speed of the sound along OB is 340 - 5 cos 45...

    I think at B the wind still moves in the direction of the sound so it still adds up?

    thx
     
  17. Jul 26, 2009 #16

    rl.bhat

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    At B the sound is moving from B to O. The component of wind along the velocity is from O to B
     
  18. Jul 26, 2009 #17
    Oh i see

    and for (c), we just neglect the wind?
     
  19. Jul 26, 2009 #18

    rl.bhat

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    No. It also adds up to the velocity of sound in both the directions.
     
  20. Jul 27, 2009 #19
    Oh i get it from the projection.

    thx a lot rl.bhat ^^
    thx also to turin and sam_goldberg ^^
     
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