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Sound Frequency

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A truck moving at 36 m/s passes a police car moving at 45 m/s in the opposite direction. If the frequency of the siren relative to the police car is 500 Hz, what is the frequency heard by an observer in the truck after the police car passes the truck? (The speed of sound in air is 343 m/s.)
    a)396 Hz
    b) 636 Hz
    c)361 Hz
    d)393 Hz
    e)617 Hz

    2. Relevant equations
    f′ = [(v + vO)/(v − vS)]f

    3. The attempt at a solution
    Using the equation above, as they are moving away from each other, i made vO = -36 m/s and vS = -45m/s. Plugging in those values, I get 396 Hz, which is option a) however the correct answer appears to be 636 Hz, which can only be obtained assuming that they are moving towards each other.
     
  2. jcsd
  3. Feb 26, 2015 #2

    SteamKing

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    The way you have specified the velocities VO and VS, both the observer and the police car are moving in the same direction. Over time, the distance separating the truck from the police car should increase.
     
  4. Feb 26, 2015 #3
    I'm sorry but I don't understand what you mean?
     
  5. Feb 26, 2015 #4
    I get 636 but I added the cars velocities because they are moving AWAY from each other.
     
  6. Feb 26, 2015 #5
    When they are moving away from each other, the equation becomes f' = [(v-vO)/(v+vs)]f. Thats what it says in our textbook, and when doing that, I get the value of 396Hz.
     
  7. Feb 26, 2015 #6
    get rid of the negative signs when you do the equations. :wink:
     
  8. Feb 26, 2015 #7

    SteamKing

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    If you make the velocity of the truck and the police car both negative, it implies that the two vehicles are traveling in the same direction, relative to a fixed point. Same deal if both velocities are positive. It stands to reason then, that if the truck and the police care are traveling in opposite directions, then their velocities must have different signs.
     
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