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Dimensional analysis

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    The speed v of sound waves in air depends on the atmospheric presure P and density. By using the method of dimensional analysis, find an expression for v in terms of P and density.



    2. Relevant equations

    What I have so far;

    V= density x P



    I found a textbook which said that M can be ignored, as such...I end up getting

    [L][T^-2]=[M^x+y] [T^-2x]

    M---> 0 = x +y
    y = x

    L---> 1= -1x-3y

    T---> T-1= -1
    T = 0


    It must be obvious by now that I am missing a key concept here...any help would be appricated!
     
  2. jcsd
  3. Sep 14, 2010 #2

    lewando

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    What units are you using for v?
    What units are you using for P?
    What units are you using for ρ?
     
  4. Sep 15, 2010 #3
    Hey lwando,

    V= [L^3]
    Pressure= F/A [M][L^-1][T-^2]
    Density= [M][L^-3]
     
  5. Sep 15, 2010 #4

    lewando

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    try:

    V (velocity) = [L][T^-1]
     
  6. Sep 15, 2010 #5

    lewando

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    Key concept: put V on the left hand side of the equation, and put different arrangements of P and ρ on the right hand side and, based on what you know about the units (dimensions), see if the units/dimensions are equivalent. It becomes a puzzle-solving problem.
     
  7. Sep 15, 2010 #6
    I think I have a bit of a hang on this now....

    [L][T^-1] = [M][L^-1][T-^2] [M][L^-3]

    We ignore [M] as it is not on the left side thus;

    [L][T^-1] = [L^-1x][T-^2x] [L^-3x]


    -1 = -2x -------> x=1/2

    1= -1x - 3y sub x= 1/2

    y= -2/3


    V= sqt pressure / density ^2/3


    does that seem like the proper way to do? This might be a very simplistic q...but how do I take care of that 2/3?

    Thanks allot! This is a second year phys course everyone seems to be flying through except for the older guy that came with a biology degree (me!)
     
  8. Sep 15, 2010 #7

    lewando

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    You are sort of on the right track . A coulple of points:

    1) Cannot simply "ignore" M on the other side. M has to go away naturally (hint: M/M or using our notation, [M][M^-1] needs to appear somehow on the RHS. Noting your earlier post:

    Pressure= F/A [M][L^-1][T-^2]
    Density= [M][L^-3]

    it would seem that some form of division needs to happen between P and ρ to make M go away.

    2) You came up with V= sqt pressure / density ^2/3 as a result of your process that I am struggling to understand (also "sqrt" of just pressure? or the whole RHS? Not clear.) Still, the units (dimensions) should balance on both sides. You should check this result. I don't think they balance.

    Have another go at it.
     
  9. Sep 15, 2010 #8

    lewando

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    One more point: I would start with an hypothesis expression involving V, P, and ρ, then substitute the units in place of the variables, simplify and see if the units balance. If not, tweak your V, P, ρ expression and repeat the process. Insight will eventually occurr. You started with:

    [L][T^-1] = [M][L^-1][T-^2] [M][L^-3]

    which is V = Pρ

    Backwards from how I suggest (but who am I to force a method)

    Clearly they don't balance because of M on RHS, so V = Pρ can't be the expression.
     
  10. Sep 16, 2010 #9
    I think I got it!

    Pressure= F/A [M^x][L^-1x][T-^2x]
    Density= [M^y][L^-3y]

    For M M^y plus M^x= 0

    For T -1 = -2x
    x= 1/2

    For L 1= -1x - 3y sub in x=1/2
    x= 1/2

    As such,

    Velocity = sqt (pressure/density)

    I hope this it? yes?

    I did not ignore or 'take care' of M, I just applied the rules and realized x=-y


    Could you verify if this method of taking care of M would also work in cases inwhich x does not equal y?
     
  11. Sep 16, 2010 #10

    lewando

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    That's correct.

    As far as verifying the method, I understand what you did and it seems reasonable. In this case, x does not equal y. I don't see why it wouldn't work for other cases.
     
  12. Sep 16, 2010 #11
    You been a great help lewando.
     
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