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Finding Propagation of Uncertainty?

  1. Aug 28, 2012 #1
    You measure the mass of the cylinder to be m = 584.9 +- 0.5 grams, and you measure the length of the cylinder to be L = 18.195 +- 0.003 cm. Just like in the lab you performed, you now measure the diameter in eight different places and obtain the following results.
    Diameter (cm)
    2.125
    2.090
    2.065
    2.240
    2.110
    2.100
    2.080
    2.240

    This gives an average of 2.131 +- 0.0695

    This makes the density = 9.01 g/cm^3 +- propagation of uncertainty

    Trying to calculate this, I have: sqrt( ((1*0.5) / 584.9)^2 + ((-2 * 0,0695) / 2.131)^2 + ((-1 * 0.003) / 18.195)^2 ) = 0.0652

    However the online grading system say's that I'm wrong. So where have I gone wrong with the uncertainty of the density. All of the other values have been graded and marked correct, so where did I mess up with the uncertainty.

    I'm not really clear where the "1", "-2", or "-1" came from in the formula above, I was basing it on my notes from class.
     
  2. jcsd
  3. Aug 28, 2012 #2

    gneill

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    Staff: Mentor

    It looks like you forgot to multiply the "sqrt" by the calculated value of the function being evaluated (the density value).
    They are values that depend upon the exponent of the variable in the function. Write out the function being evaluated in one line (promote the variables in the denominator to the numerator and adjust exponents accordingly):

    $$f(m,d,L) = \frac{4}{\pi}M^1 d^{-2} L^{-1}$$

    You can pick out the values as the exponents of the variables. Thus, for example, the value "-2" is associated with the diameter variable d.
     
  4. Aug 29, 2012 #3

    mfb

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    2016 Award

    Staff: Mentor

    You calculated the relative uncertainty, and both the formula and the result look good. Similar to gneill, I think the online grading system wants the absolute uncertainty.
     
  5. Sep 6, 2012 #4
    I've been meaning to get back here to say Thank You! Multiplying by the density was exactly what I needed to do for WebAssign (the online grading system) to accept the answer.
     
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