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Homework Help: Easy question calculating uncertainty

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the mass uncertainty, [tex]\delta[/tex]m = m+ - m, of the particle?

    2. Relevant equations
    m = [tex]\frac{R*|q|*B}{v}[/tex]

    R = (6.4[tex]\pm[/tex]0.1) cm (convert to meters so 0.064[tex]\pm[/tex]0.001)
    |q| = 1.6*10-19
    B = (2.50[tex]\pm0.02[/tex]) T
    v = (5.25[tex]\pm0.04[/tex])x107

    3. The attempt at a solution

    I figured out that m = m = 4.88*10-28kg

    When I tried to find the upper limit I got 4.95x10-28
    and the lower limit, 4.80x10-28

    My professor's answer sheet says that m+ = 5.03 x 10-28 kg
    and [tex]\delta[/tex]m = 0.2x10-28 kg.

    First of all, I don't know how he determined that [tex]\delta[/tex]m = 0.2x10-28 kg. Can someone help me?

    Second of all, if m = (4.88*10-28kg) and [tex]\delta[/tex]m = 0.2x10-28 kg then shouldn't m+ = 5.08 x 10-28 kg?? NOT 5.03 x 10-28 kg??
  2. jcsd
  3. Feb 12, 2010 #2


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    Technically, the correct formula from error analysis is

    [tex]\delta f(x_1, \cdots, x_n) = \sqrt{ \sum_{i = 1}^n \left( \frac{\partial f(x_1, \cdots, x_n)}{\partial x_i} \delta x_i \right)^2 }[/tex]
    where [itex]\delta x_i[/itex] is the uncertainty in xi.

    So in this particular case, you would get
    (\delta m)^2 =
    \left( \frac{\partial}{\partial R} \frac{R|q|B}{v} \right)^2 (\delta R)^2 +
    \left( \frac{\partial}{\partial |q|} \frac{R|q|B}{v} \right)^2 (\delta |q|)^2 +
    \left( \frac{\partial}{\partial B} \frac{R|q|B}{v} \right)^2 (\delta B)^2 +
    \left( \frac{\partial}{\partial v} \frac{R|q|B}{v} \right)^2 (\delta v)^2.
  4. Feb 12, 2010 #3


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    Sorry, you can disregard my earlier post. I missed that you posted in precalculus and using a much simpler formula.

    Note that to find the upper bound, you have to plug in the extreme values, such that the result for m will be as large as possible.
    There is a variable v in the denominator. That means that to make m large, you should make v small. For example, 1/1000 is smaller than 1/10.

    In other words, you shouldn't use 5.29 x 107 for the extremal value of v, but 5.51 x 107.
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