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Absolute Uncertainty

  1. Mar 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Given values and uncertainties of d, find the absolute uncertainty in 1/d^2 (see table attached)

    2. Relevant equations
    I know that if it's a square function the absolute uncertainty will be 2 (delta d) / d but I'm not sure if it's correct for an inverse function.

    3. The attempt at a solution
    I read somewhere that it might be 2 (deIta d)/ d^3? If I use this equation, I get the range of answers they gave in the marking scheme which are 0.03 till 1. If I use only 2(delta d)/d I get uncertainties from 0.02 till 0.07.

    Thank you
     

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  2. jcsd
  3. Mar 15, 2017 #2

    BvU

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    Good place. The rule is simple if you know how to differentiate:$$\big(\Delta f(x) \big )^2 = (f'(x) \Delta x)^2$$

    Another way to get the same:
    percent uncertainty in 1/d = percent uncertainty in d
    percent uncertainty in dn = | n | x percent uncertainty in d​
     
  4. Mar 15, 2017 #3
    So if it was d^2 it would just be 2(uncertainty in d)/d
    And for 1/d^2= 2(uncertainty in d)/d^3
    From differentiation
     
  5. Mar 16, 2017 #4

    haruspex

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    No. Δf(x)≈f'(x)Δx.
    If f(x)=x2, what is f'(x)?
     
  6. Mar 16, 2017 #5
    Is there a difference between absolute and percentage uncertainty? Because all the A'level Physics questions I've done so far have mentioned that if there's a power you just multiply the power by (change in uncertainty/actual value) so if it's d^3 for example, it will be 3 (delta d)/d. Is this just for percentage uncertainty?
     
  7. Mar 16, 2017 #6

    haruspex

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    Yes.
    If f(x)=xn then f'(x)=nxn-1, Δf(x)=f'(x)Δx=nxn-1Δx; Δf(x)/f(x)=nΔx/x.
    But it would be better to call that fractional uncertainty. To make it percentage you would multiply that by 100%.
     
  8. Mar 16, 2017 #7
    The absolute uncertainty of a quantity must have the same dimensions as the quantity. The relative uncertainty is dimensionless. Which of your expressions seems more appropriate for the absolute uncertainty?
     
  9. Mar 20, 2017 #8
    2(delta d)/d^3 has units of 1/m^2 and 2(delta d)/d has no units so the former is the absolute uncertainty and the latter is the fractional/relative/percentage uncertainty
     
  10. Mar 20, 2017 #9
    So to get the absolute uncertainty you simply do
    Delta f(x)=f'(x) (delta x)
    But for fractional/relative/percentage uncertainty you divide the above by f(x)
    Correct?
     
  11. Mar 20, 2017 #10

    haruspex

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    Yes.
     
  12. Apr 24, 2017 #11
    The fractional/percentage uncertainy for 1/d^2 is just 2(Δd)/d right?

    Thanks
     
  13. Apr 24, 2017 #12

    haruspex

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    Yes.
     
  14. Apr 24, 2017 #13
    Thank you and sorry just going back to the basics but table doesn't make sense, especially the second and third row. How are they even getting these uncertainties? This is much more basic than what I'm doing so that's why I'm confused.
     

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  15. Apr 24, 2017 #14

    haruspex

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    I'd need to see the text leading to this, but it looks to me as though the table is taking into account human visual accuity. E.g. in the first row, with 1mm markings, it is reasonable to judge a measurement to the nearest mm, but no better; in the third row, even with only 10cm markings, it should be possible to judge to within 10% of the distance between markings.
     
  16. Apr 25, 2017 #15
    There's actually nothing in the text that relates to the table. They just say here's an example of some uncertainties unfortunately!
     
  17. Apr 25, 2017 #16

    haruspex

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    In that case I would read it as possible entries in a lab report. The experimenter is recording the value measured, the gradations of the measuring instrument used, and her estimate of the uncertainty. There is no suggestion that the uncertainty is calculated by some algorithm from the other two.
    As I described in my preceding post, the entries do make sense.
     
  18. May 13, 2017 #17
    Ok thank you.

    In the picture I attached, they asked us to measure the length of the wire using a ruler and then asked for the percentage uncertainty. In the mark scheme it says
    "Percentage uncertainty in L based on absolute uncertainty of 2mm to 8mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty."

    How did they get an absolute uncertainty of 2mm to 8mm?

    Thank you
     

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  19. May 13, 2017 #18

    BvU

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    Am I right in assuming this has nothing to do with the thread thus far ?

    Anyway:
    Where exactly does the clamp hold the wire ? And where exactly at the other end is the mass effectively acting ? A millimeter uncertainty at each end for those is a fair guess; add another for measuring accuracy at each end and you are in the range of 2 to 4 mm. 8 mm seems a it pessimistic to me.
     
  20. May 13, 2017 #19

    haruspex

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    Agreed, but maybe they are allowing for a range of students' physical abilities.... Poor vision, coordination... What I don't like is that apparently they would reject an absolute error less than 2mm. At each end, a skilled operator could claim an absolute error of 0.5mm, then combine them to get a statistical total of 0.75mm.
     
  21. May 13, 2017 #20
    Thank you for your replies.

    So it's all a guess at the end? There's no definite way to find out the absolute uncertainty like we could when given an equation?
     
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