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Homework Help: Calculating error in measurements w/ uncertainty

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data
    The radius of a circle is measured to be 2.4 cm +/- 0.1 cm.
    Find the error in the area of the circle.
    Find the error in the circumference.

    2. Relevant equations
    Have no idea but I'm taking a guess it could be multiplying fractional uncertainties?

    3. The attempt at a solution
    I got 18.09 for area and 15.08 for circumference...just need a way to go about calculating error

    ...really rusty in math and learning physics now, pardon the noob question.
     
  2. jcsd
  3. Jun 30, 2011 #2

    gneill

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    Staff: Mentor

    A discussion of error analysis can be found here:

    http://teacher.pas.rochester.edu/PHY_LABS/AppendixB/AppendixB.html

    Near the end of the document is a table showing how to deal with the uncertainty values when performing various mathematical operations.

    Keep in mind that constants (like 2 or π) are exact and have zero uncertainty.
     
  4. Jun 30, 2011 #3
    Thanks but I'm still a little confused...would I have to do the 0.1/2.4 twice and add them or what?
     
  5. Jun 30, 2011 #4

    gneill

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    Staff: Mentor

    For what calculation?

    You might want to ponder entry 5 in the table, which deals with numbers with uncertainties raised to a power.
     
  6. Jun 30, 2011 #5
    Well in that scenario delta Z would be the change in 2.4 + .1 and 2.4 -.1 right? What about delta A?
     
  7. Jun 30, 2011 #6

    gneill

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    Staff: Mentor

    Z is the result. ΔZ is the uncertainty in the result. A is the number with uncertainty ΔA. So in the case of calculating Z = A2, n = 2 and
    [tex] \frac{\Delta Z}{Z} = n \frac{\Delta A}{A} [/tex]
    giving
    [tex] \Delta Z = 2 Z \frac{\Delta A}{A} [/tex]
    and since Z = A2, this yields
    [tex] \Delta Z = 2 A^2 \frac{\Delta A}{A} [/tex]
    You could reach the same result using entry 3 in the table (for Z = A*B) by setting B = A and ΔB = ΔA.
     
  8. Jun 30, 2011 #7
    Ok thanks for clarifying.
     
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