Uncertainity, Error) Calculations In Scientific Measurements

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E:\Excel files\ch 00 P111 uncertainty.doc Page 1 of 3 Dr. Fritz Wilhelm Physics 111 2/7/2004 Last printed 2/7/2004 5:37:00 PM

Uncertainty (Error) Calculations In Scientific Measurements:
Significant Figures: To be unambiguous in our notations of physical measurements we should write all numbers in terms of powers of ten. We should always do this in final results. We also must always indicate the dimensions involved, for example we write: 1.7 mm = 1.7x10-3 m. In both cases we use 2 significant figures. The number 0.0017 m also has just two significant figures, whereas the number 0.00170 m has three significant figures. Whenever a number is given without indication of the relative or absolute error we assume that the error lies in the next significant figure and is equal to 5 units of that figure. 1.7 mm means therefore 1.7mm +/- 0.05mm. When we multiply or divide numbers the term with the smallest number of significant figures prevails. We wait until the final result to round our numbers up or down. x=1.334 y=3.45 z=2.6 V=π·x·y·z =3.759223x101 cm3 which we must round to 2 significant figures or 3.8x101 cm3 Let us say that we have measured the length of a table to be 3.654cm +/-0.05cm. The number 0.005cm is called the absolute error or absolute uncertainty in the measurement of x. We write also ∆x=0.05cm. (1.1)
∆x = 0.05cm absolute error in x.

∆x is called the absolute uncertainty (error) in the measurement x, in contrast to the . The relative uncertainty is a ratio of two numbers with the x same dimension, therefore, the dimensions cancel out and the result is a fraction smaller than 1. We therefore often express the ratio as a percent ratio: 0.05 = 5% relative uncertainty
∆x

(1.2) x The value for x can also correspond to a function in several variables. Such a function could be f(x,y), for example: f(x,y) = 4x · 5y2, where x = 5.34 cm +/- 0.004 cm and

∆x

relative uncertainty in x.

E:\Excel files\ch 00 P111 uncertainty.doc Page 2 of 3 Dr. Fritz Wilhelm Physics 111 2/7/2004 Last printed 2/7/2004 5:37:00 PM y has been measured to be 33.68 cm +/- 0.008 cm. f = 1.21(147768)· 105 cm3, where I have put the last meaningless six digits in parentheses. To determine the error in this final result we would use 3 significant figures (sig. figs.), if we wouldn’t know anything else about the uncertainties. But we have in this case determined the absolute errors: This means that ∆x = 0.004 cm and ∆y = 0.008cm. These are the absolute errors or uncertainties in x and y. Note that the absolute error in y is twice the absolute error in x, but that the relative error in y, ∆y/y = 0.008/33.68 is much smaller than the relative error in x, ∆x/x = 0.004/5.34. If f is an arbitrary function in terms of powers of x, y, and z we can write: (1.3) f ( x, y, z ) = kx a y b z c ∆f dx dy dz =a +b +c f x y z

For the example above that would give us: ∆f ∆x ∆y 0.004 0.008 (1.4) = +2 = +2 = 1.22 ⋅10−3 f x y 5.34 33.68 As the absolute error in the denominator is given in terms of 1 significant figure, the final relative uncertainty should also have only one significant figure or 0.1%. Note that, even though the uncertainty itself is calculated with 1 significant figure, the result f itself must contain at least three significant figures. To calculate the absolute error in f we multiply 0.00122 by f =