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TOPIC 1: Units, Dimensions, Errors and Uncertainties
What is the difference between Errors and Uncertainties? Errors are uni-directional whereas Uncertainties are bi-directional. Therefore, Errors can be added or subtracted, but Uncertainties can only be added. (Note: this statement may sound paradoxical but the following example shall illustrate).
For example, let's use the formula for the period of pendulum:
T = 2p Ö(l/g)
The length of the pendulum, l, which is supposed to be 0.80 m, is measured as 0.86 m (with an error of +0.06m). The acceleration due to gravity, g, which is supposed to be 9.81 m s-2, is measured as 9.84 m s-2 (with an error of +0.03 m s-2)
To find the error that results in the measurement of the period T, we have to first find the fractional error of l and g.
Dl / l = 0.06 / 0.80 = 0.075
Dg / g = 0.03 / 9.81 = 0.003
Since l is in the numerator, its positive error will result in an increase in the period T. Since g is in the denominator, its positive error will result in a decrease in the period T. Therefore, the fractional error of l/g will be 0.075 - 0.003 = 0.072
Since l/g is square-rooted (ie. to the power of 1/2), the fractional error of of Dl/g will be 1/2 x 0.072 = 0.036.
Next, evaluate the period T by substituting l = 0.80m and g = 9.81 m s-2 , we find that T = 1.794s
The error of T will be its value (1.794 s) multiplied by the fractional error (0.036), the answer is 0.065, or 0.07 after rounding off to 1 significant figure. The erroneous value is therefore 1.79 + 0.07 = 1.86 s.
Let's check the answer by substituting l = 0.86m and g = 9.84 m s-2 into the equation, the answer is 1.86 s.
Using the same example, instead of a positive error of +0.06 m for the length l, we have an uncertainty of +0.06 m; and instead of a positive error of +0.03 m s-2 for the acceleration due to gravity g, we have an uncertainty of +0.03 m s-2. Then the total fractional uncertainty is 0.075 + 0.003 = 0.078.
The uncertainty of T will be its value (1.794) multiplied by the fractional uncertainty (0.078), the answer is 0.14.
Therefore, period T = 1.794 + 0.14, but since uncertainties have to be rounded upward to only have 1 significant figure, the proper presentation is T = (1.8 + 0.2) s.
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