Errors and uncertainties in experiments
Any experiment will involve a series of measurements and
each of these will be made to a certain degree of accuracy. For example the calculation of a
velocity needs both a time and distance measurement. Therefore errors and inaccuracies in each
measurement will give a final error in the result.
There are two basic types of error that may
appear in the result:
Systematic errors
These are errors which are due to
the apparatus and may result from either faulty apparatus, badly calibrated apparatus or a zero
error.
The only way to eliminate systematic errors is to re-calibrate the apparatus or change
it!
Random errors
These are errors that are due to experimenter - in
others words us! The size of these errors depends on how well the experimenter can use the
apparatus. The better experimenter you are the smaller will these errors be. The way to reduce
these errors, apart from simply being careful, is to repeat the readings and take an
average.
Precision and accuracy
It is important to understand the difference between the precision of a measurement and its accuracy. It is possible to measure something to a high degree of precision (for example a mass to a milligram using an electronic balance) but still to have an inaccurate answer because the instrument used had an offset zero error.
Quoting an answer
When you have made your set of
readings you must be careful when you quote your result. You will probably use a calculator
reading to eight digits to calculate and answer involving perhaps three of four quantities - do not
use the eight digits - the accuracy of the final answer cannot be better than the accuracy of any of
the single quantities.
A word of warning here:
small quantities may only be ignored in comparison with large ones. For example in an answer
such as 5.4000003 the 0.0000003 may be ignored in comparison to the 5.4 and the answer
quoted as 5.4 but this is not the case in an answer such as 0.0000013 where the 0.0000003 is
23% of the answer. An Analogy: If you are sat on by an elephant it is not important whether or not
the elephant has a fly on its back - you are still squashed flat. However to another fly the first fly is
important!
Calculating an error
In this section we will imagine that we
want to find the value of a quantity Q that involves the measurement of two other quantities a and
b
There are two possibilities:
(i) Q is the sum or difference of a and b or (ii) Q is the
product or quotient of a and b.
(i) Suppose that Q is the length of an object found by
taking the reading of the position of its two end a and b from a ruler Q
= a - b
Let the error in a be Δa and that in b be Δb. Then the error in Q is simply the error in a plus the error in b.
That is:
ΔQ = Δa + Δb
Then the value of Q should
be written as:
Q +/-ΔQ
Example problem
Let a = 26.3 cm +/- 0.1 cm and b = 15.8 +/- 0.1 cm then:
Q = 26.3 - 15.8 = 10.5 +/- 0.2 cm or 10.5 cm +/- 1.9%
(ii) Suppose that Q is the area of a rectangle with sides a and b
so that Q = ab (Figure 4).
Let the error in a be Δa an that in b be Δb.
Therefore the maximum value of Q is :
Q + ΔQ = (a + Δa)(b + Δb) = aΔb +bΔa +ab +ΔaΔb
But we can ignore the quantity ΔaΔb since it will be very small
compare with the other quantities in the equation and so:
ΔQ = aΔb + bΔa
and this is true whether or not Q is a product or a quotient.
If one of
the quantities, say b, is raise to a power n (i.e. Q = ab
n) the value of ΔQ becomes:
ΔQ = anΔb + bΔa
It is probably easier to understand these two
equations if they are quoted as a fractional error. In other words:
ΔQ/Q = Δa/a + Δb/b and ΔQ/Q = Δa/a + Δb/b ΔQ/Q = Δa/a + nΔb/b
or even easier as a percentage error:
%Q = %a + %b and %Q = %a + n%b
Notice that pure numbers have no errors; this
can be assumed for quantities such as
p and e
Example problems
1. Find the maximum possible error in the measurement of the force on an object of mass 4.0 = +/- 0.1 kg travelling at 5.2 +/- 0.2 ms-1 round a circle of radius 0.75 m +/- 0.01 m.
The equation is F = mv2/r.
Using fractional errors
ΔQ/Q = Δa/a + nΔb/b and so we have
ΔF/F = Δm/m + 2Δv/v + Δr/r = 0.1/4.0 + 2x0.2/5.2 + 0.01/0.75 = 0.025 + 0.077 + 0.013 = 0.115
But F = mv2/r = 4.0x5.22/0.75 = 144.2 N therefore ΔF = 144.2x0.115 = 17 N
The result should be quoted as F = 144 +/- 17 N
2. Find the maximum possible error in the density of a material in the form of a cube if the mass can be measured to +/- 5 % and the length of its sides to +/- 3.4 %.
Using percentage errors:
%Q = %a + n%b and so we have %r = %m + 3%L = 5% + 3x3.4% = 15.2 %
You can see how, although the percentage error in L is smaller than the percentage error in mass, it is more important in the answer because it appears as L3 in the formula.
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