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# Uncertainties (1)

```Year 12 Physics
Uncertainties and Errors
Uncertainty
You can't measure a physical quantity exactly because all instruments have limitations. These
limitations make any reading uncertain. However, some digital instruments appear to give more
exact measurements than the manufacturers ever intended. For example, an ammeter with a display
of 258 mA seems to be indicating that the current is exactly 258 mA, whereas it may really mean
258 ± 1 mA. A general rule-of-thumb is that the uncertainty in a reading is said to be equal to a half
scale division on the instrument. For a ruler marked in millimetres the absolute uncertainty would
therefore be ± 0.5 mm. An object measured to be 24.5 mm in length could be stated as 24.5 ± 0.5
mm. This absolute uncertainty could be also expressed as a percentage uncertainty:
absolute uncertaint y
 100%
observed measuremen t
0.5

 100%  2%
24.5
Percentage uncertaint y 
The uncertainty is a way of expressing how confident you are about the readings provided by the
instrument. It is a measure of the limitations of the instrument.
Uncertainty calculations
When you add or subtract numbers, the absolute and relative uncertainties are required.
 To add or subtract two numbers, add their absolute uncertainties.
When you multiply or divide numbers, the relative uncertainties are required.
 To multiply and divide two numbers, add their relative or percentage uncertainties.
Example 1
A container of water rises in temperature from 25.5 ± 0.5°C to 36.0 ± 0.5°C. Calculate the rise in
temperature and its percentage uncertainty.
Solution
Rise in temperature = 36.0 ± 0.5°C – 25.5 ± 0.5°C = 10.5 ± 1.0°C
1.0
 100%  9.5%
Percentage uncertainty =
10.5
Example 2
A piece of paper is measured and found to be 5.63 ± 0.05 mm wide and 64.2 ± 0.5 mm long. What
is the area of the piece of paper?
Solution
= length × width
= (5.63 ± 0.05 mm) × (64.2 ± 0.5 mm)
= (5.63 + 0.89%) × (64.2 + 0.78%) (convert to percentage uncertainty)
= 361.446 ± 1.67% (add percentage uncertainties)
= 361.446 ± 6.025 mm2 (convert percentage uncertainty to absolute uncertainty)
= 361 ± 6 mm2
(round answer to three significant figures and round the uncertainty to one significant figure as
given in the original data).
Area
Year 12 Physics
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Uncertainties and Errors
Questions
1. A cube of brass was measured and found to have a side of length 13.0 ± 0.5 mm. Determine the
volume of the cube.
2. A student made two measurements using a metre ruler calibrated in millimetres. First
measurement = 25.5 mm. Second measurement = 174.5 mm.
a. What are the absolute uncertainties for these measurements?
b. Convert these absolute uncertainties to relative uncertainties.
c. Add the two measurements and show the absolute uncertainty of the result.
d. Multiply the two measurements and show the absolute uncertainty of the result.
3. Determine the correct value for the area of a horse paddock 645 ± 5 m long and 345 ± 5 m
wide. What is the total length of fencing needed to fence this paddock?
Accuracy and precision
Students often find that despite performing an experiment as accurately as possible and reading the
instruments as best as they are able, their results are often different from the accepted or textbook
result. This difference is called the error. The error is a measure of the accuracy of a result.
Accuracy refers to the closeness of a measurement to the accepted value.
Imagine your group measured the density of water to be 1.02 g.rnl-1 when the accepted value was
given as 1.00 g/ml at that temperature. Your (absolute) error would be 0.02 g.ml-1. Hence:
 The absolute error (Ea) = |observed value - accepted value| = |O – A|.
Note: the straight lines (|) in the above equation mean the 'absolute value', that is, the sign (+/-)
of the answer is ignored.
 The relative error (Er) is the absolute error expressed as a percentage of the accepted value
(A):
Relative error ( Er ) 
Ea
 100%
A
0.02
 100%  2%
1.00
This is useful for comparing the accuracy of different experiments. Imagine that another student
measured the density of lead as 11.29 g.cm-3, while the accepted value was 11.41 g.cm-3. Which
result is the more accurate – the density of water or the density of lead? In this case you need to
compare relative errors: the error for water was 2% whereas that for lead was 1% and hence was
more accurately measured.
In the above example Er would equal
Example
Calculate (a) the absolute error and (b) the relative error in a student’s measurement of the
acceleration due to gravity. She obtained 9.73 m.s-2 whereas the accepted value at that location was
9.813 m.s-2.
Solution
(a) Ea = |O – A| = | 9.73 - 9.813 | = 0.08 m.s-2 (to the correct number of significant figures).
(b) Er 
Ea
0.08
 100% 
 100%  0.8%
A
9.813
Uncertainties in Graphs
When data is plotted on a graph, error bars can be drawn to indicate the level of uncertainty in the
data. Consider the following example:
A pendulum is constructed by attaching a pendulum bob to the end of a string. The string is set to
various lengths and the period of the pendulum measured by timing 10 oscillations with a
Year 12 Physics
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Uncertainties and Errors
stopwatch. It is believed that the team of experimenters could be as much as 0.4 s out in any one
timing. The data are presented in the table below.
Length (m)
0.20 ± 0.001
0.40 ± 0.001
0.60 ± 0.001
0.80 ± 0.001
1.00 ± 0.001
1.50 ± 0.001
2.00 ± 0.001
Time for 10
oscillations
(s)
9.8 ± 0.4
13.6 ± 0.4
16.2 ± 0.4
18.6 ± 0.4
20.5 ± 0.4
24.9 ± 0.4
28.7 ± 0.4
Period – T (s)
With absolute
uncertainties
0.98 ± 0.04
1.36 ± 0.04
1.62 ± 0.04
1.86 ± 0.04
2.05 ± 0.04
2.49 ± 0.04
2.87 ± 0.04
Period – T (s)
With relative
uncertainties
0.98 ± 3.9%
1.36 ± 3.0%
1.62 ± 2.5%
1.86 ± 2.2%
2.05 ± 1.9%
2.49 ± 1.6%
2.87 ± 1.4%
Period2 – T2 (s2)
With relative
uncertainties
0.96 ± 7.8%
1.85 ± 6.0%
2.62 ± 5.0%
3.46 ± 4.4%
4.20 ± 3.8%
6.20 ± 3.2%
8.24 ± 2.8%
Period2 – T2 (s2)
With absolute
uncertainties
0.96 ± 0.08
1.85 ± 0.11
2.62 ± 0.13
3.46 ± 0.15
4.20 ± 0.16
6.20 ± 0.20
8.24 ± 0.23
These data can be plotted, the uncertainty of each can be represented with an error bar, and a line
drawn through the error bars.
Period squared vs Pendulum length
9
8
Period Squared (s^2)
7
6
5
4
3
2
1
0
0
0.5
1
1.5
2
2.5
Pendulum length (m)
The two lines on the graph above both pass through the error bars, but have quite different
equations. One passes through the points (0, 0.1) and (2, 8.4) and so has an equation
T2 = 4.2 l + 0.1
The other passes through the points (0, 0.25) and (2, 8.0) and so has the equation
T2 = 3.9 l + 0.25
Year 12 Physics
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Uncertainties and Errors
Therefore, it can be said that the gradient of the line is 4.05 ± 0.15 and the intercept 0.175 ± 0.075.
These figures would be rounded to gradient = 4.1 ± 0.2; intercept 0.18 ± 0.08.
Another method
An alternative approach is to use Excel or your graphics calculator to find the correlation
coefficient (r) and the gradient (m) and intercept (c) of the line of regression. There are a number of
ways to estimate the uncertainty in m and c. The way recommended by the University of
Queensland Physics department is as follows:
Uncertainty in m = ∆m =
m tan(cos 1 r )
Uncertainty in c = ∆c = ∆m
s
n2
2
x

 x
2
Where
m  The gradient of the regression line
c  The intercept of the regression line
r  The correlatio n coefficien t
s x  The standard deviation of the x values
x  The mean of the x values
n  The number of points plotted
This process can be illustrated with this segment
Period squared (s^2)
0.2
0.4
0.6
0.8
1
1.5
2
0.96
1.85
2.62
3.46
4.2
6.2
8.24
Mean of length =
Std Devn of Length =
Correlation coeff =
Intercept =
3.932857143
0.634335047
0.999911529
4.01295858
0.206538462
Using the formulae from above:
m 

m tan(cos 1 r )
n2
4.01296  tan(cos 1 0.999912)
72
 0.024


c  m s x2  x
2
 0.024  (0.6343 2  3.9329 2 )
 0.096
Year 12 Physics
10
9
Period squared (s^2)
Length (m)
Period squared vs Pendulum length
8
7
6
5
4
3
2
1
0
0
0.5
1
1.5
2
Length (m )
With appropriate rounding, this approach gives
m = 4.01 ± 0.02
and the intercept A
c = 0.2 ± 0.1
The two methods do not give identical results –
the second method gives far less uncertainty in
the gradient because it is highly unlikely that all
uncertainties will be in the same direction – but
both give some idea of the reliability of the
results.
Page 4
Uncertainties and Errors
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