Download Project Learning in Secondary School Physics

Measurement
Measurement
Level: Secondary 4 / 6
Project aims:
The main objectives of this project are to understand the importance of measurement in our daily life, to
explore different methods for measurement, to understand that measurements can never be exact and to
compare the accuracy of different measuring methods.
In the project tasks, you are asked to design your own methods for measuring particular object(s),
building(s) and distance(s). You have to explain how you make the measurements, identify the possible
sources of error of each method, and compare and estimate the accuracies of your measurements.
Learning objectives:
1. Understand the importance of measurement in science.
2. Gain familiarity with a variety of measuring instruments and methods.
3. Understand that measurements are never exact through error analysis.
4. Recognize the limitation of measuring instruments used.
Task 1: Measuring length
Measure the length of an object (shorter than one foot long recommended), say a pencil, using a one feet
ruler without finer scale.
You may use other tools in your measurement when necessary. However, the tools should not contain any
scale or mark, which can directly be used to measure the length of the object.
Veriner caliper may be provided in measuring the length of a small object.
Task 2: Measuring height
Decide two methods to measure the height a given point of our school.
At least one method should make use of a stop-watch.
Metre rule can only be used in measuring the length of a string.
Task 3: Measuring distance
Measure the distance between any two particular places, without the use of any electronic equipment or
professional devices e.g. metre ruler only can be used in indirect measurement e.g. measuring the length
of the tile.
Apparatus/Materials:
Any appropriate apparatus and materials proposed are welcome. Typical examples include meter ruler
(scale covered with paper), identical coins (many), protractor, beaker, balance, measuring tape, tennis ball
(as a falling object), stop watch, string / rope with a weight, meter wheel and so on.
Pre-knowledge:
Measurement is of central importance in our daily life. It is one of the concrete ways we deal with our
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Measurement
world. Measurement is particularly important in science. Science is concerned with the description and
understanding of nature and measurement is one of its most fundamental tools. One may say that science
cannot exist without measurement and we start learning physics by learning how to measure a few
physical quantities. Physics attempts to describe nature in an objective way through measurement.
However, in every measurement, there is uncertainty associated with it – there is no such thing as a
perfectly accurate measurement. There are uncertainties arise from the measuring instrument, the people
doing the measurement, the method of making the measurement and the characteristics of the systems
being measured. We need to learn to due with these uncertainties in measurement through error analysis.
Questions:
1. Which method of measurement is more accurate? Can you account for this?
2. How could you deal with the difficulty in measurement involving slope and gradient, like staircase?
3. How can you measure the distance between two very far points, for example, the distance from earth
to moon?
Appendix:
Propagation of error in different forms
In many cases, a particular result will be obtained by combining a number of measurements of different
quantities in some ways.
For instance, when two independent measurements (A  A) and (B  B) are combined to give a result
X, the associated error terms X will be given by:
Combinations of A and B
X  A B
X  A B
X  AB
A
X 
B
Error of X
X  A  B
X
A B


X
A
B
X  kA ( k is a constant)
X
A

or X  k A
X
A
X  An
X
A
n
X
A
X  AB n
X
A
B

n
X
A
B
The combinations involve addition and multiplication with more variables or more terms can be resolved
into combinations listed above.
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Measurement
Example 1: Combinations of error involving addition and multiplication
A student finds the constant speed of a moving trolley with a stopwatch. The equation used is v 
s
. The
t
distance s is measured with a meter ruler, the time t is measured with a stopwatch. Their values are s =
(4.01 ± 0.01) m; t = (3.2 ± 0.2) s.
The speed v
= 4.01 / 3.2 ms-1
= 1.253 ms-1 (calculator display)
= 1.3 ms-1
This result is corrected to 2 significant figures since the time measured is only known to 2 significant
figures.
The estimated error in speed, v
 s t 
= v


t 
 s
 0.01 0.2 
= 1.253

 ms-1
 4.01 3.2 
= 0.081437 ms-1 (calculator display)
= 0.1 ms-1
Since v has 1 decimal place, the error term ∆v should also has 1 decimal place, i.e. v = 0.1 ms-1.
Therefore the speed of the trolley is 1.3 ± 0.1 ms-1.
Handling the trigonometric functions of an angle
In the tasks students may measure an angle and apply trigonometric functions of the angle for their
analysis. Here shows two examples to deal with the tangent of an angle.
Example 2: Handling the trigonometric functions with multiple measurement
Students used a protractor to measure the angle of elevation to a building and apply the formula
h  d tan  to calculate the height of the building. If d  (8.23  0.005) m and θ = 27º, 29º, 28º, 31º
Then the average value of height is
h
 27  29  28  31 
 8.23 tan 

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

 4.515m
Since the tangent of an angle is an increasing function, the extreme values of the height are
hmax  8.235 tan 31 = 4.948m
hmin  8.225 tan 27 = 4.190m
hmax  h = 4.948 – 4.515 = 0.433m
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Measurement
h  hmin = 4.515 – 4.190 = 0.325m
So, maximum error = 0.433m
After rounding off to have the same decimal place with d, h  (4.52  0.43) m
Example 3: Handling the sine of the angle with multiple measurement
If a student has only one measurement for the angle, and have result d  (8.23  0.005) m,
  (29  0.5)
Then the mean and extreme values are:
h
 8.23 tan 29 = 4.561m
hmax  8.235 tan 29.5 = 4.659m
hmin  8.225 tan 28.5 = 4.466m
hmax  h = 4.659 – 4.561 = 0.098m
h  hmin = 4.561 – 4.466 = 0.095m
So, maximum error = 0.098m
After rounding off to have the same significant figure with D, h  (4.56  0.10) m
Example 4: How error analysis can help you to analysis the coins or cube method.
For measuring a Length of about 19.8cm
let N be the number of coins and D be diameter of the coins.
L=NxD
L N D


L
N
D
For Very Small Coins :
D = 5 mm measuring the coins by metre ruler. D = 1mm
N = 1
N = 40
L = N x D = 40 x 0.5cm = 20cm
L N D
L
1 1mm
L




 (0.025  0.2) =>L = 4.5cm
=>
=>
L
N
D
20cm 40 5mm
20cm
L = 205cm
For Small Coins :
D = 1cm
N = 1
measuring the coins by metre ruler. D = 1mm
N = 20
L = N x D = 20 x 1cm = 20cm
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Measurement
L N D
L
1 1mm
L




 (0.05  0.1) =>L = 3cm
=>
=>
L
N
D
20cm 20 1cm
20cm
L = 203cm
For Large Coins :
D = 4cm measuring the coins by metre ruler. D = 1mm
N = 1
N=5
L = N x D = 5 x 4cm = 20cm
L N D
L 1 1mm
L


 
 (0.2  0.02) =>L = 4.4cm
=>
=>
L
N
D
L 5 5cm
20cm
L = 204cm
Questions
Q1. How to reduce the uncertainty of the diameter ?
Q2.How to reduce the uncertainty of the counting. ?
Example 5 For counting of two types of tiles.
Let
L = n1 T1 + n2T2 =>
let n be the number of coins and T be length of the tile .
Let X1 = n1 T1
and X2 = n2T2
L = X1 + X2
then L = X1 + X2
Since
And
X 1 n1 T1


=>
X1
n1
T1
X 1  (
n1 T1

) X1
n1
T1
X 2 n2 T2
n T


=> X 2  ( 2  2 ) X 2
n2
T2
X2
n2
T2
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Measurement
Further Example
(This method is provided for students with good mathematical background.)
For any function f(x), we can find the error by
d

 f x    f x  x
 dx

f x  x 
 f x 
f x 
 f x 
f x  x
x x
x  x x x  x
we have
 f x  1 f x  x   f x  x 

x
2
x
for an infinitesimally small error, that is, x  0 ,
lim
 f  x  df  x 

x
dx
d

 f x    f x  x
 dx

or, for a small value of error,
Example 6 : Handling error by calculus
A student measures the angle of elevation of a building. He finds the angle is   34  0.5 . Find the
value of sin  and its error.
34  34 

180
0.5  0.5 
rad  0.593

180
rad  0.00872
sin   sin 0.593  0.559
 d

sin    
sin     cos 0.593  0.00872  0.00723
 d

sin   sin    0.559  0.007
* Notice that we should use radian scale because we are using calculus.
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