Chapter 1 Measurement
1/6
Measurement
Objectives:
Identify the base quantities in the SI system. Name the most frequently used prefixes for SI units.
Change units (here for length, area, and volume) by using chain-link conversions.
Explain that the meter is defined in terms of the speed of light in vacuum.
Change units for time by using chain-link conversions. Use various measures of time, such as for
motion or as determined on different clocks.
Change units for mass by using chain-link conversions.
Relate density to mass and volume when the mass is uniformly distributed.
What is Physics ? phys·ics /ˈfiziks/
The branch of science concerned with the nature and properties of matter and energy. The subject matter of physics,
distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity,
magnetism, and the structure of atoms.
Physics is the natural science that studies matter and its motion and behavior through space and time and that studies the
related entities of energy and force. Physics is one of the most fundamental scientific disciplines, and its main goal is to
understand how the universe behaves.
Physics is the science of nature, or that which pertains to natural objects, which deals with the laws and properties of matter
and the forces which act upon them.
Physics is “the study of matter, energy, and the interaction between them”
Physics is a natural science based on experiments, measurements and mathematical analysis with the purpose of finding
quantitative physical laws for everything from the nanoworld of the microcosmos to the planets, solar systems and galaxies that
occupy the macrocosmos.
Measurement in Physics Physics is based on measurement of physical quantities. Certain physical quantities have been
chosen as base quantities (such as length, time, and mass); each has been defined in terms of a standard and given a unit of
measure (such as meter, second, and kilogram). Other physical quantities are defined in terms of the base quantities and their
standards and units.
SI Units The unit system emphasized in this book is the International System of Units (SI). The three physical quantities
displayed in Table 1-1 are used in the early chapters. Standards, which must be both accessible and invariable, have been
established for these base quantities by international agreement. These standards are used in all physical measurement, for both
the base quantities and the quantities derived from them. Scientific notation and the prefixes of Table 1-2 are used to simplify
measurement notation.
Changing Units Conversion of units may be performed by using chain-link conversions in which the original data are
multiplied successively by conversion factors written as unity and the units are manipulated like algebraic quantities until only
the desired units remain.
Physical Quantities We measure each physical quantity in its own units, by comparison with a standard. The unit is a unique
name we assign to measures of that quantity—for example, meter (m) for the quantity length.
The standard corresponds to exactly 1.0 unit of the quantity.
Base quantities and standards must be both accessible and invariable. The demand for precision in science and engineering
pushes us to aim first for invariability. We then exert great effort to make duplicates of the base standards that are accessible to
those who need them.
In 1971, the 14th General Conference on Weights and Measures
picked seven quantities as base quantities, thereby forming the
basis of the International System of Units, abbreviated SI from its
French name and popularly known as
the metric system.
Chapter 1 Measurement
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Length (meter) Length The meter is defined as the distance traveled by light
during a precisely specified time interval.
1792 the meter was defined to be one ten-millionth of the distance from the north
pole to the equator.
1889 the distance between engraved fine lines in a platinum–iridium bar became
the standard meter bar kept at the International Bureau of Weights and Measures in
Paris.
1960 the length equal to 1 650 763.73 wavelengths in vacuum of the radiation
corresponding to the transition between the levels 2p10 and 5d5 of the krypton 86
atom.
1983 this was updated to the current definition: "the length of the path travelled by
light in a vacuum in 1 / 299,792,458 of a second”.
Time (second) Time The second is defined in terms of the oscillations of light
emitted by an atomic (cesium-133) source. Accurate time signals
are sent worldwide by radio signals keyed to atomic clocks in standardizing
laboratories.
The second is the SI base unit of time, commonly understood and historically
defined as 1⁄86400 of a day.
1956 the second was redefined as "the fraction 1⁄31,556,925.9747 of the tropical
year for 1900 January 0 at 12 hours ephemeris time“.
1967 The second is defined to be exactly "the duration of 9,192,631,770 periods of
the radiation corresponding to the transition between the two hyperfine levels of the
ground state of the caesium-133 atom.
An atomic clock is a clock device that uses an electron transition frequency in the
microwave, optical, or ultraviolet region of the electromagnetic spectrum of atoms
as a frequency standard for its timekeeping element. Atomic clocks are the most accurate time and frequency standards known,
and are used as primary standards for international time distribution services, to control the wave frequency of television
broadcasts, and in global navigation satellite systems such as GPS.
Mass (kilogram) The kilogram is defined in terms of a platinum–iridium standard mass
kept near Paris. For measurements on an atomic scale, the atomic mass unit, defined in
terms of the atom carbon-12, is usually used.
1795 The gram, 1/1000 of a kilogram, was provisionally defined as the mass of one cubic
centimeter of water at the melting point of ice.
1875 The kilogram had a mass equal to the mass of 1 dm3 of water under atmospheric
pressure and at the temperature of its maximum density, which is approximately 4 °C.
The kilogram is the base unit of mass in the International System of Units (SI) defined by a
platinum alloy cylinder, the International Prototype Kilogram (informally Le Grand K or
IPK), manufactured in 1889, and carefully stored in Saint-Cloud, a suburb of Paris.
October 2010, the CIPM voted to submit a resolution for consideration at the General
Conference on Weights and Measures (CGPM), to "take note of an intention" that the
kilogram be defined in terms of the Planck constant, h (which has dimensions of energy
times time)
November 2018 The International Committee for Weights and Measures (CIPM) approved a proposed redefinition of SI base
units that defines the kilogram by defining the Planck constant to be exactly 6.62607015×10−34 kg⋅m2⋅s−1. This approach
effectively defines the kilogram in terms of the second and the meter.
Chapter 1 Measurement
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Dimensional Analysis
Dimensional or unit analysis is particularly useful in setting up the relationship of an unknown quantity against the base
quantities. Use the “[ ]” notation for units, and let the base units; M stand for mass, L for length, and T for time.
For example, we can write a speed as [v] = L/T, a force as [F]=ML/T2 and the gravitational constant as [G] = L3/(MT2).
A mass m hangs from a massless string of length and swings back and forth in the plane of the paper. The acceleration due to
gravity is g. What can we say about the period of oscillations, T?
We can make a guess based on observation or experience that the period [T] is proportional (related) to the following
quantities; the length of pendulum L, the mass of the pendulum M, and gravity g.
period  T   La M b g c
L 
2 
T
 
T   La M b 
where a,b,c are to be determined
c
Since the unit of period is [T] then the exponent b of M must be 0.
b  0  M 0 1
L 
2 
T 
T   La 
c
The length exponents a and c must also cancel.
La Lc  La c  1  a  c  0  a  c
1 
2 
T 
T   Lac 
c

1 
2 
T 
T   
c
The exponent c is finally determined to be;
c
 1 
 2c
 2   T  T   2c  1  c  1 / 2
T 
a  c  a  1 / 2
b0
Putting the exponents together, we get;
period  T   L1/ 2 M 0 g 1/ 2
T  
L
g
period  constant
Other units of time:
Jiffy (light to travel one centimeter in a
vacuum (approximately 33.3564
picoseconds),
Flick (1.42 × 10−9 s),
Shake (10ns),
Moment (40s)
A flick is a unit of time equivalent to exactly
1/705,600,000 of a second. The figure was
chosen so that frequencies of 24, 25, 30, 48,
50, 60, 90, 100 and 120 Hz, as well as
1/1000 divisions of all those, can be
represented with integers. The unit was
launched in January 2018 by Facebook.
Why Did Facebook Invent A New Unit Of
Time? The "Flick" Explained With Math.
YouTube video
A flick is approximately 1.42 × 10−9 s,
which makes it larger than a nanosecond but
much smaller than a microsecond.
L
g
We just shown thru dimensional analysis (and good guessing) that the period of a pendulum does not depend on the mass and
dependent only on the length L and gravity. By performing experiments on various lengths, the proportionality constant can
readily be obtained.
Uncertainty
The uncertainty estimate associated with a measurement should account for both the accuracy and precision of the
measurement.
Accuracy is the closeness of agreement between a measured value and a true or accepted value. Measurement error is the
amount of inaccuracy.
Precision is a measure of how well a result can be determined (without reference to a theoretical or true value). It is the degree
of consistency and agreement among independent measurements of the same quantity; also the reliability or reproducibility of
the result.
measurement = best estimate ± uncertainty
Chapter 1 Measurement
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Experimental uncertainties should be rounded to one significant figure.
If the uncertainty starts with a one, some scientists quote the uncertainty to two significant digits (example: ±0.0012 kg).
Wrong: 52.3 cm ± 4.1 cm
Correct: 52 cm ± 4 cm
Always round the experimental measurement or result to the same decimal place as the uncertainty.
Wrong: 1.237 s ± 0.1 s
Correct: 1.2 s ± 0.1 s
Uncertainty estimates are crucial for comparing experimental numbers.
Are the measurements 0.86 s and 0.98 s the same or different? The answer depends on how exact these two numbers are.
If the uncertainty too large, it is impossible to say whether the difference between the two numbers is real or just due to sloppy
measurements. That's why estimating uncertainty is so important!
If the ranges don't overlap, the measurements are discrepant:
0.86 s ± 0.02 s
and
0.98 s ± 0.02 s
If the ranges overlap, the measurements are consistent:
0.86 s ± 0.08 s
and
0.98 s ± 0.08 s
Uncertainty Propagation: Upper Lower bounds
The basic idea of this method is to use the uncertainty ranges of each variable to calculate the maximum and minimum values
of the function. Think of this procedure as examining the best and worst case scenarios.
For example, the Area of rectangle with length of 1.2±0.2 m and the width is 1.3±0.3 meters.
Area  l  w  1.2 1.3  1.56 m 2
The minimum area would be using the "minimum" measurements;
l (min)  1.2  0.2  1.0m and
w(min)  1.3  0.3  1.0m
Area(min)  l (min)  w(min)  1.0m  1.0m  1.0m 2
Likewise for the maximum area,
l (max)  1.2  0.2  1.4m and
w(max)  1.3  0.3  1.6m
Area(max)  l (max)  w(max)  1.4m  1.6m  2.24m 2
The estimated Uncertainty is halfway of the max (best) and min (worst) scenarios.


Uncertainty  12  Area(max)  Area(min)  12 2.24m 2  1.0m 2  0.62m 2
Thus, the best estimate for the measured Area,
Area  l  w  1.2  1.3  1.56  0.62 m 2
Using the rule#1 of Significant figures:
Area  l  w  1.2  1.3  1.5  0.6 m 2
To increase the confidence in experimental data, the same measurement can be repeated in (N) number of times.
Taking multiple measurements also allows better estimate of the uncertainty in the measurements by checking how
reproducible the measurements are. The best estimate would then be the mean (or average) of the multiple measurements.
x
i
i
mean  x 
N
.
The standard deviation (SD) of the sample mean is equivalent to the standard deviation of the error in the sample mean with
respect to the true mean, since the sample mean is an unbiased estimator. Therefore, the standard error (SE) of the mean can
also be understood as the standard deviation of the error in the sample mean with respect to the true mean (or an estimate of
that statistic).
2
 x  x 
i
SD 
i
N
and
SE 
SD
N
Chapter 1 Measurement
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Example: Ch1-18 Because Earth’s rotation is gradually slowing, the length of each day increases: The day at the end of 1.0
century is 1.0 ms longer than the day at the start of the century. In 20 centuries, what is the total of the daily increases in time?
Increase per century  1.0ms / century
Change in length of day at the end of 20 centuries : ( 20 centuries) x ( 1.0ms / century)  20ms
 0  20ms 
Average increase in the length of day : 
  10ms
2


Total change in the length of day for 20 centuries : Average increase in the length of day xtotal days 
 10ms 
 100 y   365.25days 
 x20century x
 x
  7305s
 
day
1y


 1century  

approximately 2hrs 
Example: Ch1-19 Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a
stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height H=1.70 m, and stop the watch
when the top of the Sun again disappears. If the elapsed time is t=11.1 s, what is the radius r of Earth?
From the triangles in the Figure provided :
2
d 2  r 2  r  h   r 2  2rh  h 2

h2 
h

  2rh1  
d 2  2rh  h 2  2rh1 
2
rh
2
r



h
since r  h leads to
0
2r
d 2  2rh1  0 
tan  
d
r
r tan 2   2h  r 

360 o
r

 d 2  r 2 tan 2   2rh
 d  r tan 
t
24hrs
2h
tan 2 

2h
tan 2 
 24hrs 
  
t
o 
 360 
 t  11.1s
21.70m
 24hrs 
tan 
11.1s 
o 
 360 

 5.2 × 10 6 m
2
Accepted value  6.3781 × 10 6 m
Example: Ch1-25 During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope,
and 2.0 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the
valley measuring 0.40 km by 0.40 km and that mud has a density of 1900 kg/m3. What is the mass of the mud sitting above a
4.0 m2 area of the valley floor?
Mud volume in the slope  Vslope  l  w  h  2500m 800m 2.0m   4.6  106 m 3


Mud volume in the valley  Vvalley  l  w  h  400m 400m h   h 1.6  10 4 m 3
Vslope  Vvalley


 4.6  10 6 m 3  h 1.6  10 4 m 3

 h
4.6  10 6 m 3
 25m
1.6  10 4 m 3



V floor  area   h   4.0m 2 25m   100.0m 3
 mud 
m floor
V floor



 m floor   mudV floor  1900kg / m 3 100.0m 3  1.9  10 5 kg
Chapter 1 Measurement
6/6
Ch1-26 One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 mm.
For that range, give the lower value and the higher value, respectively, for the following. (a) How many cubic meters of water
are in a cylindrical cumulus cloud of height 3.0 km and radius 1.0 km? (b) How many 1-liter pop bottles would that water fill?
(c) Water has a density of 1000 kg/m3. How much mass does the water in the cloud have?



2

(a) Volume of cylindrical cloud  Vcloud   r 2  h   1000m  3000m    3.0  10 9 m 3  9.42  10 9 m 3

50drops  50drops   10 6 cm3  50  10 6 drops
 

  
 Dmin 
3
3
cm3
m3
 cm   m 
number of drops

density of water drop  D 
 
6
3
6
volume
 D  500drops   10 cm   500  10 drops
max
3
3
3



cm
m
 m 

number of water drop  N  density of drop   cloud volume 
N  D  Vcloud
 50  10 6 drops 
  9.42  10 9 m3  4.71  1018 drops
N min  Dmin  Vcloud  
m3


 500  10 6 drops 
  9.42  109 m3  4.71  1019 drops
N max  Dmax  Vcloud  
m3





 






volume of water drop  Vdrop 

3
4
4
 r 3   10  10 6 m   1.33  10 15 m 3  4.19  10 15 m 3
3
3




volume of water  Vwater  number of water drop   volume of water drop 
Vwater  N  Vdrop


 
 


min
 Vwater
 N min  Vdrop  4.71  1018 drops  4.19  10 15 m 3  19,739.21m 3
 
max
19
15
3
3
 Vwater  N max  Vdrop  4.71  10 drops  4.19  10 m  197,392.1m
 m3

  0.001m 3
(b) volume of bottle  Vbottle  1liter  1liter   
1000
liter





 


volume of water Vwater
number of bottles  B 

volume of bottle Vbottle
(c) density of water   water  1000kg / m3 
M water   waterVwater

 



Bmin 
min
Vwater
19,739.21m 3

 2  10 6 bottles
Vbottle
0.001m 3
Bmax 
max
Vwater
197,392.1m 3

 2  10 7 bottles
Vbottle
0.001m3






M
mass of water
 water
volume of water Vwater






min
M min   waterVwater
 1000kg / m3 19,739.21m3  2  10 6 kg
max
M max   waterVwater
 1000kg / m 3 197,392.1m 3  2  10 7 kg
1kg water  1liter
Example: Ch1-30 Water is poured into a container that has a small leak. The mass m of the water is given as a function of time
t by m  5.00t 0.8  3.00t  20.00 , with t≥0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b)
what is that greatest mass? In kilograms per minute, what is the rate of mass change at (c) t=2.00 s and (d) t=5.00 s?
m  5.00t 0.8  3.00t  20.00
dm
 0.85.00t 0.81  3.00t 11  0  4t 0.2  3.00
dt
dm
set the derivative to zero to get the maxima;
 0  4t 0.2  3.00  3.00  4t 0.2
dt
taking the derivative with respect to time;
1
3.00  0.2

0  4t 0.2  3.00   t 0.2 

4 

0.8
mmax  m(t  4.21s )  5.004.21s 
1
 3.00  0.2
 t 
  4.21s
 4 
 3.004.21s   20.00  15.79  12.63  20.00  23.16 g
dm
0.48 g  1kg   60 s 
0.2
2
  
 42.0s   3.00  0.48 g / s 
 
  2.89  10 kg / min
dt t 2.0 s
s
 1000 g   1 min 
dm
 0.101g  1kg   60 s 
 0.2
3
  
 45.0 s   3.00  0.101g / s 
 
  6.05  10 kg / min
dt t 5.0 s
s
 1000 g   1 min 