Chapter 1 Physical Quantities and Units

1.1 Basic Quantities and International System of Units (SI units)1.2 Dimensions, and Physical Quantities1.3 Scalar and Vectors1.4 Metrology

Introduction

What is physic ?

1. Definition of physics - derives from Greek word means nature.

2. Each theory in physics involves:

(a) Concept of physical quantities.

(b) Assumption to obtain mathematical model.

(c) Relationship between physical concepts.

- proportional

- inversely proportional

- exponential

(d) Procedures to relate mathematical models to actual measurements from experiments.

(e) Experimental proofs to devise explanation to nature phenomena.

1.1 Quantities and International System of Units (SI units)

Learning outcome:

(a) list base quantities and their SI units: mass (kg), length (m), time (s), current (A), temperature (K) and quantity

of matter (mol);

(b) deduce units for derived quantities

Physical quantity

A physical quantity is a quantity that can be measured.

Physical quantity consist of a numerical magnitude and a unit.

Example:

250 ml (magnitude and unit)

Basic quantity

Basic Quantity is cannot be derived from other quantities.

This quantity is important because it

- can be easily produced

1

- does not change its magnitude

- is internationally accepted

SI units

SI unit is the unit of a physical quantity is the standard size used to compare different magnitudes of the same

physical quantity.

Systems of units

Several systems of units have been in use. Example:

- The MKS (meter-kilogram-second) system

- The cgs (centimetre-gram-second) system

- British engineering system: foot for length, pound for mass and second for time.

Today the most important system of unit is the Systems International or Sl units.

Physical Quantity and the SI Base Units1. Physical quantities can be divided into two categories:

a) basic quantities and

b) derived quantities.

2. The corresponding units for these quantities are called base units and derived units.

Basic Quantities1. In the interest of simplicity, seven basics quantities, consistent with a full description of the

physical world, have been chosen.

Basic quantity Symbol

Dimension(base

quantity symbol)

Definition SI units

Length l Llength most commonly refers to the

longest dimension of an objectMeter

Mass m M

Mass , more specifically inertial mass, can be defined as a

quantitative measure of an object's

resistance to acceleration

Kilogram

Time t T

Time is a dimension in which events

can be ordered from the past through

the present into the future, and also

the measure of durations of events

Second

2

Basic quantity Symbol

Dimension(base

quantity symbol)

Definition SI units

Electric

current I A

Electric current is a flow of electric

charge through a conductive mediumAmpere

Thermodynamic

temperature T

Temperature is a physical property of

matter that quantitatively expresses

the common notions of hot and cold.

Kelvin

Amount of

substances,

Quantity of

matter

n N

Amount of substance is a

standards-defined quantity that

measures the size of an ensemble of

elementary entities, such as atoms,

molecules, electrons, and other

particles

Mole

Luminous intensity Iv J

luminous intensity is a measure of

the wavelength-weighted power

emitted by a light source in a

particular direction per unit solid angle

Candela

Table 1- 1

Base UnitsThere are only seven base unit in SI system.

SI Base units Symbol Definition

Metre m"The metre is the length of the path travelled by light in vacuum during a time

interval of 1/299 792 458 of a second."

17th CGPM (1983, Resolution 1, CR, 97)

Kilogram kg"The kilogram is the unit of mass; it is equal to the mass of the international

prototype of the kilogram."

3rd CGPM (1901, CR, 70)

Second s

"The second is 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."

13th CGPM (1967/68, Resolution 1; CR, 103)

"This definition refers to a caesium atom at rest at a temperature of 0 K."

(Added by CIPM in 1997)

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SI Base units Symbol Definition

Ampere A

"The ampere is that constant current which, if maintained in two straight parallel

conductors of infinite length, of negligible circular cross-section, and placed 1

metre apart in vacuum, would produce between these conductors a force equal to 2

× 10−7 newton per metre of length."

9th CGPM (1948)

Kelvin K

"The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the

thermodynamic temperature of the triple point of water."

13th CGPM (1967/68, Resolution 4; CR, 104)

"This definition refers to water having the isotopic composition defined exactly by

the following amount of substance ratios: 0.000 155 76 mole of 2H per mole of 1H,

0.000 379 9 mole of 17O per mole of 16O, and 0.002 005 2 mole of 18O per mole of 16O."

Mole Mol

"1. The mole is the amount of substance of a system which contains as many

elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol

is 'mol.'

2. When the mole is used, the elementary entities must be specified and may be

atoms, molecules, ions, electrons, other particles, or specified groups of such

particles."

14th CGPM (1971, Resolution 3; CR, 78)

"In this definition, it is understood that unbound atoms of carbon 12, at rest and in

Candela cd

"The candela is the luminous intensity, in a given direction, of a source that emits

monochromatic radiation of frequency 540 × 1012 hertz and that has a radiant

intensity in that direction of 1/683 watt per steradian."

16th CGPM (1979, Resolution 3; CR, 100)

Table 1- 2

Prefixes•For very large or very small numbers, we can use standard prefixes with the base units.

Prefix tera giga mega Kilo deci centi mili micro nano pico

Factor 1012 109 106 103 10-1 10-2 10-3 10-6 10-9 10-12

Symbol T G M K d c m µ n P

4

Table 1- 3

Example 1.1:

Write 2 x 10-7 in a suitable prefix.

Solution:2 x 10-7 ---- 2 x 10-6 x 10-1 ---- 2 x 10-1 – 0.2

Derived quantities and derived units

Derived Quantity is derived from basic quantities through multiplication and division.

•For example,

Derived quantity Derive from base quntity of Derived unit

Area length x length m2

Volume length x length x length m3

Densitymass

volumekg m-3

Velocitylt

m s-1

Accelerationvelocity

timem s-2

Frequency 1T

s-1/hz

Momentum Mass x velocity Kg ms-1

Force Mass x acceleration Kg ms-2

Pressure forceArea

N m-2

Energy12

mv2 Kg m2 s-2

Table 1- 4

The derived unit change

Example 1.2 :

7854 kg m-3 change into g cm-3.

Solution:

5

7854 kg1 m3

----

7854×103 g100 cm×100 cm×100 cm -----

7854×103

106 gcm−3

-- 7.854 g cm-3

1.2 Dimensions and Physical Quantities

(a) use dimensional analysis to determine the dimensions of derived quantities;

(b) check the homogeneity of equations using dimensional analysis;

(c) construct empirical equations using dimensional analysis;

The dimension of a physical quantity is a product of the basic physical dimensions each raised to a rational

power.

1. Each derive quantity in physic can be represent by basic quantity.The dimension of a physical

quantities is the relation between the physical quantity and the base quantities2. The Bracket ‘[ ]’ meant The dimension of (pronounce its loudly) or the power of base quantity of

Example :

[v] “the dimension of velocity” , this means that the power of base quantities in the velocity.

Example 1.3

Write the dimensions for the following physical quantity

(a) Acceleration

Solution:

(a) [a ]=[ v−u

t ]=LT−1−LT−1

T=LT−2

Use of dimensions•To check the homogeneity of physical equations

Concept of homogeneous

•The dimensions on both sides of an equation are the same.•Those equations which are not homogeneous are definitely wrong.

•However, the homogeneous equation could be wrong due to the incomplete or has extra terms.

•The validity of a physical equation can only be confirmed experimentally.

•In experiment, graphs have to be drawn then. A straight line graph shows the correct equation and the

non linear graph is not the correct equation.

•Deriving a physical equation

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•An equation can be derived to relate a physical quantity to the variables that the quantity depends on.

Example 1. 4

Determine the homogeneous of the equation v2 =u2 +2as.

Solution:Left hand side :

[v2] = [v]2 = (LT-1)2 = L2 T-2

Right hand side :

[u2 + 2as] = L2 T-2 + L T-2 . L = L2 T-2

Conclusion ; the RHS dimension as same as the LHS dimension, meaning that the above equation is

homogenic.

Derivation of Physical EquationFrom observations and experiments, a physical quantity may be found to be dependent on a few other physical

quantity. To find this relationship we use dimension method.

Example 1.5

From the observations speed of sound in medium maybe affected by density d, wavelength , and Young

Modulus E. Derive an equation for the speed of sound in the medium. ([E] =ML-1T-2)

Solution:

It’s observe that vαda λb Ec suppose the a,b and c are dimensionless constant.

Then v=kda λb Ec

Assume that k is the dimensionless constant.

LHS : [v ]=LT−1

RHS :

[da ]=(ML−3 )a=M a L−3a

[ λb ]=Lb

[ Ec ]=(ML−1T−2)c=M c L−c T−2c

[da λb Ec ]=M a+c L−3 a+b T−2c

So LHS=RHS

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LT−1=M a+c L−3 a+b T−2c

Pairing the similar physical quantities Dimension indices of both sides.

LHS RHS

M0 Ma+c

L1 L-3a+b

T-1 T-2c

Table 1- 5

M: 0=a+c

L : 1=−3 a+c

T : −1=−2c

Solve the above equation :

a=−12 ,b=0 and

c= 12

So the equation v=kda λb Ecwill become :

v=kd−

12 λ0 E

12

Meaning that the equation of v is

v=k √ Ed

Example 6

(a) Given below are the equation of the liquid flow inside the horizontal pipe.

(1) p+Aρv2=W

(2) p+BTg

v2=X

(3) p+Cg ρv=YWhere;

W,X,Y have the dimension as same as pressure

A,B,C are the constant without dimension.

g represent gravitational acceleration.

T represent liquid surface tension (it’s dimension is MT-2)

represent liquid density

v represent liquid velocity

p represent the pressure change

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Determine the homogeneity of the above equation.

(b) Below are the reading for p and v :

p (Nm-2) 2.0 103 1.5 103 1.2 103 0.7 103 0.3 103

v (m s-2) 1.0 1.4 1.6 1.9 2.1

Table 1- 6

Using the above reading ,

(i) Determine the correct equation.

(ii) Determine the constant for the correct equation using the information below.

[ = 1.0 103 kg m-3, T = 7.4 10-2 N m-1]

Solution :(a) Equation (1),

has no dimension so it’s should be term P and Av2 have the same dimension.

[P] = M L-1 T-2

[v2] = M L-3 x L2 T-2 = M L-1 T-2

Equation (2)

X has no dimension so it’s should be term P and

Tgv2

have the same dimension.

[Tgv2 ]=MT−2×LT−2× 1

L2T−2=ML−1T−2

Equation (3)

Y has no dimension so it’s should be term P and gρv have the same dimension.

[ gρv ]=LT−2×ML−3×LT−1=ML−1 T−3

It’s meant that equation (3) is dimensionally wrong.

(b) Verily we can conclude that equation (1) and equation (2) are dimensionally correct.

(i) Equation (1) and equation (2) have to be rearrange into y = mx + c

(1) p=−Aρv 2+W meaning graph p vs v2 is should be a straight line

(2) p=−BTg

v2+X

meaning graph p vs

1v2

is should be a straight line

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y = -0.5011x + 2.5

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

equation (1)equation (2)Linear (equation (1))Log. (equation (2))

2

1v

P, Pressure(103)Nm-2

To confirm which equation is correct the we have to plot both of the graph according to the data given in

the table.

p (Nm-2) 2.0

103

1.5

103

1.2

103

0.7

103

0.3

103

v (m s-2) 1 1.4 1.6 1.9 2.1

 v2 1 1.96 2.56 3.61 4.41

 

1v2 1.000 0.510 0.391 0.277 0.227

Table 1- 7

Graph 1

From the graph it’s confirm that the equation (1) is correct because it is a linear graph with negative gradient.

(ii) to find A ;

From the graph the gradient is -0.5011.

From the equation (1) the gradient is −Aρ

∴−Aρ=−0 .5011

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A=0 . 5011

ρ = 1.0 103 kg m-3

A= 0 . 50111 . 0×103

=5×10−4

And

W=2. 5×103N m-2, W is the intercept of the graph.

1.3 Scalar and Vectors

(a) determine the sum, the scalar product and vector product of coplanar vectors;

(b) resolve a vector to two perpendicular components;

> A scalar quantity is a physical quantity which has only magnitude.

For example, mass, speed , density, pressure, ....

> A vector quantity is a physical quantity which has magnitude and direction.

For example, force, momentum, velocity , acceleration ....

In most cases in physic, the physic quantity is express in vector. If the number(magnitude) can be operated

through Subtract, Add, multiplication and fraction. Then the vector also can be threat the same way except

fraction, but it’s have to follow the rule that govern them.

Graphical representation of vectors•A vector can be represented by a straight arrow,

The length of the arrow represents the magnitude of the vector.

The vector points in the direction of the arrow.

Basic principle of vectors

• Two vectors P and Q are equal if:

a) Magnitude of P = magnitude of Q

(b) Direction of P = direction of Q

• When a vector P is multiplied by a scalar k, the product is k P and the direction remains the same as P.

The vector -P has same magnitude with P but comes in the opposite direction.

Principles of vectors

Figure 1- 1

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(a) Substitute of Vector (i) To show the relative velocity

Let us look at two cases:

Lets the two object VA = 10 ms-1 (faster) VB = 3 ms-1. (slower), moving in the same axis-x

Case one

The velocity of A relative to B = (VA - VB) (comparing faster toward slower)

= (10- 3) ms

= 7 ms -1 (in forward direction).(mean that A is 7 ms -1 faster than B)

Case two

The velocity of B relative to A = (VB - VA)

= (3 - 10) ms

= -7 ms -1 (in backwards direction).

We observe that(VB - VA) and (VA - VB) are same magnitude but different direction.

(ii) to show change of velocity

Lets the same object VA = 10 ms-1 (faster), then after a while it’s decrease into VA = 3 ms-1. (slower),

So the change of the velocity is 3 ms-1 subtract with 10 ms-1 then it’s -7 m s-1 which negative sign mean

that the object is reducing it’s velocity

(b) Sum of vectors (Resultant of) If there are two or more vector , these vector can be add to form a single vector called a Resultant vector.

To solve the problem involving vectors in two dimension, we usually used any one of these method depend on

the information given.

Method 1: Parallelogram of vectorsIt’s the drawing method. The drawing of the parallelogram need to be draw according scale and angle

given in the question. The instrument used for this drawing are:

(a) ruler

(b) protractor

(c) sharp pencil

It two vectors O A and O B are represented in magnitude and direction by the adjacent sides OA and

OB of a parallelogram OABC, then OC represents their resultant.

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Figure 1- 2

This method is used when there are information about angle and magnitudes of the vector.

Method 2: Triangle of vectors and polygon of vectorIt’s the drawing method. The drawing of the vectors need to be draw according scale and angle given in

the question. The instrument used for this drawing are:

(d) ruler

(e) protractor

(f) sharp pencil

•Use a suitable scale to draw the first vector.

•From the end of first vector, draw a line to represent the second vector. (attaching the head with the it’s

tail)

•Complete the triangle/polygon. The line from the beginning of the first vector to the end of the second

vector represents the sum in magnitude and direction.

Figure 1- 3

Example 7

A kite flies in still air is 4.0 ms-1. Find the magnitude and direction of the resultant velocity of the kite when the air

flows across perpendicularly is 2.5 ms-1. If the distance of the kite is 30 m from the player, what is the time taken

for the kite to fly? Calculate the height of the kite from the ground.

Solution:

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Draw a straight line from A to B with the length of 4 cm, (1 cm : 1 m s-1). And another line B to C with the length

of 2.5 cm. the angle of ABC is 90 degree. The resultant of the vector can be measured from A to C, 4.72 cm.

This answered can be converted into 4.72 ms-1.

Using the protractor the angle of DAC is 58 degree. Meaning that the kite is moving at speed of 4.71 m s-1 and

58 degree from the ground.

The distance from A to C is 30m, the time taken from A to C is :

t= sv=30 m

4 .71 ms−1=6 . 37 s

And the height of the kite is :

DCAC

=sin θ

DC=30 m×sin 58=25 . 44 m

Example 8

Figure 1- 5

Five coplanar forces act on a particle, as drown in Figure above. Draw a scaled force polygon for these forces.

State the magnitude and direction of the resultant of these forces.

Solution:

Figure 1- 4

D

CB

A

Scaled to 1 cm : 1 m s-1

4.72 cm

2.5 cm

4 cm

14

Draw the polygon using a scaled 1 cm : 1N, refer to figure1- 6. The angle of the vector must be referred to the

figure 5 while connecting the head and tail of every vector.

The resultant of the vector can be measured using metre ruler from A to F and it’s length is 3.5 cm, meaning that

3.5 N

And it’s direction or angle, is 21 degree from the ground.

Method 3 : Component Method

G

Figure 1- 6

45

45

2 cm

5 cm4 cm

4 cm

3 cm

F

E

D

C

BA

Scaled to 1 cm: 1 N

15

Resultant,R

Ry: component of vector R on y-axis (+ )

Rx: component of vector R on x-axis (+)

It’s is a calculation method , because every vector can be replace into x-component and y-component.

Replacing a single vector into it’s components is called Resolving.

To determine the resultant of the vector using this method, it’s need to follow these four keyword carefully.

1. Axis

2. Resolve vector

3. add vector component

4. Resultant

Axis

Need to be determine before resolving the vector.

•Resolving vectorThe vector that is not on any axis have to be resolve into it’s component. Resolving vector mean resolving :

(a) magnitude

(b) Direction

A vector R can be considered as the two vectors. R refers to the resultant vectors. There are two mutually

perpendicular component Rx and Ry

Figure 1- 7

Add Vector Component

∑ F x=Fx 1+Fx 2+Fx 4 . .. .. . .. .. . .. . and ∑ F y=F y 1+F y2+F y 4 . . .. .. .

Only the same axis component can be added.

Resultant

Magnitude, R=√(∑ Fx )

2+(∑ F y )

2

and Direction of R, θ=tan

∑ F y

∑ Fx

Example 9

The figure 8 shows 3 forces F1, F2 and F3 acting on a point O. Calculate the resultant force and the direction of

resultant.

16

Figure 1- 8

Solution:

Step one : Draw the axis x and y

Step two : resolve the vector that is not on any axis into two component, x and y

Step three : determine their angle

Figure 1- 9

17

Step four : tabulate the component according their axis.

Force Component X Component Y

F1

Mag: 3N

Dir : to the Right (+)+3N

No Component0

F2

Mag : 5N Cos 30= 4.33N

Dir : to the left (-)-4.33N

Mag : 5N sin 30 = 2.5 N

Dir : upward (+)+2.5N

F3

Mag: 4N Cos 60= 2N

Dir : to the left (-)-2N

Mag: 4N sin 60

Dir : downward (-)-3.46N

∑ F x -3.33N ∑ F y -0.96N

Table 1- 8

Step Five : Calculate magnitude and direction of the resultant force.

Magnitude, R=√(∑ Fx )

2+(∑ F y )

2

R=√(−3 . 33 N )2+(−0 .96 N )2

R=3. 47 N And

Direction of R, θ=tan

∑ F y

∑ Fx

θ=tan −0 . 96N−3 .33 N

=196°

(d) Multiplication of vectorIt’s have been discuss about subtraction and addition of the vector. From subtraction and addition of vector we

can explain most of the physical quantity. Now is about multiplication of vectors. When two vectors were multiply

the result is called product.

There are two kind of product produced :

1. Dot Product

2. Cross Product

Dot Product :The dot product is fundamentally a projection.

18

S = 5M

F = 5N

= 60

Figure 1- 10

The dot product of a vector with a unit vector is the projection of that vector in the direction given by the unit

vector. This leads to the geometric formula

v .ω=|v||ω|cosθFurthermore, it follows immediately from the geometric definition that two vectors are orthogonal if and only if

their dot product vanishes, that is

v . ω=0

Example 10

Calculate the scalar product of vector F and s below.

Figure 1- 11

Solution:

F⋅S=(5 N )(5 m)cos60 °=12. 5 Nm

19

v

vv

v

Cross Product : The cross product is fundamentally a directed area.

Figure 1- 12

whose magnitude is defined to be the area of the parallelogram. The direction of the cross product is given by

the right-hand rule, so that in the example shown v×ω points into the page.

|v×ω|=|v|sinθ|ω|To determine the direction of the cross product we used the right hand rule. In mathematics and physics, the

right-hand rule is a common mnemonic for understanding notation conventions for vectors in 3 dimensions. It

was invented for use in electromagnetism by British physicist John Ambrose Fleming in the late 19th century.

Figure 1- 13

Example 11

20

b = 18 unit

a = 12 unit

b

a

ba

There are two vector a and b, calculate the a b.

Figure 1- 14

Solution:

Magnitude,|a× b|=(12 )(18 )sin 90 °=216Direction :

Figure 1- 15

1.4 Metrology (Uncertainties in measurements)

Learning outcome:

(a) calculate the uncertainty in a derived quantity (a rigorous statistical treatment is not required);

(b) write a derived quantity to an appropriate number of significant figures.

Metrology is the science of measurement and its application.

Terminology related to measurement uncertainty is not used consistently among experts. To avoid further

confusions lets refer to BIPM-VIM(International Vocabulary of Basic and General Terms in Metrology) and

GUM (Guide to the expression of uncertainty in measurement).

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1.4.1 ErrorVIM define the error as below:

error (of measurement) [VIM 3, 2.16] - measured quantity value minus a reference quantity value

There are two type of error

(a) Systematic ErrorCharacteristics of systematic error in the measurement of a particular physical quantity:

-Its magnitude is constant.

-It causes the measured value to be always greater or always less than the true value.

Corrected reading = direct reading - systematic Error

Sources of systematic Error:

- Zero Error of instrument.

- Incorrectly calibrated scale of instrument.

- Personal error of observer, for example reaction time of observer.

- Error due to certain assumption of physical conditions of surrounding for example, g = 9.81 ms-2

Systematic error cannot be reduced or eliminated by taking repeated readings using the same method,

instrument and by the same observer.

(b) Random ErrorCharacteristics of Random Error :

- It's magnitude is not constant.

- It causes the measured value to be sometimes greater and sometimes less than the true value.

Corrected reading = direct reading ± Random Error

The main source of random Uncertainty is the observer.

The surroundings and the instruments used are also sources of random error.

Example of random Error:

- Parallax Error due to incorrect position of the eye when taking reading

Parallax Error can be reduced by having the line of sight perpendicular to the scale reading.

- Error due to the inability to read an instrument beyond some fraction of the smallest division

Reading are recorded to a precision of half the smallest division of the scale.

Random Error can be reduced by taking several readings and calculating the mean.

Error contributes to but is different from Uncertainty

1.4.2 The Uncertainty of the Instrumental VIM define the Uncertainty as below

22

uncertainty of measurement [VIM 3, 2.6] non-negative parameter characterizing the dispersion of the quantity values being attributed to a measurand (quantity intend to measure), based on the information used and it’s have a statistical concept of standard deviation means.

Instrumental Measurement

When handling the experiment the reading is given by the apparatus used, these apparatus have their own

uncertainty.

instrumental measurement uncertainty(VIM 3, 4.24) - the amount (often stated in the form ±x) that along with the measured value, indicates the range in which the desired or true value most likely lies. Instrumental measurement uncertainty is used in a Type B evaluation of measurement uncertainty

Here the magnitude of ±x is called the absolute Uncertainty.

Absolute Uncertainty is the smallest scale of the instrument or half of the smallest scale if it’s can be determine

“easily”.

Instruments Absolute Uncertainty Example of readings

Millimetre ruler 0.1 cm (50.1 ± 0.1)cm

Vernier calliper 0.01 cm (3.23 ± 0.01)cm

Micrometer screw gauge 0.01 mm (2.63 ± 0.01)mm

Stopwatch (analogue) 0.1 s (1.4 ± 0. 1 )sStopwatch(Digital) 0.01 s (1.452 0.01)s

Thermometer 0.5 °C (28.0 ± 0.5)°C

Ammeter (0 - 3A) 0.05 A (1.70 ± 0.05)A

Voltmeter (0 - 5V) 0.05 V (0.65 ± 0.05)V

Table 1- 9

The smaller absolute uncertainty of the instrument is contribute to the high accuracy, precision and sensitivity of the measuring system of the experiment.

1.4.3 Analysing Uncertainty of the data

- specifically Uncertainty analyzing is refer to Uncertainty that cause by repetition measurement to

produce more accurate data.

- Meaning that if we want to measure a mass of cube, of course we cannot just used a single

measurement then we will get the answer. We have to measure the mass with the triple balance beam

more than one time for example 3 time.

- While doing the measurement actually we have continually increasing the Uncertainty.

- It is a good idea to mention the Uncertainty for every measurement and calculation.

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- In this subtopic we deal with the repetition reading or data. It’s known that if we have more than one

reading so the true value is the mean of the reading.

- Mean value for a is ⟨a⟩=

a1+a2 . .. . .. .+an

n

- Mean value of Uncertainty of a, Δ ⟨a ⟩should be calculated this way

1. Calculated the deviation of every data given:

s1=|a1−⟨a ⟩|

s2=|a2−⟨a ⟩|.

.

.

sn=|an−⟨a⟩|2. Find the sum of deviation

s=∑n

1

sn=s1+s2+.. .. .+sn

3. find the mean of deviation ⟨s ⟩= s

nIt’s known that the mean deviation is equally the same as the Uncertainty of the mean

value(true value).

Or

Δ ⟨a ⟩=⟨s⟩

Working example on a single quantity :1. Aim : to determine the diameter, d of a wire

2. Theory : used outer jaw of vernier caliper

3. Precaution : measure more than one reading

4. Choosing Apparatus and Determine the absolute uncertainty:

Instruments Uncertainty

(Absolute/actual)Vernier caliper 0.01 cm

Table 1- 10

5. Manage the reading/data:Diameter ,d of a wire was measured several time to reduce the Uncertainty and the reading is given in the

table below. Find the true value(mean value) and the Uncertainty of the diameter.

24

i ii iii iv v vi

(d0.01)/

cm

1.55 1.52 1.54 1.53 1.54 1.53

Table 1- 11

6. Determine the quantity and it’s uncertainty a. Calculating the true value of diameter (mean value) <d>:

⟨d ⟩=1 .55+1. 52+1. 54+1 .53+1. 54+1 . 536

=1. 54 cm

b. Calculating the uncertainty of diameter:

Δ ⟨d ⟩=|1 .55−⟨d ⟩|+|1 . 52−⟨d ⟩|+|1. 54−⟨d ⟩|+|1.53−⟨d ⟩|+|1 .54−⟨d ⟩|+|1 .53−⟨d ⟩|6

=0. 01cm

So the diameter of a wire should be written (1.54 0.01)cm

Note: calculating the uncertainty this way is refer to a single quantity and not involving with the graph.

1.4.4 Primary data and secondary data1. Primary data are raw data or readings (taken directly from apparatus) in an experiment. Primary data

obtained using the same instrument have to be recorded to the same degree of precision i.e to the same

number of decimal places.

2. Secondary data are derived from primary data. Secondary data have to be recorded to the correct

number of significant figures. The number of significant figures for secondary data may be the same (or

one more than) the least number of significant figures in the primary data. Measurement play a crucial role

in physics, but can never be perfectly precise.

3. It is important to specify the Uncertainty or Uncertainty of a measurement either by stating it directly

using the ± notation, and / or by keeping only correct number of significant figures.

Example: 51.2 0.1

Processing significant figures• Addition and subtraction

25

When two or more measured values are added or subtracted, the final calculated value must have the

same number of decimal places as that measured value which has the least number , of decimal

places.

Example

1. a = 1.35 cm + 1.325 cm

= 2.675 cm

= 2.68 cm

2. b = 3.2 cm - 0.3545 cm

= 2.8465 cm

= 2.8 cm

3. c =

1. 15+1. 13+1 .16+1. 14+1 .135

cm

= 1.142 cm

= 1.14 cm

Multiplication and division When two or more measured values are multiplied and/or divided, the final calculated value must have

as many significant figures as that measured value which has the least number of significant figures.

Example1. Volume of a wooden block = 9.5 cm x 2.36 cm x 0.515 cm

= 11.5463 cm3

= 12 cm3

2. If the time for 50 oscillations of a simple pendulum is 43.7 s, then the period of oscillation = 43.7 ÷ 50

= 0.874 s

3. The gradient of a graph

9 .15−3 .000 .450−0 .050

= 6 .150 . 400

=1. 5375=1 . 54

Note: Sometimes the final answer may be obtained only after performing several intermediate

calculations. In this case, results produced in intermediate calculations need not be rounded off. Round

only the final answer.

1.4.5 Analysing Uncertainty of a derive quantity.

1. Actual Value - is in the scale reading (pointer reading) of an instrument.(single reading)

Or

- is in the mean value.(of the repetition reading)

2. Fractional and percentage Uncertainty,

26

(a) The fractional Uncertainty of R : R=ΔR

R

(b) The percentage Uncertainty of R : R=ΔR

R×100%

3. Consequential Uncertainties/Uncertainty- to state the Uncertainty of a derive quantities Given

R 1 R1 = Data Absolute Data Uncertainty = 51.2 0.1

R 2 R2 = Data Absolute Data Uncertainty = 30.1 0.1

(a) Addition

W = R1 + R2 = 51.2 + 30.3 = 81.3

W = R1 + R2 = 0.1 + 0.1 = 0.2

So W W = 81.3 0.2

(b) Subtraction

S = R1 - R2 = 51.2 - 30.3 = 21.1

S = R1 + R2 = 0.1 + 0.1 = 0.2

So S S = 21.1 0.2

(c) Product

P = R1 R2 = 51.2 30.3 =1541.12

From

ΔPP

=ΔR1

R1+

ΔR2

R2

∴ ΔP=( ΔR1

R1+

ΔR2

R2)P

ΔP=( 0 . 151. 2

+ 0 .130 . 1 )1541. 12

∴ ΔP=7 . 71

P P = 1541.12 7.71

27

(d) Quotient

Q=R1

R2=51.2

30.1=1. 70

From

ΔQQ

=ΔR1

R1+

ΔR2

R2

∴ ΔQ=( ΔR1

R1+

ΔR2

R2)Q

ΔQ=( 0 . 151. 2

+ 0 .130 .1 )1. 7

∴ ΔQ=0. 01

Q Q = 1.70 0.01

Working example:1. Aim : to determine the value of B

2. Theory :

B is given by

B=( a−b )d2

q √T3. Precaution : B have a combine uncertainty from various apparatus (quantity)

4. Choosing Apparatus and Determine the absolute uncertainty:

Quantity Instruments Uncertainty

(Absolute/actual)a,b meter ruler 1 cm

q Stopwatch(Digital) 0.01 s

Table 1- 12

5. Manage the reading/data:After the measuring and calculating the uncertainty of the quantity a,b,d,q and T(refer 1.4.2).

The true value (mean value) and the uncertainty of the quantities are witten as below :

a =(1.83±0.01)m,

b=(1.65 ±0.01) m,

d=(0.00106±0.00003)m,

28

q = (4.28 ± 0.05) s

T = (3.7 ± 0.1) x 103 s.T is

6. Determine the quantity and it’s uncertainty

(a) Find B use the equation given

B=( a−b )d2

q √T

=(1 .83 m−1. 65 m)0 .00106 m2

4 .28 s√3 .7×103 sB = 7.8 x 10-11 m3 s-

(b) Find the uncertainty of B

1. Fisrt check the equation for addition and subtraction, by applying 1.4.3 no 3 (b) , subtraction

so (a - b) = (0.18±0.02)m

2. Second calculate the percentage uncertainties in each of the 4 terms:

Term Magnitude and uncertaintyFractional

Uncertainty

Uncertainty

percentage

(a - b) = (0.18±0.02)m0 .020 .18

11%

d = (0.001 06 ± 0.000 03) m0 .000030 .00106

3%

q = (4.28 ± 0.05) s0 .054 .28

1.2%

T = (3.7±0.1) x 103 s0 .1×103

3. 7×103 3%

Table 1- 13

- The Uncertainty in (a - b) is now very large, although the readings themselves have been taken

carefully. This is always the effect when subtracting two nearly equal numbers.

- The percentage Uncertainty in d2 will be twice the percentage Uncertainty in d;

- The percentage Uncertainty in √T will be half the percentage Uncertainty in T because a

square root is a power of

12 .

29

This gives:

Uncertainty percentage in B = 11% + 2(3%) + 1.2% +

12 (3%) = 19.7% ≈ 20%

This gives B = (7.8 ± 1.6) x 10-11 m3 s-1.

the rules for uncertainties therefore :

Operator Uncertainty

addition and subtraction ADD absolute uncertainties

multiplication and division ADD percentage uncertainties

powers Multiply the percentage Uncertainty by the power

Table 1- 14

1.4.6. Uncertainty of a Linear graph

Figure 1- 16: where n is the number of points plotted.

1. The usual quantities that are deduced from a straight line graph are

(a) the gradient of the graph m, and the intercept on the y-axis or the x-axis

(b) the intercepts on the axes.

First calculate the coordinates of the centroid using the formula

30

(∑ xi

n,∑ y i

n ) where n is the number of sets of readings4,5.

2. The straight line graph that is drawn must pass through the centroid Figure . The best line is the

straight line which has the plotted points closest to it. This line will give m the best gradient

together with c.

3. Two other straight lines, one with the maximum gradient mmax and another with the least

gradient mmin , are then drawn. For a straight line graph where the intercept is not the origin , the

three lines drawn must all pass through the centroid. Here also we can find cman and cmin

4. To find the Uncertainty for the gradient and intercept used this equation

Δm=(mmax−mmin )

2 and Δc=

(cmax−c min)2

Working Example1. Aim To determine the acceleration due to gravity using a simple pendulum.

2. Theory : the theory of the simple pendulum, the period T is related to the length l, and the

acceleration due to gravity g by the equation

T 2=4 π2 lg

Hence, the acceleration due to gravity, g=4 π2 l

T2

A straight line graph would be obtained if a graph of T2

against l is plotted.

3. Precaution :

The time t for 50 oscillations of the pendulum is measured for different lengths l of the pendulum.

The period T is calculated using

T= t50

4. Choosing Apparatus and Determine the absolute uncertainty:

Instruments Uncertainty

(Absolute/actual)Millimeter ruler 0.1 cm

31

Stopwatch (analogue) 0.1 s

Table 1- 15

5. Manage the tableNote the various important characteristics when tabulating the data as shown in Table 15

Length

(l 0.05) cm

Time for 50 oscillation (t 0.1)sPeriod

T(s)

T= t50

T2(s2)

(i) (ii) Average

90.00 94.7 94.9 94.8 1.90 3.61

80.00 88.5 88.5 88.5 1.77 3.13

70.00 84.0 83.8 83.9 1.68 2.82

60.00 78.4 78.6 78.5 1.57 2.47

50.00 70.1 69.9 70.0 1.40 1.96

40.00 63.2 63.0 63.1 1.26 1.59

30.00 55.8 55.8 55.8 1.12 1.25

20.00 44.7 44.9 44.8 0.896 0.803

10.00 31.9 32.0 32.0 0.640 0.410

Table 1- 16

(a) Name or symbol of each quantity and its unit are stated in the heading of each

column. Example: Length and cm, and T(s). The Uncertainty for the primary data,

such as length and t time for 50 oscillations, is also written. Example: (l 0.05) cm

and (t 0.1)s.

(b) All primary data, such as length and time, should be recorded to reflect the precision

(absolute uncertainty) of the instrument used.

For example, the length of the pendulum l is measured using a metre rule. hence it

should be recorded to two decimal places of a cm, that is 10.00 cm, and not 10 cm

or 10.0 cm.

The time for 50 oscillations t is recorded to 0.1 s, that is 32.0 s and not 32 s.

The average value of t is also calculated to 0.1 s. The average value of 31.9 s and

32.0 s is recorded as 32.0 s and not 31.95 s.

(c) The secondary data such as T and T2, are calculated from the primary data.

Secondary data should be calculated to the same number of significant figures as I

hat in the least accurate measurement. For example, T and T2, are calculated to

three significant figures, the same number of significant figures as the readings of t.

32

(d) For a straight line graph, there should be at least six point plotted. If the graph is a

curve, then more points should be plotted, especially near the maximum and

minimum points.

From the graph we can determine the intercept and the gradient, both of them also have their own uncertainty. In

order to find the uncertainty of intercept and gradient , it’s have to calculate the centroid point. Centroid point is

the average reading of both in x-axis and y-axis4.

The x-coordinate of the centroid =

∑ li

n

=

19

(10+20+30+40+50+60+70+80+90 ) cm

= 50 cm

The y-coordinate of the centroid =

∑ T i2

n

=

19

(0 .410+0.803+1 .25+1. 59+1.96+2. 47+2.82+3 .13+3. 61 ) s2

= 2.00s2

The coordinate for the centroid is (50cm, 2.00s2)

Figure 1- 17

33

from the equationg=4 π2 l

T2

T 2=4 π2 lg

Hence a graph of T 2 against l is a straight line, passing through the origin, and gradient,

m=4 π 2

g

From the graph,

gradient of best line,

m=2 . 00 s2

0 . 50 m¿4 .00 s2 m−1

Maximum gradient,

mmax=3 .05 s2

0.75 m¿4 .07 s2m−1

Minimum gradient,

mmin=2. 35 s2

0. 60 m¿3 .92 s2 m−1

Absolute Uncertainty in the gradient,

Δm=(mmax−mmin )

2=( 4 .07−3.92 )s2 m−1

2=0 .15

2=0 .075 s2 m−1

Fractional Uncertainty in the gradient

Δmm

=0. 0754 .00

=0. 01875≈0 . 0188

percentage Uncertainty in gradient

Δmm

×100 %=1. 88 %

34

Acceleration due to gravity, g= 4 π 2

m= 4 π 2

4 . 00=9 . 870 ms2

Hence the percentage Uncertainty in g is the sum of the percentage Uncertainty in m only because

42 is a constant.

Therefore percentage Uncertainty in gravity,g = Uncertainty percentage = 1.88% according

to above equation

Hence acceleration due to gravity,

Written in percentage Uncertainty

g = (9.8701.88%) m s2

also can be write in absolute Uncertainty

Δg=1 .88 %×9 .870=0 .2ms2

g = (9.9 0.2) m s2 Since there is Uncertainty in the second significant figure, the value of g is

given to two significant figures.

35