Summary Part Summary Part 11

Measured ValueMeasured Value = True Value + Errors= True Value + Errors

Errors = Random Errors + Systematic Errors = Random Errors + Systematic ErrorsErrorsHow to minimize RE and SE: How to minimize RE and SE: (a)(a) RE – by taking RE – by taking MORE MORE measurements!measurements!(b)(b) SE – by SPOTTING! SE – by SPOTTING! Precision Precision Accuracy Accuracy

How to present the measured value ?

In general, the result of any measurement of a quantity x is stated as following

units

(Best estimate Uncertainty) units

If possible x is represented by the standard error!

( )x x

How to present the measured value ?

Rule for stating Uncertainties: Experimental uncertainties should almost always be rounded to one significant figure.

Example: x = 0.000153 = 0.0002 m

How to present the measured value ?

Rule for stating Answers: The last significant figure in any stated answer

should usually be of the same order of the same order of magnitude (in the same decimal position) as uncertainty.

Example: x = 12.23452 m x = 0.0002 m

x ± x = (12.2345 ± 0.0002) m

Statistical Formulas

1

1 n

ii

x xn

xn

2

1

1( )

1

n

ii

x xn

Mean Mean Standard DeviationStandard Deviation

Standard Error Standard Error

Error PropagationError PropagationHow to estimate the error of the quantity How to estimate the error of the quantity RR from the errors associated with the primary from the errors associated with the primary measurement of measurement of xx, , yy, , zz, …, respectively?, …, respectively?

22 22 2 2 ...R x y z

R R R

x y z

Summary Part 2 Summary Part 2 Presentation of final answer: Presentation of final answer: (1.2 (1.2 0.3) m 0.3) m

MeanMean: average of the data: average of the data

Standard deviationStandard deviation: measures the spread : measures the spread of the data about the meanof the data about the mean

Standard errorStandard error: standard deviation of the : standard deviation of the mean mean

Use propagation of error formula to Use propagation of error formula to compute the combined error! compute the combined error!

Example 1Example 1In the determination of gravity,In the determination of gravity, g g by means of by means of a simple pendulum (i.e. a simple pendulum (i.e. TT = 2 = 2 (ℓ /(ℓ /gg) ), the ) ), the following data are obtained:following data are obtained:

Length of the string, ℓ: 99.40, 99.50, 99.30, Length of the string, ℓ: 99.40, 99.50, 99.30, 99.45, 99.35 cm.99.45, 99.35 cm.Time for 20 oscillations, Time for 20 oscillations, tt: 40.0, 39.8, 40.2, : 40.0, 39.8, 40.2, 39.9, 40.1 sec.39.9, 40.1 sec.Calculate Calculate gg and its error. and its error.

Solution for Example 1Solution for Example 1For ℓ, For ℓ, nn = 5 = 5

Mean: Mean:

Standard deviation of ℓ: Standard deviation of ℓ:

Standard error: Standard error:

5

1

199.40cm

5 ii

52

1

1( ) 0.07906 cm

5 1 ii

0.03536 cm5

Solution for Example 1Solution for Example 1For time of an oscillation For time of an oscillation TTii = t = tii/20, /20, nn =5 =5

Mean: Mean:

Standard deviation:Standard deviation:

Standard error:Standard error:

5

1

12.00sec

5 ii

T T

52

1

1( ) 0.007906 sec

5 1T ii

T T

0.003536 sec5T

T

Solution for Example 1Solution for Example 1(99.40 0.04) cm

(2.000 0.004)sec

TT

2 224 981.039 cm / secgT

The estimated uncertainty of g:

29.81039m / secg

2 22 2 2g T

g g

T

Solution for Example 1Solution for Example 1

2

24

T

g

3

28

TT

g

224T

gFrom this functionFrom this function

Take the partial derivative of Take the partial derivative of gg w.r.t. w.r.t. ℓℓ and and TT

Solution for Example 1Solution for Example 1

2 22 2 2 3.486 cm/sec

g T

g g

T

2

24),(

T

Tg

3

28),(

TT

Tg

2(9.81 0.04) m / secg

g

Which is the BEST line?

Least Squares Method Least Squares Method (LSM)(LSM)

Finding the best straight line

ybest = mbest x + cbest

to fit a set of measured points (x1, y1), (x2, y2), …, (xn, yn).

The following assumptions are made to find the best line: 1.The uncertainty in our measurements of x

is negligible but not in y2.The uncertainty in our measurements of y

is the same (i = )

Least Squares Method (LSM)

Let coordinates (Let coordinates (xxii , , yyii) is the ) is the i-i-th data point th data point

that you have plotted in the graph and that you have plotted in the graph and

The best fit of a straight line takes theThe best fit of a straight line takes thefollowing form: following form: yybestbest((xxii) = ) = mmbestbest x xii + + ccbestbest

n

i 1

Least Squares Method (LSM)

21

'best i i i i ic x y x x y

1

'best i i i im n x y x y

22' ii xxn

Least Squares Method (LSM)

22

'bestc ix

2

'bestm n

21

2 i best best iy c m xn

Least Squares Method (LSM)

ybest(x) = mbest x + cbest

Example 2Example 2In the determination of In the determination of gg by means of a simple pendulum by means of a simple pendulum (i.e. (i.e. TT = 2 = 2 (ℓ /(ℓ /gg)), the following data are obtained:)), the following data are obtained:

ℓℓ11 = 40.00 cm = 40.00 cm tt11 = 26.3, 25.5, 25.9 sec = 26.3, 25.5, 25.9 sec

ℓℓ22 = 60.00 cm = 60.00 cm tt22 = 31.8, 32.3, 32.7 sec = 31.8, 32.3, 32.7 sec

ℓℓ33 = 80.00 cm = 80.00 cm tt33 = 36.9, 36.5, 37.1 sec = 36.9, 36.5, 37.1 sec

ℓℓ44 = 100.00 cm = 100.00 cm tt44 = 42.6, 41.5, 41.7 sec = 42.6, 41.5, 41.7 sec

ℓℓ55 = 120.00 cm = 120.00 cm tt55 = 43.8, 44.8, 45.3 sec = 43.8, 44.8, 45.3 sec

ℓℓ66 = 140.00 cm = 140.00 cm tt66 = 48.4, 48.9, 49.1 sec = 48.4, 48.9, 49.1 sec

ℓℓ77 = 160.00 cm = 160.00 cm tt77 = 51.2, 51.5, 51.9 sec = 51.2, 51.5, 51.9 sec

ℓℓ88 = 180.00 cm = 180.00 cm tt88 = 54.5, 54.8, 54.3 sec = 54.5, 54.8, 54.3 sec

where where ttii is the time for 20 oscillations. Determine is the time for 20 oscillations. Determine gg and and its error.its error.

Solution for Example 2Solution for Example 2

2880.00cm/secii

x 2 2 2113600.00 cm /secii

x 8n

-237.075475 secii

y 24765.85778 cm/seci ii

x y

x yT 2

2 -4( ) ( ) 0.05645254 seci best ii

y x y x

Solution for Example 2Solution for Example 2slope, mbest = 0.040926 sec2/cm y-intercept, cbest = 0.132583 sec2 m_best = 0.000748 sec2/cmc_best = 0.089178 sec2/cm

mbest m_best = (0.0409 0.0008) sec2/cm

cbest c_best = (0.13 0.09) sec2

The best fit of a straight line,T2

best = 0.0409 ℓ + 0.13

Solution for Example 2Solution for Example 242 / g = Tbest

2 / ℓ 42 / g = mbest = 0.0409 sec2/cm

gbest = g = 965.2425 cm/sec2

How To Compute, g ?

Solution for Example 2Solution for Example 2

2

2

4

best best

g

m m

How To Compute, g :From 42 / g = mbest g = 42 / mbest

where

g = 18.88005 cm/sec2

g = 6.67510 cm/sec2 g g = (9.65 0.07) m/sec2

2

2 2

bestg mbest

g

m