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Physics Factsheet April 2003

Number 50

Errors and Uncertainties Although these two names actually mean the same thing, the word “uncertainty” is often the better term since it cannot be confused with the word “mistake”.

When you are using a digital meter do not be fooled into believing that the uncertainty is the least significant digit. The uncertainty will depend on the tolerance of the electronic components in the meter. A good rule of thumb is to say that the uncertainty is at the level of the next to least significant digit – since you are likely not to know its precision unless you have the meter’s calibration certificate.

Whenever you try to measure something you may not really measure its true value because even you are human and the instruments that you use are not perfect (nor can they ever be). In recognising this it is helpful for you to calculate an uncertainty in your value – since you will then be able to see how accurate your answer is likely to be. This is a measurement of how close you judge your answer is to the true value. This is not simply a matter of your guessing; there are several techniques that you can use to make sure that you calculate a sensible value for the uncertainty in your reading.

Example: A digital ammeter shows value of 1.56 A. Write down the value that you should quote for the current. Answer: You should quote this as (1.6±0.1) A – explaining that you don’t know the precision of the ammeter and you are estimating it as being 1 decimal place.

There are two types of uncertainty: called systematic and random uncertainties. • Systematic uncertainties are those that are due to the system of measurement that is being used. For example, if a meter has not been calibrated correctly it will never give the true value even if it could be read perfectly. Systematic uncertainties shift all your readings in the same direction but cannot easily be detected without using a second set of apparatus – then, of course, you don’t know which is right or wrong! The systematic uncertainty determines the accuracy of your readings (how far they are from the true value) and is not helped by repeated readings using the same apparatus. •

Your reading is likely to be better than this but you are flagging to the examiner that you recognise that there are reasonable uncertainties even when using digital meters. In a similar way, when you use a digit stopwatch to time 30 oscillations as taking 23.54 seconds – don’t fool yourself that this precision is valid! Your reaction time is likely to be of the order of 0.1-0.2 s and so you should use this as your uncertainty and quote the time as being (23.5 ±0.2) s. 0.2s is called the absolute uncertainty in the time; 0.2 × 100% = 0.85% is called the % uncertainty in the time. 23.5

Random uncertainties are those that are due to several possible causes and are distributed around a mean value. They usually result from the your inability to take the same measurement in exactly the same way to get exactly the same number. Examples of sources of random uncertainties are: using an instrument with a scale of low precision (very spaced out scale divisions that force you to estimate a reading between them) or having a slow reaction time. Random uncertainties can be reduced by taking a series of readings and finding the average (mean) value. These uncertainties affect the precision of your answer (the number of significant figures that you are justified in quoting).

Incidentally, by measuring the time for thirty oscillations you are helping yourself to come up with a precise value for the period because you should divide both the value and the uncertainty by 30 to give a good value: 0.2 24 .3 23.5 = 0.7833 s; uncertainty = = 0.007 s. You are 30 30 justified in quoting the period to be (0.783 ± 0.007) s – a high level of precision. Period, T =

Example (a) Check your value for the period if you had miscounted by one and found a time of 24.3 s for 30 oscillations. (b) Quote a measured value for the period with its uncertainty. (c) Comment on how important it is to count a large number of oscillations.

When you carry out an “error” analysis, you are assessing the random uncertainties not the systematic ones.

Estimating Uncertainties It is important that your uncertainty limits are large enough for you to include the true value of the quantity that you are measuring. For analogue meters, rulers, thermometers etc. it is probably best to quote the uncertainty as being the smallest scale division (graduation). You are likely to estimate the reading to better than this but what you are saying is that my answer is definitely within this uncertainty range. For example when you read a single value on a ruler to be 85 mm you should quote the value as being (85±1) mm This says that the reading is definitely between 84 mm and 86 mm.

Answer 24.3 (a) T = = 0.81s 30 (b) absolute uncertainty in T =

0.2 = 0.007 s so I would quote 30

T = either (0.810 ± 0.007) s or (0.81 ± 0.01) s error period with miscount 0.03 ≈ ≈ 4% period with correct counting 0.78

Exam Hint - Always quote your uncertainty to the same number of decimal places as the quantity that you are measuring.

- this demonstrates the value of counting a significant number of oscillations – the error (right word!) introduced is acceptable at 4%

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Physics Factsheet

Errors and Uncertainties Absolute uncertainties have units while percentage uncertainties do not.

Example The density of the material in Fig 1 used to make a block of material is to be calculated by measuring the lengths of its sides with a 30 cm ruler and its mass using a top pan balance with a precision of 0.1 g. Calculate the density of the material and estimate the uncertainty in your value.

Improving the level of uncertainty The example for the period shows that by taking a series of readings you are increasing the probability that your average value is closer to the true value (assuming no systematic errors are present). This will reduce your level of uncertainty and increase your confidence that you are measuring well.

Fig 1

Some examination boards like you to quote a simple standard uncertainty as being:

mass = 95.2 g 2.1 cm

maximum reading – minimum reading number of readings taken

6.6 cm 2.1 cm

Answer m m ρ= = 2 V wl this gives a value of ρ =

(This is an approximation to a statistical formula but it’s ok for A level physics!). When you measure a potential difference to be 1.54 V, 1.58 V and 1.56 V, the simple standard uncertainty for these values is

95.2 = 3.27 g cm -3 2.12 × 6.6 taking the precision of the ruler to be ± 1 mm

(1.58 −1.54 )

= 0.01 V 3 so you should quote the mass as being (1.56 ± 0.01) V.

∆ρ ∆ w ∆l   ∆m =± +2 +  m w l  ρ  0.1 0.1  ∆ρ  0.1 =± +2 +  2.1 6.6  ρ  95.2

Combinations of uncertainties 1. Formulae There are formulae for calculating uncertainties but you do need to be very careful whether you use the absolute or percentage uncertainty in each case. I am not making any attempt to prove these formulae – you would never be expected to do this at AS or A level. When W, X, Y and Z are measured quantities with uncertainties ∆W, ∆X, ∆Y, and ∆Z. A,B, C, D and E are quantities that are calculated from the measure values.

this makes the % error in density 11.1% and so we should quote ρ = (3.27 ± 0.36) g cm-3 2. Numerical Methods An alternative to using the formulae for combining uncertainties to use a numerical method in order to calculate uncertainties just by using your value and your estimates of the absolute uncertainty in each value. In measuring the resistivity, the current in and voltage across a wire are measured to be (6.6 ± 0.1) A and (2.6 ± 0.1) V. The length of the wire is (0.800 ±0.002) m and its diameter (0.50 ± 0.01) mm. Using these values the resistivity can be calculated together with a value for the uncertainty in the measurement.

• Addition If A = W + X + Y then the uncertainty in A is ∆A = ±(∆W + ∆X + ∆Z). • Subtraction If B = W – X then the uncertainty in B is ∆B = ±(∆W + ∆X) yes you add them!

ρ=

• Multiplication If C = W × X then the percentage uncertainty in C

RA V π d 2 = ! 4I !

substituting the values into this we get a value for ρ

ρ=

is ∆C = ±  ∆W + ∆X    C X   W you would then find ∆C by multiplying the bracketed term by C.

×× (0.50 2.6 10××m) 2.6 V V ×× ππ(0.50 10-3−3m)22 = 9.6 × 10-8 Ω m 4 × 6.6 A × 0.80m

A second, larger, value that would be just possible to obtain for ρ can be found by making each of the quantities on the top of the equation larger by its maximum uncertainty and by making each of the quantities on the bottom of the equation smaller by its uncertainty:

• Division If D = W ÷ X then the percentage uncertainty in D is ∆D = ±  ∆W + ∆X    D X   W yes you add the percentage uncertainties, you would then find ∆D by multiplying the bracketed term by D.

ρ big =

-3 2 ×× (0.51 ×× m) 2.7 10 2.7 V V××ππ(0.51 10−3m)2 = 10.6 ×10 -8 Ωm 4 × 6.5 A × 0.798m

A third, smaller, value that would be just possible to obtain for ρ can be found by making each of the quantities on the top of the equation smaller by its maximum uncertainty and by making each of the quantities on the bottom of the equation larger by its uncertainty:

• Raising a quantity to a (constant) power If E = Wn , where n is a constant (which may or may not have units) then the percentage uncertainty in E

ρ small =

is ∆E = ± n  ∆W    E  W 

-3 2 ×× (0.49 2.5 10××m) 2.5 V V ×× ππ(0.49 10−3m)2 = 8.7 × 10 -8 Ωm 44××6.7 6.7A 0.802 m A ××0.802

The average difference between the ρ value calculated assuming no uncertainties and that calculated assuming maximum uncertainties (in the same overall direction) is approximately 1.0 × 10-8 Ωm and so the value that you should quote is: ρ = (9.6 ±1.0) ×10-8 Ωm.

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Physics Factsheet

Errors and Uncertainties Questions

Exam Workshop

1. Explain the difference between systematic and random uncertainties.

This is a typical weak student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner’s answer is given below.

2. Three readings of current are taken as 1.2 A, 1.5 A and 1.6 A. Calculate the mean value, the simple standard uncertainty and the percentage uncertainty.

The circuit in the diagram is used to find the resistance, R of a light Ω ± 5%. dependent resistor (LDR). The resistor has a value, R, of 5.0 kΩ The voltmeter reading, V, is (0.029 ± 0.005) V and the emf of the cell, E, is (1.5 ± 0.1) V. resistor

3. ∆x is the absolute uncertainty in x and ∆y is the absolute uncertainty in y; m is a constant. Explain how you would calculate the uncertainties in: (a) x + y; (b) x – y; (c) xy; x (d) ; y (e) xm .

LDR

V S

4. In measuring the acceleration of a toy car the speeds are measured to be (0.20 ± 0.02) ms-1 at time (25 ± 1) s and (0.27 ± 0.02) ms-1 at (40 ± 1) s. Calculate the acceleration of the car and give the uncertainty in your value.

E = 1.5 V

(a) Calculate the resistance, R L , of the LDR using the following equation: RV [2] RL = E -V 5000 × 0.029 1/2 = 98.6 !" (1.5 - 0.029 )

5. The rate of discharge of a capacitor through a resistor can be used to measure the capacitor’s capacitance C using the equation: 1

V = V0 e RC

At time 0 s the voltage across the capacitor (V0) is measured to be (1.5 ± 0.1) V. After (20±1) s it is measured to be (0.8 ± 0.1) V. The resistor has a value of (4700 ± 500) Ω.

Fine but no unit – Ω missed out! mark dropped!

(b) Write down the absolute uncertainty in the value for V. 0.024 < V < 0.034!

[1] 1/1

Using the numerical technique calculate a value for the capacitance of the capacitor and the uncertainty in your value.

Not expected answer, but the student has given the range of possible values for V and this has gained the mark (c) Calculate the absolute uncertainty in E – V ⇒ ± 0.095 " ± (0.1 - 0.005) =>

Answers 1. See Factsheet.

[1] 0/1

2. 1.43 A; 9.1%. 3. (a) ∆x + ∆y;

No, you add absolute errors when you are subtracting quantities (d) Calculate a value for the percentage uncertainty in RL. ∆RL ∆R ∆V ∆( E − V ) = + + RL R V E −V 250 0.1 0.105 ⇒ + + = 0.188 Or 19% !!" 5000 1.5 1.471

(b) ∆x + ∆y; [3] (c)

 ∆x ∆ y  (d) ∆a =± +  a y   x

2/3

(e) ∆a  ∆x  = ± m  a  x 

Nearly right – the student has, in fact, calculated the % uncertainty in E not V in the second part of the calculation – the correct answer [using the correct ∆(E-V)] is 29.4% Why has the student gone to the trouble of converting 5% into an absolute uncertainty in order to change it back into the % uncertainty for R? This was a waste of time and effort

4. 0.0047 ms-2; 70% or (0.0047 ± 0.0033) ms-2 – since we are not told whether it is % or absolute uncertainty that is required. 5. ±4.5 × 10-3 F or ± 2.4 × 10-3 F depending upon whether you choose greater over smaller or smaller over greater (log is not a linear relationship).

Examiner's Answers (a) RL =

 ∆x ∆ y  ∆a =± +  a y   x

! ! 5000 × 0.029 RV = 98.6 Ω = (1.5 - 0.029 ) E -V

(b) 0.005 V ! (c) ± (0.1 + 0.005) = ± 0.105 ! (d)

∆RL ∆R ∆V ∆ ( E − V ) ! = + + RL R V E −V 0.005 0.105 ! + = 0.05 + 0.029 1.5 − 0.029

Acknowledgements: This Physics Factsheet was researched and written by Mike Bowen Jones. The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

= 0.294 = 29.4% !

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