16.1 introduction -...

116.1 Introduction

16.1 Introduction Physics is a very practical subject, and experimental work should form a significant part of your A-level Physics course. It is commonly acknowledged that it is easier to learn and remember things if you have actually done them rather than having read about them or been told about them. That is why this book, like Book 1, is illustrated throughout by experiments, usually with a set of data for you to work through and questions to answer.

Nothing is like the ‘real thing’, however, so you should be carrying out many of these experiments for yourself in the laboratory. In particular you will need to play careful attention to the experiments designated as Core Practicals in the Edexcel specification. These will form the basis of questions on practical work in Paper 3 of your A-level examination. You may be asked to describe these experiments in the examination and/or answer questions based on them. Nine Core Practicals were fully described in the Year 1 Student’s Book. Seven more Core Practicals based on A-level work are described in this book, with a set of typical results and the sort of questions that you may be asked in the examination.

Many other experiments are also described in detail. Some are included to illustrate and give you a better understanding of the theory, whilst some are particularly designed to help you develop your practical skills.

The Edexcel specification identifies particular key practical skills that should be developed through teaching and learning and will form the basis of practical assessment in the written examination:

● Independent thinking● Use and application of scientific methods and practices● Numeracy and the application of mathematical concepts in a practical

context● Instruments and equipment.

Full details of the requirements for practical work, including a comprehensive appendix on Uncertainties and practical work can be found in the specification posted on the Edexcel website: www.edexcel.com.

Information about the Science Practical Endorsement can also be found on the Edexcel website. Whilst carrying out the practical work described in this book will help you develop the skills necessary for the Endorsement, it is beyond the scope of the book to provide full details of the requirements. Your practical competency is assessed by your teacher and reported on your certificate either as ‘pass’ or ‘fail’. It does not contribute to your overall A-level grade.

16 Practical skills

16 Practical skills 2

Before we look at the above aspects of practical work, however, we need to be reminded of some of the terminology used in physics, particularly precision, accuracy and errors, which we discussed in Year 1.

In the following sections, the text in orange indicates in more detail the skills that will be assessed in the examinations.

16.2 Errors: accuracy and precision The Edexcel specification emphasises that it is important that the words used have a precise and scientific meaning as distinct from their everyday usage. Edexcel has used the terminology adopted by the Association for Science Education, full details of which can be found in the specification. In the following text, these key terms are expressed in italics. One of the specified requirements is that you should

● consider margins of error, accuracy and precision of data

In everyday English, accuracy and precision have similar meanings, but in physics this is not the case. Their meanings are not the same and the difference must be understood.

An error is the difference between a measured result and the true value, i.e. the value that would have been obtained in an ideal measurement. With the exception of a fundamental constant, the true value is considered unknowable. An error can be due to random or systematic effects and an error of unknown size is a source of uncertainty.

When you repeat a measurement you often get different results. There is an uncertainty in the measurement you have taken. It is important to be able to determine the uncertainty in measurements so that its effect can be taken into consideration when drawing conclusions about experimental results.

A measured value is considered to be accurate if it is judged to be close to the true value – it cannot be quantified and is influenced by random and systematic errors.

The term precision denotes the consistency between values obtained by repeated measurements - a measurement is precise if the values ‘cluster’ together. Precision is influenced only by random errors and can be quantified by measures such as standard deviation.

This can be illustrated by considering a digital speedometer in a car. In this situation, precision means the resolution of the instrument, which will normally be 1 mph (or 1 kph). Accuracy means is this actually the correct speed? If your speedometer reads 50 mph, you could argue that your speed was 50 ± 1 mph, i.e. an uncertainty of 2%. But is it? The answer is ‘No’. Manufacturers, by law, must not provide a speedometer that reads less than the true speed. This means that they invariably overstate the speed by several mph to be on the safe side, perhaps as much as 4 or 5 mph. Thus a reading of 50 mph is inaccurate as it has a systematic error, introduced by the manufacturer, of perhaps as much as 10%. You can get a much better indication of your true speed using your sat nav (but don’t try this if you are driving!).

316.3 Planning, making measurements and recording data

In physics experiments:

● systematic errors can be minimised by taking sensible precautions, such as checking for zero errors and avoiding parallax errors, and by drawing a suitable graph

● random errors can be minimised by taking the average of a number of repeat measurements and by drawing a graph that, in effect, averages a range of values.

Consider a mechanical micrometer screw gauge. Such an instrument has high resolution as it can measure to a very small interval – it can be read to 0.01 mm – which is the source of uncertainty in a single reading.

Furthermore, with careful use, it can make measurements with high precision – the measurements are repeatable, i.e. very similar when several readings are taken. However, its accuracy will depend on how uniformly the pitch of the screw has been manufactured and whether or not there is a zero error. You cannot do much about the former, although misuse, such as over tightening, can damage the thread. You should always check for zero error before using any instrument and make allowance for it if any is present. This is particularly true for digital instruments, which we often take for granted will be accurate and thus do not bother to check for any zero error. You should also understand that the accuracy of digital instruments is determined by the quality of the electronics used by the manufacturer.

We will take this further in section 16.3


You will be expected to:

● use online and offline research skills, including websites and textbooks● select appropriate equipment and measurement strategies in order to

ensure suitably accurate results.● obtain accurate, precise and sufficient data● consider margins of error, accuracy and precision of data● identify uncertainties in measurements and use simple techniques to

determine uncertainty when data are combined by addition, subtraction, multiplication, division and raising to powers.

Tip An uncertainty for a single reading, or repeated readings that are the same, will be the resolution of the instrument. For repeated readings that are different, the uncertainty can be taken as being half the range of the readings.

Example

A student makes some measurements of the time period T of the oscillations of a wooden metre rule in order to find a value for the Young modulus of the wood, using the arrangement shown in Figure 16.1.

She finds in a textbook that the Young modulus E for the wood is given by the formula

E = 16 π2 ML3

wt3T 2

where M is the mass suspended from the rule at a horizontal distance L from where

stopclock

L

M

Figure 16.1


the rule is clamped, and w and t are the width and thickness of the rule respectively.

She plans to use:

l two 100g slotted masses attached with elastic bands to either side of the rule, near the end, to make L as long as possible – with the slots in the masses perpendicular to the rule so that she can read the scale to determine the position of the centre of mass;

l a digital stopwatch to time the oscillations – one that can record to a precision of 0.01 s, which she considers to be more than adequate for timings that are likely to be of the order of 10 s;

l vernier callipers to measure the width and thickness of the rule, at different points along its length; the callipers can read to a precision of 0.1 mm;

l a pin secured to the metre rule with a small piece of Blu-Tack to act as a fiducial mark to help her judge the centre of the oscillations.She records the following results:

M = 200 g

L = 937 − 40 = 897 mm

w = 28.5, 28.0, 28.6, 28.9 mm; average 28.5 mm

t = 6.6, 6.7, 6.6, 6.6 mm; average 6.6 mm

20T = 8.59, 8.54, 8.18, 8.61 s; average 8.48 s

1 Explain why a value of 8.58 s for the average time should be used rather then the value stated by the student.

2 Show that this data gives a value of the order of 1010 Pa for the Young modulus of the wood.

3 Explain the number of significant figures to which the final value should be given.

4 Suggest a suitable uncertainty for the value of the period T.

5 Discuss whether the selection of vernier calipers, which could only be read to a precision of 0.1 mm, to measure the width and thickness of the rule was an appropriate strategy. Justify your answer by estimating the percentage uncertainty that these measurements introduce into the value obtained for the Young modulus.

6 To what extent would using a micrometer screw gauge, or digital callipers, reading to 0.01 mm reduce the percentage uncertainty in the value obtained for the Young modulus?

Answer1 Looking at the four values for 20 T, it would appear

that the 8.18 reading is anomalous as it is about 0.5 s smaller than the other values (suggesting only 19 oscillations may have been counted!). This value should therefore be ignored (with a note to this effect) when determining the average value.

2 Care needs to be exercised with units – all of the measurements need to be converted to SI units:

M = 0.200 kg t = 6.6 × 10−3 m

L = 0.897 m T = 8.58 ÷ 20 = 0.429 s

w = 28.5 × 10−3 m

⇒ E = 16 π2 ML3

wt3T 2

= 16 π2 × 0.200 kg × (0.897 m)3

28.5 × 10−3 m (6.6 × 10−3 m)3 × (0.429 s)2

= 1.5 × 1010 Pa ∼ 1010 Pa

3 The value should be quoted to 2 SF, 1.5 × 1010 Pa, because this is the number of significant figures of the least precise measurement – the thickness of the rule, 6.6 mm, has been measured to only 2 SF.

Tip Always record measurements with appropriate units and record all your measurements, not just the average values.

Tip Results that are clearly anomalous should be ignored when calculating the average of several values and the fact that you have done this should be stated. Note that one advantage of taking repeat readings is that an anomalous reading can be spotted and allowed for.


4 Although the time taken for 20 oscillations has been recorded to a precision of 0.01 s, the range of values is only (8.61 − 8.54) s = 0.07 s. This is probably slightly less than the human reaction time. A more realistic uncertainty in 20T would be 0.1 s. This would give an uncertainty for the value of T of 0.1 ÷ 20 = 5 ms. We can then say T = 0.429 ± 0.005 s.

5 The measurements recorded for the width were:

w = 28.5, 28.0, 28.6, 28.9 mm; average 28.5 mm

The range of values is therefore (28.0 − 28.9) mm = 0.9 mm. If we use half the range (0.45 mm) as our uncertainty we get

% uncertainty in w = 0.45 mm28.5 mm × 100% = 1.6%

Using a vernier scale reading to a precision of 0.1 mm to measure a width of 6.6 mm means that there is a percentage uncertainty in the measurement of

0.1 mm6.6 mm × 100% = 1.5%

As the term t3 occurs in the formula for the Young modulus, the percentage uncertainty introduced by this measurement will be 3 × 1.5% = 4.5%.

The combined contribution of the measurements for width and thickness is therefore:

% uncertainty = 1.6% = 4.5% = 6.1 %

This is a not insignificant uncertainty. A vernier only reading to 0.1 mm to measure the thickness was not a good choice.

6 If an instrument reading to 0.01 mm had been used, the percentage uncertainty in the thickness would have been reduced by a factor of 10, giving an uncertainty due to this measurement of less that 0.5%, which is very acceptable. However, due to the relatively large spread of values in the measurement of the width, using a more sensitive instrument would not have had much effect on the previously calculated uncertainty of 1.6%. The overall uncertainty would therefore be about 2%, which is a considerable improvement.

Example

The student’s teacher in the previous question suggests that a better value for the Young modulus might be obtained if measurements were taken with different masses and a suitable graph was plotted.

1 For a given metre rule, the equation for the Young modulus contains three variables. State what they are and, following the teacher’s suggestion, which one must be kept constant? How would you ensure this?

2 Explain what should be plotted in order to get a linear graph.

3 Show how the graph could be used, together with the rest of the experimental data, to get a value for the Young modulus.

4 Suggest why a graph is likely to reduce the uncertainty in the value obtained for the Young modulus.

You will also be expected to

● correctly follow instructions to carry out experimental techniques or procedures

● identify and control significant quantitative variables and plan approaches to take account of variables that must be controlled

● plot and interpret graphs.

Tip Remember:

● if a quantity is raised to a power in an equation, the % uncertainty it contributes is the power × the % uncertainty in the quantity

● if two or more quantities are multiplied together (or divided), their percentage uncertainties are added.


Answers1 For a given rule, the three variables are:

● the mass M,

● the period T

● the length L.

Of these, L should be kept constant and T should be found for different values of M.

To ensure L remains constant:

● the masses should be secured tightly

● the distance should be checked before each timing.

2 From the equation

E = 16 π2ML2

wt3T 2 ⇒ T 2 = 16 π2L3

wt3 × M

Therefore a graph of T2 against M should be a straight line through the origin having

gradient = 16 π2L3

wt3

3 Substituting this value of the gradient back into the equation gives us

E = 16 π2L3

wt3 × gradient

4 Plotting a graph would

● give an average of a number of readings and

● reduce both random and systematic errors.

Example

A student decides to find out more information about the absorption of gamma radiation and discovers on the Internet that knowledge of the absorption coefficient μ can be used to find the energy of the gamma rays incident on the absorber. A property of an absorber, called its ‘mass absorption coefficient’, is given by μ/ρ, where ρ is the density of the material of the absorber. The mass absorption coefficient varies with the energy of the incident gamma radiation. The student finds the information in Table 16.1 in a radiation data book.

Table 16.1

Energy/MeV 0.80 1.00 1.20 1.40 1.60

Mass absorption coefficient/10−3 m2 kg−1

8.75 7.04 6.13 5.42 5.01

The student then determines the density of a lead absorber and finds its value to be 1.06 × 104 kg m−3.

Questions1 The lead absorber used was a disc of diameter

about 25 mm and thickness about 6 mm. Describe in detail how the density of the absorber could be determined. Justify your choice of measuring instruments and explain any particular techniques that you would use to reduce the uncertainty in the measurements.

2 Suggest why it is a good idea to repeat the measurements using the other three absorbers.

3 Calculate the value of the mass absorption coefficient for the lead absorbers, using a value for μ of 0.0590 mm−1. (Hint: be careful with units.)

4 Plot a suitable graph to determine the energy of the gamma rays.

Tip Practise so that you are confident in arranging complex formulae to give straight-line graphs of the form y = mx + c and determining which of the terms are variables and which are constant.

In Chapter 14 we investigated the absorption of gamma radiation by lead in Core Practical 15. We saw that the absorption of gamma radiation by lead was an exponential function of the thickness of absorber of the form N = N0e−μx and that by plotting a graph of ln (N/min−1) against x/mm we were able to obtain a value for the absorption constant μ. An interesting extension to this investigation is shown in the following example, which illustrates some of the other skills you will be expected to demonstrate:

● applying investigative approaches and methods to practical work● evaluating results and drawing conclusions with reference to measurement

uncertainties and errors● correctly citing sources of information.


Answers1 If digital calipers reading to 0.01 mm were available,

these would be suitable for measuring both the diameter and the thickness. The uncertainty in measuring the thickness would be 0.01 mm in 6 mm, which is less than 0.2%. If not, vernier calipers for the diameter and a micrometer for the thickness would be adequate.

A balance reading to 0.1 g (or better) would be suitable for the mass. An uncertainty of 0.1 g in 200 g is only 0.05%.

The following steps should be taken to determine the density:

● The calipers should first be tested for zero error by closing the jaws and noting the reading.

● The diameter D should then be measured at least three times, rotating the disc about 60° in between readings, in case the disc is not perfectly circular.

● The thickness x should be determined by taking several measurements in different parts of the disc to check for uniformity.

● The values for the diameter and thickness should then be averaged.

● The volume of the disc can be found from V = πd 2x

4● The balance is used to find the mass M.

● The density ρ can then be calculated from ρ = mass

volume = 4M

πd2x

2 Measurements using the other three discs should be taken to:

● check that the discs are all made of the same material, and

● determine an average value for the density.

3 The first thing we have to do is to put the value for μ into base SI units, that is:

μ = 0.0590 mm−1 = 59.0 m−1.

Then:

mass absorption coefficient μρ = 59 m−1

1.06 × 104 kg m−3

= 5.57 × 10−3 m2 kg−1

4 Your graph should look like Figure 16.2 below.

From the graph we can read off that the energy of the gamma radiation is 1.35 MeV. The nature of the experiment is such that we can only just about justify 3 SF, so a final answer of 1.35 ± 0.05 MeV would be sensible.

If you look up the details of cobalt-60 on the ‘Web’ (e.g. en.wikipedia.org/wiki/Cobalt-60), you will find that it decays by β−-emission of maximum energy 0.31 MeV, and then the excited nucleus decays in two stages, emitting γ-rays mainly of energy 1.33 MeV and 1.17 MeV. It would appear, then, that the value we have obtained of 1.35 ± 0.05 MeV is reasonable when we take into account the experimental uncertainty caused by the random nature of radioactivity.

1.35 MeV

mass absorption coefficient/

10–3m2kg–1

energy/MeV

9.0

8.0

6.0

7.0

5.0

00 0.80 1.00 1.20 1.40 1.60

5.57 � 10 �3 � 2 �g �1

Figure 16.2

Tip Always justify your argument with numerical data if at all possible. An answer such as ‘A digital vernier would be suitable as it has high precision’ will not gain full marks.

Tip Note the techniques employed here to determine the diameter and thickness. Merely stating that you would take repeat readings and average them will not get full marks.


As way of revision, and to emphasise some other skills you may be assessed on:

● carrying out techniques or procedures methodically, in sequence and in combination, identifying practical issues and making adjustments when necessary

● using appropriate analogue apparatus and interpolating between scale markings

● safely using a range of practical equipment and materials, identifying safety issues and making adjustments when necessary.

It is worth repeating the Example that we looked at in Year 1.

Example

A student planned to investigate a property of a prism. He wanted to find how the deviation, δ, produced by the prism depends on the angle of incidence, i, of a laser beam striking the prism.

laser

A

�

id

Figure 16.3

The student placed the prism on a sheet of white paper and drew around it. He removed the prism and drew lines at angles of incidence of 30°, 40°, 50°, 60°, 70° and 80° using a protractor. He then replaced the prism and shone the laser beam along each of the lines in turn, taking particular care not to look directly into the laser beam. For each angle of incidence, he marked the direction of the emerging beam and determined the respective deviation. He recorded the data as shown in Table 16.2.

Table 16.2 Results

I/° 30.0 40.0 50.0 60.0 70.0 80.0

δ/° 47.0 39.0 38.0 39.5 42.0 49.5

Note that the student took his measurements methodically by initially drawing angles of incidence at 10° intervals and recorded them in a suitable table with units. He attempted to measure the angles to a precision of 0.5° by interpolating between divisions on the protractor. This is reflected in the table by the student giving all of the angles to an appropriate number of significant figures.

1 Plot a graph of the deviation, δ, against the angle of incidence, i.

2 Use your graph to determine the angle of minimum deviation, δmin.

3 Suggest extra readings that could be taken to improve the graph.

4 The student looked in a book on optics and found that the refractive index, n, of the material of the prism is given by the equation:

n = sin A + δmin

2

sinA2

where A is the angle of the prism. The student measured A and found it to be 60.0°. Use this information to find a value for the refractive index.

Tip Wherever realistically possible, try to interpolate between scale divisions, e.g. 0.5 mm on a millimeter scale, 0.5° for a protractor and 0.5°C or better on a mercury thermometer calibrated in degrees.

916.4 Analysis and evaluation

Answer1 Your graph should look like that in

Figure 16.4

2 The minimum deviation is 37.9°.

3 The angle of minimum deviation could be determined with more certainty if extra readings were taken for i = 45° and i = 55°.

4 n = sin A + δmin

2

sinA2

= sin 60.0 + 37.9min

2

sin60.02

= sin 49sin 30 = 1.51

You should always take an appropriate number of measurements over as wide a range as possible. In the above example, this is achieved by taking readings between 30° and 80° (readings with i less than about 30° are not achievable as total internal reflection occurs) and by subsequently taking extra readings at i = 45° and i = 55°.

In consideration of safety, you should note that the student took particular care not to look directly into the laser beam. Even a laser pointer is a very hazardous device and could cause serious eye damage. It is not advisable to use lasers in a darkened room as your eye irises will be dilated. The utmost care should always be taken when working with lasers.

You always need to be aware of the potential dangers of the apparatus with which you are working and act accordingly.

16.4 Analysis and evaluation You are required to develop and use a range of mathematical skills, details of which you can find on the Edexcel website. Your attention is also drawn to Chapter 18 in Book 1 ‘Maths in Physics’. You should note that more advanced mathematical skills are required for A level. These are shown in bold in Chapter 18 of Book 1. In particular, you will need to be confident in handling exponential functions and drawing and analysing logarithmic graphs. You will be expected to:

38

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��

��

��

�8

��

3� �� 8�i��

d��

Figure 16.4


● process and analyse data using appropriate mathematical skills as exemplified in the mathematical appendix of the specification

● solve problems set in practical contexts and apply scientific knowledge to practical contexts

● evaluate results and draw conclusions with reference to measurement uncertainties and errors.

It goes without saying that you should be able to produce a graph with sensible scales and with appropriately labelled axes, including correct units, and then plot points correctly and draw the straight line, or smooth curve, of best fit.

The graph in Figure 16.4 is an example of when the most suitable scale can be achieved by not starting the axes at the origin. You should always choose a scale that occupies at least half the graph paper in both the x and y directions (or else the scale could be doubled!) and one that avoids awkward scales such as scales having multiples of 3.

Points should be plotted accurately, with interpolation between the scale divisions, and marked with a neat cross or a small dot with a ring round it. In more advanced work the plots should be plotted as error bars. The length of the bars should reflect the uncertainty of the plotted value. When drawing the line of best fit, you should remember that not all functions in physics are linear. If your points clearly lie on a curve, you must draw a smooth curve through them. Furthermore, not every straight line will necessarily pass through the origin, even though you may be expecting it to. If this is the case, you need to discuss possible reasons as to why the line does not pass through the origin.

Tip First plot your points in pencil. If they look right, with no apparent anomalies, ink them in. Then draw your line in pencil. If you are not happy with your line, you can easily rub it out and have another go without also rubbing out the plotted points.

Example

Let us go back to the experiment with the metre rule. Another student observes that as the length L projecting beyond the bench is reduced, the time period T also gets less. He thinks that T and L may be related by an expression of the form

T = kLn

where k is a constant and n is a numerical power.

The student finds that with a mass of 200 g, the oscillations are too fast to time. He therefore increases the mass to 400 g. He then records the following data (for simplicity, his values for the time for 20 oscillations, repeated, have been omitted):

Table 16.3L/m 0.900 0.850 0.800 0.750 0.700 0.650

T/s 0.61 0.57 0.52 0.47 0.42 0.38

Questions1 Explain how a linear graph can be obtained from the

expression T = kLn in order to find values for n and k.

2 Copy Table 16.3 and add values for ln(T/s) and ln(L/m).

3 Plot a graph of ln(T/s) against ln(L/m).

4 Use your graph to find the vales of n and k. What are the units of k?

5 Discuss the extent to which the experiment confirms that T2 is proportional to L3.

Tip Note that the student has identified a practical issue (with 200 g the oscillations were too fast to time) and made adjustments where necessary (by making the mass 400 g).

1116.4 Analysis and evaluation

Answers1 From T = kLn, if we take logs to base e we get:

ln(T/s) = n ln(L/m) + ln(k)

2 Table 16.4L/m 0.900 0.850 0.800 0.750 0.700 0.650

T/s 0.63 0.57 0.52 0.47 0.42 0.38

ln(T/s) −0.462 −0.562 −0.654 −0.755 −0.868 −0.968

ln(L/m) −0.105 −0.163 −0.223 −0.288 − 0.357 −0.431

3 Your graph should look like that shown in Figure 16.5. Note that as both quantities being plotted have negative values, the axes have been drawn so that the points can be plotted in the negative quadrant. You could get round this by plotting, say, ln(T/ms) against ln(L/mm) but you would have to be careful of the units for k. You would actually get the same values for n and k if you plotted the points in the positive quadrant (try turning the graph upside down!) but it is better to choose the axes correctly.

4 n = gradient = −(0.30 −1.00)−(0.00 −0.45) = 1.56 ≈ 3/2

Applying our scientific knowledge to this practical context we could reasonably assume that n = 3/2 as powers are usually integers or simple fractions.

Reading off intercept when ln(L/m) = 0 gives ln(k) = −0.30 from which k = 0.74.

As k is a physical quantity we must check its units. We can now write our formula as

T = k L32

The LHS has units of s.

For the RHS to also have units of s, the units

of k must therefore be s m− 32.

5 The value we obtained for n was 1.56, which differs from 3/2 by

(1.56 −1.50)1.50 × 100% = 4%

From T = kLn ⇒ T = k L32

Squaring both sides gives

T 2 = k2L3

We can therefore say that within the experimental uncertainty of 4%

T 2 ∝ L3

0

–1.0

–0.9

–0.8

–0.7

–0.6

–0.5

–0.4

–0.3

–0.1–0.2–0.3–0.4–0.5

In(T/s)

In(L/m)

Figure 16.5

Tip Note the choice of scale. A larger scale can be obtained by not including the origin for ln(T/s). However, Including the origin for ln(L/m) enables the intercept – ln(k) – to be read off directly without compromising the scale too much.


You are reminded that at A-level Physics you will be internally assessed for a Practical Endorsement. For this endorsement you will need to show practical competency in the skills highlighted in this chapter, together with evidence of the ability to:

● apply investigative approaches and methods to practical work● keep appropriate records of experimental activities● use appropriate software and/or tools to process data, carry out research

and report findings● use online and offline research skills, including websites and textbooks● correctly cite sources of information.

Examples of these skills are included throughout this book in the Core Practicals and in the Activities, where you will find questions of the type you will meet in your examination, particularly in Paper 3. Some questions that might typically be asked in Paper 3 can be found in the ‘Preparing for the exam’ chapter.

In the meantime, here are a couple of questions for you to have a go at now!

Tip Remember:

● the % difference between two experimental values is given by:

% difference = difference between the valuesaverage of the two values × 100%

● the % difference between an experimental value and a stated or known value is given by:

% difference = difference between the values

stated values × 100%

13Exam practice questions

Exam practice questions 1 In the last experiment, we showed that T 2 = k2L3 where the units for k

were s m− 3

2. Show that this is consistent with the units for k given by the formula

E = 16 π2ML3

wt3T 2

which we had in the first experiment with the oscillating rule. [Total 4 marks]

2 A student plans to investigate how the frequency of vibration of air in a conical flask depends on the volume of air in the flask. She blows directly into the neck of the flask and listens to the sound of the air vibrating. She then pours water into the flask until it is approximately half filled and blows into the flask as before. She notices that the pitch (frequency) of the vibrating air is higher when the flask is half full of water.

She thinks that there might be a relationship between the natural frequency of vibration f of the air in the flask and the volume V of air in the flask of the form f ∝ V n, where n is a numerical constant.

She sets up the arrangement shown in Figure 16.6.

a) The student increases the frequency of the signal generator until the air in the flask vibrates very loudly.

i) Explain why this happens at the natural frequency of vibration of the air.

ii) Describe a technique by which the uncertainty in determining this frequency could be reduced.

iii) Describe how she could vary, and measure, the volume of air in the flask.

iv) Explain how plotting a graph of lnf against lnV would enable her to test whether f ∝ V n and would enable her to find a value for n.

b) The student obtained the following data:

i) Calculate values of ln(V/cm) and ln( f/Hz) from the data above.

ii) Plot a graph of ln( f/Hz) against ln(V/cm).

iii) Use your graph to determine a value for n.

iv) Explain qualitatively whether your value for n is consistent with the student’s initial observations.

Table 16.5

V/cm3 f/Hz

554 219

454 242

354 274

254 324

204 361

154 415

calibrated signal generator

loudspeaker

Figure 16.6

16.1 introduction -...

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