A2 Further Practical Skills

8
T h e pr ac ti ca l work in th e s ec o nd yea r of y our A l eve l co u rse bu i lds on wh a t y ou have cov e r e d in th e fir s t yea r . Th e exa min a tion will t es t you on t wo ar e as : pl a nnin g ex p r im er ıt s, a nd an a l ys i s and eva lu a ti o n of y ou r res ult s. Planning Al rh oug h yo u s h o uld b e t hinkin g of ex p e rim e nt al p roce dur es f r o m as soo n as yo u e mb a r k on a n A l eve l co u rse, i f n o t b efore, b y th e tim ey ou c o mpl e r e th e co ur se yo u should b e c omp e t e nt in plannin g . ex p rim e nt a l proc e dur es. T h e r ea r e differ e nt s t a g es in plannin g an e x p e rim e nt . Defining the problem it m ay see m o b v i o u s, but t he first thin g i s t o i d er ı tify t h e p ro bl e m . To do t h at yo u mu s t id e nr i f y : th e independent var iable in th e ex p e rim e nt th e dep endent variable in th e ex p e rim e nt th e va ri a bl e s that are to b e controll e d. F or in s t a n ce y ou ma y b e investi ga tin g th e r e spon se t im e o f a th e rmo c oupl e w h e n it i s u se d to m eas u re ch ang in g te mp e r at u tes, as s ho w n in F i g ur e Al. I . W h at a r e t h e va ri a bl es h e r e? p r o b e ju c tion wa ter b a th i =-- - I ~~in s f ix ed j u n ct i o n m elting ice Fi g u r e A 2 . 1 In ves tiga ti ng th e respon s e time o f a therm oc oupl e th e rmom ete r . i I B I Th e ind e p e nd e nt v a ri a bl e would b e th e temper a tu re w h ic h i s t o b e m eas ur e d. T h e d e p e nd e nt va ri a bl e wo uld b e th e time i t t a k es f o r th e rea din g on th e th er moc o upl e m e t e rt o r eac h a s t ea d y va lu e. T h e r e may b e severa l o th e r va ri a bl es th a tc o uld af f ect th e r es ul t . O n e mi g ht b e th e st a rtin g t e mp e r a tur e of th e t h e rrn os t a t ' p ro b e jun c tion ', ar ı o th er i s th e Ap pendix A2 F u r th e r practical skills - - ~------------------ -- ' (f i xe d ) t e mp e r at u re of th e seco nd jun c t ion of t h e t h e rrnocoupl e. Data coUeetion T h e n ex t t as k is t o th i n k a b out h ow you are going to ca rr y th e ex p erimen t o ut . Once you have a metho d in mi nd y ou n ee d t o : D eser ibe t h e m et h o d t o b e u se d to vary the i nd e p e nd e nt va ri a bl e. D es er i be ho w th e ind e p e nd e n t varia bl e i s to be m eas u red. D es eribe how th e depend e nt va ri a bl e i s t o b e m eas ur e d . D es eribe ho w o th e r v ari a bl es a r e t o b e co ntr o ll ed. D esc ribe , w it h t h e ai d o f a c lear, l a b e ll e d d i agram, t h e a r range m e nt of a pp ara tu s f or t h e ex p e ri ment a nd t h e proce dur es t o be fo ll owe d . In th e ex p e rim ent to in vest i gate the response time of a th e rmo co upl e yo u may d eci d e t h at t h e sirnp l est way o f v a ry in g th e ind e p e nd e n t va ri a bl e is to h ave a wa ter bath a nd t o v ary i ts te mp e r a tur e. W h e n th e t e mp e ratur e h as reac hed th e re qui re d l eve l , l u n ge th e ' pr o b e j uncti o n ' int o th e wa t e r . T h e pro ce du re wo uld b e: Meas ur e th e t e mp e r a tu re of t h e water wit h a l abo rat o r y th er m o m eter . 2 T h e r ea din g f r o m th e th err n oco upl e wi ll be d is pla y ed o n a milli vo l t m ete r a nd th e depen d e n t va ri a ble, th e t im e tak e nf or th e r ea din g to reac h i t s s t ea d y valu e, m e a s ured w ith a S top watc h . 3 T h e r e ar e se ve ra l diff e r e n t va ri a bl es t h at need to b e c ontr o ll ed. In pr ac t ice yo u w ill o nl y need to co n s id e r o n e o r tw o . T h e seco n d j u n c ti o rı of t h e t h er m oco u p l e s h ou l d b e ke pt a t a co n sta n t tempera tur e b y keepi n g it i n c orıt ac t w ith m e ltin g ice. T he pr o b e jun c tion s h o uld b e coo l ed to t h e same t e mp e ratu re ea ch tim e, aga in b y holdin g it in th e m e ltin g i ce . Y o ur d es cripti o n of th e ex p eri m e n t s h o ul d be c l ea r a nd co n c i se. It some t i m es h e l ps t o give t h e desc r ipt i on in t h e f o rm of bull e tp o in ts or a n um bered l ist . A l a b e ll e d d i a gram wi ll a l ways c l arify yo u r descriptio n . In th e ex p e rim e n t d ese r i b e d h ere, t h e d i agram shown i n Fig ur e Al. ı wo uld be s u f f icie n t .

Transcript of A2 Further Practical Skills

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The practical work in the second year of your A level

course builds on what you have covered in the first year.

The examination will test you on two areas: planning

experimerıts, and analysis and evaluation of your results.

PlanningAlrhough you should be thinking of experimental

procedures from as soon as you embark on an A

level course, if not before, by the time you complere

the course you should be competent in planning. experimental procedures.

There are different stages in planning an experiment.

D efin in g th e p ro ble m

it may seem obvious, but the first thing is to iderıtify

the problem. To do that you must idenrify:

• the independent variable in the experiment

• the dependent variable in the experiment

• the variables that are to be controlled.

For instance you may be investigating the response

time of a thermocouple when it is used to measure

changing temperatutes, as shown in Figure Al. I.

What are the variables here?

probe juction

water bath

i=---I~~ins

fixed junction

melting ice

FigureA2.1 Investigating the response time of a thermocouple

thermometer.

I

The independent variable would be the temperature

which is to be measured.

The dependent variable would be the time it takes

for the reading on the thermocouple meter to reach a

steady value.

There maybe severalother variables that could affect

the result. One might be the starting temperature

of the therrnostat 'probe junction', arıother is the

Appendix A2 F u rth er p ra c ti ca l s k ill s

- - ~ - - - - - - - - - - - - - - - - - - - - '

(fixed) temperature of the second junction of the

therrnocouple.

D a ta c o U ee tio n

The next task is to think about how you are going to

carry the experiment out. Once you have a method in

mind you need to:

• Deseribe the method to be used to vary the

independent variable.

• Deseribe how the independent variable is to be measured.

• Deseribe how the dependent variable is to be measured.

• Deseribe how other variables are to be controlled.

• Describe, with the aid of a clear, labelled diagram, the

arrangement of apparatus for the experiment and theprocedures to be followed.

In the experiment to investigate the response time

of a thermocouple you may decide that the sirnplest

way of varying the independent variable is to have

a water bath and to vary its temperature. When the

temperature has reached the required level, plunge the

'probe junction' into the water.

The procedure would be:

1 Measure the temperature of the water with a

laboratory thermometer.2 The reading from the therrnocouple will be

displayed on a millivoltmeter and the dependent

variable, the time taken for the reading to reach its

steady value, measured with a Stop watch.

3 There are several different variables that need to

be controlled. In practice you will only need to

consider one or two.

• The second junctiorı of the thermocouple should

be kept at a constant temperature by keeping it

in corıtact with melting ice.• The probe junction should be cooled to the same

temperature each time, again by holding it in themelting ice.

Your description of the experiment should be clear

and concise. It sometimes helps to give the description

in the form of bullet points or a numbered list. A

labelled diagram will always clarify your description.

In the experiment deseribed here, the diagram shown

in Figure Al.ıwould be sufficien t.

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th od o f an alys isrequires that you deseribe how the data should be

d in order to reach a conclusion, including details

erived quantities to be cakulated. In our example,

could plot a graph of the temperature of the water(x-axis) against the response time (y-axis).

ow would we use the graph? If the graph gives

raight line through the origin, therı we see that

response time is linear across the temperatlire

e tested.

Figures A2.2a and A2.2b show two possiblesets of results for the therrnocouple experiment.

For each, state how the response time depends

on the temperature.

Time Time

Temperature Temperaturea b

FigureA2.2 Graphs showing how the response time

of a thermocouple might depend on the temperature

being measured.

need to assess the risks of the experiment anderibe precautions that should be taken to keep

s to aminimum.

our example it is sufficient to mention simple

s such as wearing goggles to protect the eyes when

ting liquids, or ensuring that the water bath is

e and cannot be easily knocked over. The use of

s to put the thermocouple junction into the bath

other example.

A d d itio n al d e ta ils

This tests your experience of doing practical work.

Have you had suflicient experience to see things which

will improve the experiment?

Here are some ideas that might be incorporated in

the general description of the experiment:

• Stir the water in the bath so that it all reaches a unitorm

temperature.

• Use an oil bath to give a wider range of temperatures.

• Replace the water bath with different substances at rheir

melting or boiling points so that the temperatures are

more repeatable.

• Check the reading on the therrnocouple volrrneter at

the melting or boiling points of the different substances

before carrying out the main experiment.• Use a digital voltıneter so that it is easy to spot when

the thermocouple reaches the steady temperature.

M ath em atic al an a ly s is o f d ata

In Appendix Al, we saw how to rearrange simple

equations into the form y= mx+ c and how to use

a straight-lirıe graph to hnd the constants m and c.

However, you also need to be able to deal with

quantities related by equations of the form y= ax"

and y = ae", For these, you need to be able to uselogarithms (logs).

In handling data, our aim is usually to process

the data to obtain a straight line graph. Then we

can deduce quantities from the gradient and the

intercepts. Table A2. ishows the graph which should

be plotted in each case, and the quantities which can

be deduced from the graph.

Relationship Graph Gradient Intercept because ...on j-axis

r==« y against x m c

y=axnInyagainst

n InaIny=

Inx nlnx+lna

y=aekx Iny against x k Ina

Iny=kx+

Ina

Table A2.1 Choice ofaxes for straight-line graphs.

IILppendix A2 Fu rth e r p ra c tic a l s kil ls

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D

A rela tionsh ip of th e form v = a x "

Abali falls under graviry in the absence of air

resistance. It falls a distance s in time t.The results

are given in the firs t two columns ofTable Al.2.Notice that the unit for the logarithm is written as

In ( s / m) and not In ( s ) / m or In ( s ) / In (m).

Time tlsDistanee

in (tl s) in (sım)fallen sım

0.20 0.20 -L.61 -L.61

0.40 0.78 -0.92 -0.25

0.60 L.76 -0.51 0.57

0.80 3.14 -0.22 1.14

L.OO 4.90 0.00 L.59

1.20 7.05 0.18 L.95

Table A2.2 Results for Example 1.

8.0

6 . 0

E

- -c~~ 4.0

< L Iuc

2.~o

2.0

o~~--,-----,------r------o 0.80 1.20

Time/s

FigureA2.3 A distance-time graph plotted using the data

in Table A2.2.

A graph of the distance fallen against time gives the

curve shown in Figure Al.3. This, being a curve, tells

us little about the relationship between the variables.

If, however, we suspect that the relationship is of the

Appendix A2 Fu rt he r p ra c tic a l s k ills

- - ~ - - - - - - - - - - - - - - - - - - -

form y= ax", we can test this idea by plotting a graph

of In s against In t (a 'log-log plot'). Table Al.2 shows

the values for I n s and Int,

and the resulting graph isshown in Figure Al.4. (Notice that here we are using

'natural logs', bur we could equally well use logs to

base 10,written as 'lg'.)

In (sım)

< i . s In (t/s )

2.0

1.0

-15 -1 -0·5

-1.0

-2.0

lRgure A2.4 A log-log plot for the data shown in Table A2.2.

From this graph the gradient is equal to the valueof n, the power of t.

. (-1.55-0.5)n = gradıent = -----------

(- 1.50 - 2.55)

= 1.98 ",,2.0

So the equation is of the form s = a t? The intercept

on the y-ax:is is equal to In a, so:

Ina = 1.6

By taking the antilogarithm we ger:

a= 4.95 ms-2"" 5.0 rns?

If we think of the equation for free fall 5= 1 - g t ? - , the

constant a= 1 - g. But g = 9.8 m S-2which is consistent

with the value we get for our constant.

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A re la t ionsh ip o f th e fo rm y= ae"

Acurrent flows from a charged capacitor when it is

connected in a circuit with a resistor. The current

decreases exponentially with time (the same pattemwe see in radioactive decay).

Figure A2.S shows the circuit and Table A2.3

shows typical values of current iand time t from

such an experiment.

Current I/mA Time t/s ln(I/mA)

ıo.OO 0.00 2.303

6.70 0.20 1.902

4.49 0.40 1.5023.01 0.60 1.102

2.02 0.80 0.703

1.35 1.00 0.300

Table A2.3 Results from a capacitor discharge experiment.

c =10IlF

L o =10 m A

R= 20.0 kQ

Figure A2.5 A circuit for investigating the discharge of a

capacitor.

The graph obtained from these results (Figure

A2.6) shows a typical decay curve, but we cannot

be sure that it is exponential. To show that the

curve is of the form 1= 10 e k " we plor Iniagainstt (a 'log-linear plor'). Values of ını are included

in TableA2.3. (Here, we must use logs to base e

rather than to base 10.)

The graph oflnI against t is a straight line (Figure

A2.?), confirming that the decrease in current

followsan exponential pattern. The negative gradient

showing exponential decay, rather than growth.

r ı / m A

5

10

0+------,-----,,-----,------

o 0.4 0.8

Time / s

1.2

FigureA2.6

(

2.0

InI

1.0

o+------,------,---~,------o 0.8

Time/s

1.2

. J

Figure Az.z

The gradient of the graph gives us the value of the

constant k.

. (O- ı. 16)k=gradıent= =-0.S04çl::::-0.SOçl

(2.30 - O)

From the graph, we can also see that the intercept

on the y-axis has the value 2.30 and hence (taking

the inverse log) we have 10= 9.97 = 10.OmA.Hence we can write an equation to represent the

deereasing current as follows:

1= 10.Oe-o .50'

We could use this equation to calculate the current

at any time t.

11lppendix A2 Fu rth e r p ra c tic a l s kills

~ - - - - - - - - - - - - - - - - -

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2 In the expressions that follow x and y are

variables in an experiment. All the other

quantities in the expressions are constants.

In each case, state the graph you would plot to

produce a straight line. Give the gradient of each

line in terms of the constants in the expression.

a y= k X 3/2

b v = c . x 'l

c m= 8x

B i

dy= yoekx

e R - (Y- Y o )- 2

x

3 The period of oscillation T of a small spherical

mass supported by a length i of thread is given

by the expression

T=2n f - fwhere g is the acceleration due to gravity.

Design a laboratory experiment using this

expression to determine the acceleration due

to gravity. You should draw a diagram showingthe arrangement of your equipment. In your

account, you should pay particular attention to:

a the procedure to be followed

b the measurements to be taken

c how to analyse the data to determine g

d any safety precautions that you would take.

T re atm e n t o f u n c erta in tie sAll results should indude an estimate of the absolute

uncertainty. For example, when measuring the time for

a runner to cornplete the 100 m you may express this as

12.1 ± 0.2 s. This can also be expressed as a percentage

uncertainty (see Appendix Al); the percentage

uncertairıry is equal to 0.2 x 100% = 1.65%. so we12.1

write the value as 12.1 s ± ı. 7%.

- Appendix A2 Fu rt he r p ra c ti ca l s kil ls

- - - - . - - - - - - - - - - - - ~

C o m b in in g u n c e r ta in t ie s

When quantities are added or subtracted, their

absolute uncertainties are added. A simple example

is that when measuring the length of a stick using a

millimetre scale there is likely to be an uncertainty

of 0.5 mm at both ends, giying a total uncerta irıtyof 1.0 mm.

To combine uncertainties when quantities are

rnultiplied or divided is a little more complex. You

can only add or subtract quantities if theyare in the

same units; you cannot subtract current from voltage!

However, you can multiply or divide them. To find

the combined uncertainry in a cas e like this we add

the percentage uncertainties.

Wo rk ed e xample

1 The poterıtial difference across aresistor is

measured as 6.0 ± 0.2 V, whilst the current is

measured as 2.4 ± O . 1A.

Calculate the resistance of the resistor and the

absolute uncertainty in its measurement.

Step 1 Find the percentage uncertainty in each

of the quantities:

percentage uncertainty in p.d. = 0.2 x 100%6.0

=3.3%

percentage uncertainty in current

= 0.1 x 100%=4.2%2.4

Step 2 Add the percentage uncertainties

= (3.3 + 4.2)% = 7.5%

Step 3 Calculate the resistance value and findthe absolute uncertairıry

R= V = 6.0 =2.50i 2.4

7.5% of2.5=0.1875===0.20

The resistarıce of the resistor is 2.5 ± 0.2 O.

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en you calculate the uncertainty in the square

a quantity rherı, since this is an example of

ltiplicatiorı you should double the percentage

. For example if A =2.0 + 0.2 cm

n A has a percentage uncertainty of 10% so

=4.0 m ' ± 20%; or using the absolute uncertai nty=4.0 ±0.8 crn".

4 You measure the following quantities:

A= 1.0m±0.4m B=2.0m±0.2m

C=2.0ms-l±O.5mçl D=0.20s±0.01s.

Calculate the answer and the uncertainry forthe following expressions. You may express

your uncertainty either as an absolute value or

as a percentage.

aA+B

bB-A

c CxD

d l LD

e A2

f 2xA

g the square root of (A xB)

5 A rifie bullet is photographed in Hight using

rwo Hashes of light separated by a time interval

of 1.00 ms ± 0.02 ms. The nrst image of the

bullet on the photograph appears to be at a

position of 22.5 ±0.5 cm on a scale underneath

the Highr path. The positiorı of the second

image is 37.5 ±0.7 cm on the same scale.

Find the speed of the bullet and the absolute

uncertainty.

ta in tie s a n d lo g ar ith m s

hen a log graph is used and we need to include

r bars (see uncertainties and graphs) we must nnd

logarithm of the measured value and either the

ithm of the largest or smallest possible value. The

inty will be the difference between the two.

Wo rk ed e xample

2 The resistance of aresistor is given as 47 ±5o.The value of In (R i O) is to be plotred on a

graph. Calculate the value and uncertainty in

In (RIO).

Step ıCalculate the logarithm of the given value:

In (RIO) =ln47 =3.85

Step 2 Calculate the logarithm of the

maximum value:

maximum value =47 + 5=520

In 52 =3.95

Step 3 The uncertainty= 3.95 - 3.85 =O.10

Thus In (RIO) =3.85 ± O.10

U n c e rta in tie s a n d g ra p h s

We can use error bars to show uncertainties on

graphs. Table A2.4 shows results for an experiment on

stretching a spring.

Load/NLength of

Extension icmspring/cm

O 12.4±0.2 0.0

1.0 14.0±0.2 1.6±0.4

2.0 15.8±0.2 3.4±0.4

3.0 17.6±0.2 5.2±0.4

4.0 18.8±0.2 6.4±0.4

5.0 20.4±0.2 8.0±0.4

Table A2.4 Results from an experiment on stretching a spring.

When plotting the graph the points are plotted as usual,

and then theyare extended to show the extreme values,

as shown in Figure A2.8. Then the best nt line is drawn.

To estimate the error in the gradient we draw not

only the best nt line but also the 'worst acceptable

line'. This line is the worst line which goes through all

the error bars, and is shown in Figure A2.9.

EDppendix A2 Fu rth er p ra ctic al s kills

~ - - - - - - - - - - - - - -

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- : : , - . : , : ~ ~:~, ' - -

- -

Extension i cm

8.0

6.0

2. 0

o~------r------.------.-------

o 2 64

Load i N

FigureA2.8 A graph representing the data in Table A2.4, with

error bars and a line of best fit drawn.

Extension i cm

8.0

6.0

2. 0

o~-----,r------.------.-------

o 4 6Load i N

FigureA2.9 The same graph as in Figure A2.8, with a 'worst

acceptable' line drawn (dashed).

The gradients for both best fit and worst fir lines are

calculated and the error is the difference in their gradients:

D

error = (gradient of best fi t line) - (gradient of

worst acceptable line)

Appendix A2 Further pract ica l skills

~ - - - - - - - - - - - - - - - - - - - - '

In our experiment, the gradients are:

(8 . 2 - O Jline ofbest fit: gradient = cmN-I

5.0 - °= 1.58 cm N-I", 1.6 N cm N-i

(8.4-0J

line of worst fi t: gradient = cm N-i5.0 - °

= 1.4cmN-I

So the uncertain ty in the gradient = ı.6 - ı.7=±O.l cm N:'

The gradient is therefore: 1.6 ± O.1cm N-I.

6 Suggest why there are no uncertainties

induded in the measurements of the load.

7 it İs suggested that R and r are related by the

equation:

R = al' where a and b are constants.

a A graph is plotted with In R on the y-axis and

In r on the x-axis. Express the gradient and

y-intercept in terrns of a and b.

b Values of R and r are given in Table A2.5.

r ı mm RıO.

2.0±0.1 175.0

3.0±0.1 77.8

4.0±0.1 43.8

5.0±0.1 28.0

6.0±0.1 19.4

Table A2.5 Results for Test yourself Q 7.

Calculate and record values of In (R iQ) and

In (rı mm) in the table and indude the absolute

uncertainties in In (rı mm).

c Plot a graph ofln (RI Q) and In (rı mm).

Indude error bars for In (rı mm).

cont inued ...~ .

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d Draw the line of best nt and a worst

acceptable straight line on your graph.

e Determine the gradient of the line of best nt.

Include the uncertainty in your answer.

f Using your answer to e, determine the value

of b .

g Determine the value of a and its uncertainty.

c lu s io ns a nd e va lu atio nre s u ltsthe previous experiment we can conclude that

extension/load for the spring in this example

.6 ± 0.2 cm N-I. If a hypothesis is rnade that the

extensiorı is proportional to the load therı there

is enough evidence here for the conclusion to be

supported, as a straight line can be drawn from the

origin through all the error bars. If this is not possible

therı the hypothesis is not validated.

Now, suppose that the hypothesis is that the springobeys Hooke's lawand stretches by 5.8 cm when a

load of 3.5 N is applied. The firsr part is validated for

the reasons given above. However, an extension of

5.0 cm for a load of 2.5 N gives a value of 2.0 cm N-I

for the gradient. This is clearly outside the range

allowed for by the uncertainty in our measurements,

and therefore the hypothesis is not supported .