6B Revision Notes Extra

8/6/2019 6B Revision Notes Extra

1/15

002 Oscillating Spring

Apparatus:Spring and cork, tall retort stand/clamp/boss, set of 100g masses with holder, mirror, long pin,

small piece of plasticine,stop watch, plumb line, one metre rule, card with line.

Procedure:1. Hang the spring vertically from a stand and attach the optical pin pointer onto the end of the spring

using a small piece of plasticine. Clamp the metre ruler vertically (check with the plumb line) next to

the spring.

2. Note the initial level (A/m) of the optical pin against the ruler, use the mirror to check that your eye is

at the same level as the pin.

3. Hang a mass, m = 200g(100g plus the mass holder) and then note the new pin level (B/m) and socalculate the spring extension, x, where x = B - A

4. Now displace this mass from its equilibrium position and release. Measure the time for N oscillations

(N should be at least 5). Use the line on the card, placed at the mid-point of the oscillations, when

counting oscillations. Repeat and so calculate the average time for N oscillations.

Finally calculate the period for one oscillation, T/s and also T2

/ s2.

5. Repeat stages 2 to 4 for four other values ofm, (maximum 700g).

6. Tabulate ALL of your measurements & results.

7. Plot the following graphs:

(a) mass, m/ kg against extension, x / m. (b) T2

/ s2

against m /kg.

Both graphs should be straight lines.

8. The extension, x caused by a mass, m of weight, mg are related by: mg = kx (equation 1)The period of oscillation of a mass hung from a spring is given by: T = 2 (m/k) (equation 2)(where, in both cases, k = the spring constant, = load required to cause an extension x of one metre)

When squared the second equation becomes: T2

= 42 m/kThe gradient of your first graph, G1 is equal to k / g.

It can be shown that the gradient of the second graph, G2 is given by: G2 = 42 / (g G1 )Hence: g = 42 / (G1 G2)Use your graph gradients, and the above relationship, to find a value for the gravitational field

strength, g and for the spring constant, k.

9. (a) How, if at all, would your graphs and results be different if you were to perform this experiment

on the moon?

(b) Show how the expression, g = 42 / (G1 G2) is obtained starting from equations 1 & 2.

spring

card with line

long pin

plasticine

metre

rule

mirror


2/15

004 Oscillation of a simple pendulum

Apparatus:Cotton thread, small pendulum bob, protractor, split cork, metre ruler, stopwatch, piece of card with a drawn

line, retort stand/clamp/boss.

Diagram:

Procedure:1. Set up a small simple pendulum, as shown in the diagram.

2. Keeping the angle of swing, less than 10o, for at least 8 lengths of the pendulum, L between20cm and 100cm, measure the period of swing, T (see 'Timing Swings' below).

3. Plot a graph ofT2

againstL.

4. For a small angle ; T = 2(L/ g) where g is the acceleration due to gravity.The gradient of your graph is equal to 42 / g.Measure your gradient and use it to calculate a value ofg.

5. Estimate the uncertainty in your result.

6. Explain why the gradient of your graph is equal to 42 / g.7. For one of the lengths you used in stage 2 investigate how (if at all) the period of oscillation,

T varies if the initial angle of oscillation is increased to at least 90o.

Timing Swings:

At the end of a swing the bob is stationary for a brief instant before swinging back again. It is therefore

difficult to identify the precise instant of the end of a swing since the bob is far out for a long time. At the

centre of the swing the bob is moving rapidly, and the instant that it passes the central point is more

precisely defined. Technically timing the swings from the central position is better. It is common to use a

mark to assist in identifying the instant when the bob passes the centre position (e.g. a line on a piece of

card). To find the time of one swing you should time a number of swings so that the total time taken is at

least 30 seconds and you should repeat this measurement at least once more.

L

split cork

bob

protractor


3/15

137 Measurement of capacitance using current-time graphs

Apparatus:Two electrolytic capacitors (470F & 1000F), 100k resistor, 2 x digital multimeters (one of these should be set on 200A DC

the other on 20V DC), Push-to-make switch, Power supply set on 9V, Stop watch and wires.

Diagram:

Procedure:

1. Set up the circuit above using the 1000F capacitor and the 100k resistorBe careful to connect the capacitor the right way around.

2. Once you switch on, the ammeter should register a current that starts to fall from an initial value

of about 100A (0.100mA) to nearly zero over a period a few minutes. If you close the switch,

momentarily, this current returns to its initial value of about 100A before decaying again. The voltmeter will read

initially zero and then rise up to about 10V in the same time. Both current and voltage changes are

exponential.

3. Take current readings over a five minute period (e.g. every 30 seconds) in order to be able to draw a graph of current (in

micro-amperes) against time (in seconds).

At the five minute point note the voltage, V of the capacitor.

4. Repeat with:

(a) the 470F capacitor and

(b) the 1000F capacitor connected in series with the 470F capacitor.

Note: With the combination, the voltmeter should be connected to read the total voltage of the capacitors.

5. Tabulate all your results.

6. ON THE SAME AXES plot graphs of current (in micro-amperes) against time (in seconds)

for the three above current variations.

7. In each case estimate (nearest cm2 will do) the area between each curve and the time axis.

This area will equal the charge Q (in micro-coulombs) transferred to the capacitor(s) after fiveminutes.

(Example of area calculation: If your axes scales are 1cm = 10A & 1cm = 30s then each

square cm on your graph will represent a charge of 10A x 30s = 300C)

8. Use this charge value and the capacitor voltage to calculate the capacitance of the capacitor or capacitor

combination using the formula:

C = Q / V9. Compare your calculated values with the marked values and with those expected by using the formulae for

capacitors in series and parallel (see below).

For two capacitors (C1 & C2), connected in series, the total effective capacitance, CTis given by::

connected in series: 1 = 1 + 1CT C1 C2

A

Supply

set at 9V

multimeter on 0.2mA (200A) range

push to

make

switch

100k resistor

X

Y

++

V


4/15

138 The Force On A Current Carrying Conductor

Apparatus:

Top pan electric balance, shaped length of copper wire, clamp-stand etc.,

digital multimeter set on 10A dc, rheostat, 6V PSU, wires, push-to-make switch, protractor, ruler.

Diagrams:

(a) side view: (b) from above:

Set up procedure:

Clamp the shaped piece of copper wire so that its lower horizontal section lies between the poles of a set ofmagnadur magnets on the pan of the top-pan balance as shown in the diagrams above. Connect the copper

wire to a circuit containing the push-to-make switch, rheostat, ammeter and 6V power supply. You will use

the rheostat to control the current, I through the copper wire.

Investigations:According to theory, the force, F exerted on a straight conductor of length, L carrying a current, I is given

by:

F = B x I x L x sinwhere B = magnetic flux density (a measure of the strength of the magnetic field)

and = the angle between the conductor and the field direction (90o in the above diagrams).With your apparatus investigate how well the above formula describes how the force exerted on the wire

varies with:

(a) the current flowing, I &

(b) the sine of the angle between the horizontal wire and the field direction.

electric top panbalance

conductor

0.23 g

T

L


5/15

306 Absorption of Beta Radiation

Apparatus:GM-tube, counter and power supply, aluminium absorbers, special source and absorber holder, micrometer,

beta source (strontium 90), local rules sheet, spill tray.

Procedure:

1. Read the sheet Local rules for the use of sources of ionising radiations by students

2. Connect the GM-tube to the digital counter/power supply and switch on.

3. Without the radioactive source present find the background count B (in counts per second) by noting

the reading of the counter over a 100 second period.

4. Set up the apparatus as in the diagram (with the thinnest any aluminium absorber in place) with the

VERY FRAGILE & EXPENSIVE front of the GM tube at a known constant distance (e.g. 8cm)

from the source. All of the above should be placed on the spill tray.

5. Record the count received by the counter over a 100 seconds period.

Calculate the count rate (per second).

Subtract the background count rate B and so obtain a corrected count rate with the absorber C (s1).

6. Use the micrometer to find the average thickness, x (mm) of the aluminium sheet.

7. Repeat stages 4 to 6 for other thicknesses of aluminium.Note: (a) You can use more than one sheet at a time.

(b) Do not use any of the lead absorbers.

8. Plot a graph of corrected count rate C/s1 against aluminium thickness x (mm).

9. Use your graph to determine the thickness of aluminium required to reduce the number of beta

particles passing by half.

This is known as the half-thickness x1/2.

to counter

beta source aluminium absorber(s)

GM tube at a

constant distance

from the source


6/15


7/15

Experiment: Measuring the half-life of protactinium-234.

Record a number of count rates with no source present and obtain an average background count. Shake the protactinium generator to transfer the protactinium compound from the lower water-based layer to the upper organic layer. When the layers re-establish, place the GM tube alongside the top layer (below). Record the count rate at intervals of 10s for 5mins. Plot a graph of corrected count rate against time. From the graph determine how long it takes for the count rate at any given time to halve its value.

The GM tube monitors the decay of the protactinium in the top layer.

The half life of this isotope is about one min.
http://www.thestudentroom.co.uk/wiki/File:Halflife_graph_2.jpghttp://www.thestudentroom.co.uk/wiki/File:Half_life_experienment.jpg


8/15

CCD - Charged Couple Devices

Advantages Over Film:

Higher quantum efficency ~70% (number of bits of info recorded / number of photons x 100) More linear response (output is more directly proportional to light intensity received) No need to replace Can be used repeatedly Faster OperationDisadvantages

Smaller field of view Need to be kept cool (about 120K) Not portable Resolution is not as goodMade from slices of silicon that store electons from photons and build an image (in pixels) from this. Pixels are from 5-50 micrometers, and

pixels where larger quantities of light land (higher light intensity) have more electons.

A voltage is applies to pixels so that the computer can "read" the image.


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