In this experiment, you will investigate the motion of a pendulum. You have been provided with a cylinder and a pendulum. (a) Use adhesive putty to attach the string to the cylinder as shown in Fig. 1.1. P is the point at which the string is attached to the cylinder. The distance between P and the centre of the bob is $L$. Adjust the adhesive putty and string so that $L$ is approximately $45\text{ cm}$. Measure and record $L$. $L$ = \_\_\_\_\_\_ (b) Set up the apparatus as…

(a) Example: L = 0.450 m · (b) Example: T = 1.36 s · (c) See working (student-dependent table with L, T, T^3, L^2) · (d)(i) Graph of T^3 vs L^2 (plotted) · (d)(ii) Straight best-fit line drawn · (d)(iii) Example: gradient = 12.0 s^3 m^-2, y-intercept = 0.10 s^3 · (e) Example: E = 12.0 s^3 m^-2, F = 0.10 s^3

Physics 9702/33 — May/June 2023

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will investigate the motion of a pendulum.

You have been provided with a cylinder and a pendulum.

(a)

Use adhesive putty to attach the string to the cylinder as shown in Fig. 1.1.

Fig. 1.1

P is the point at which the string is attached to the cylinder.
The distance between P and the centre of the bob is $L$ .
Adjust the adhesive putty and string so that $L$ is approximately $45\text{ cm}$ .
Measure and record $L$ .

$L$ = ______

DifficultyEasy

Worked solution

Measure the distance from point $P$ to the centre of the bob using a metre rule (to the nearest $1\ \text{mm}$ or $0.1\ \text{cm}$ ).

A typical reading:

L = 0.450\ \text{m}\ \left(= 45.0\ \text{cm}\right)

Final answer

Example: L = 0.450 m

Detailed explanation

Background Concept

Length measurements in practical physics must be (i) taken between the correct reference points and (ii) recorded with appropriate precision. A metre rule typically reads to $1\ \text{mm}$ , so the recorded value should reflect that resolution (e.g. $45.0\ \text{cm}$ rather than $45\ \text{cm}$ ).

Understanding the Question

You are told that $L$ is the distance between the attachment point $P$ on the cylinder and the centre of the bob. The task is to adjust the set-up so that $L$ is about $45\ \text{cm}$ , then measure and record $L$ .

Approach

Identify the two endpoints: point $P$ on the cylinder and the centre of the bob.
Use a metre rule placed alongside the string to measure the distance.
Record the value with a unit and sensible precision.

Step-by-Step Reasoning

First adjust the putty/string position on the cylinder until the length looks close to $45\ \text{cm}$ .
Place a metre rule as close as possible to the string (reduce parallax by looking square-on).
Read the distance from the mark corresponding to $P$ (or a fixed reference aligned with $P$ ) to the centre of the bob.
Record $L$ to the metre rule precision (commonly nearest $0.1\ \text{cm}$ or $1\ \text{mm}$ ).

Key Takeaways

Always measure between the points defined in the question.
Match your recorded precision to the instrument resolution.
Include units.

Common Mistakes

Measuring to the bottom/top of the bob instead of its centre.
Measuring from the edge of the cylinder rather than point $P$ .
Writing no unit, or writing an over-precise value (e.g. $45.023\ \text{cm}$ with a metre rule).

Things to Be Careful About

Parallax error: ensure your eye is directly above the scale marking.
If the string is not perfectly straight, measure along the string line (keep the rule aligned with it as well as possible).
Be consistent with units (cm vs m) for later calculations of $L^2$ .

Techniques used

identify the correct length to measure between two specified pointsuse a metre rule to take a length reading with appropriate precisionrecord a measured value with a unit and sensible significant figures

(b)

Set up the apparatus as shown in Fig. 1.2.

Fig. 1.2

Move the bob a short distance away from the stand, as shown in Fig. 1.2.
Release the bob. The bob will oscillate.
Determine the period $T$ of the oscillations of the bob.

$T$ = ______

DifficultyMedium-Easy

Worked solution

Time $N$ oscillations (e.g. $N=10$ ) with a stopwatch and divide by $N$ .

Example:

10T = 13.6\ \text{s}\ \Rightarrow\ T = \frac{13.6}{10} = 1.36\ \text{s}

(Repeat and take mean.)

Final answer

Example: T = 1.36 s

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. Stopwatch reaction time makes timing a single oscillation unreliable, so you reduce the fractional uncertainty by timing many oscillations:

T = \frac{t}{N}

where $t$ is the total time for $N$ oscillations.

Understanding the Question

You displace the bob slightly away from the stand and release it. The bob oscillates. You must determine the period $T$ .

Approach

Choose a suitable number of oscillations (commonly $N=10$ or more).
Start timing as the bob passes a fixed reference point.
Count $N$ complete oscillations and stop timing when it returns to the same reference point for the $N$ th time.
Calculate $T=t/N$ and repeat to improve reliability.

Step-by-Step Reasoning

Displace the bob by a small angle (small amplitude helps keep the period more constant).
Select a reference position (often the equilibrium position) and always start/stop timing at that same position and direction.
Count oscillations carefully: one oscillation means returning to the same position moving in the same direction.
If you measure $10T = 13.6\ \text{s}$ then:

T = \frac{13.6\ \text{s}}{10} = 1.36\ \text{s}

Repeat the timing (e.g. take two or three values of $10T$ ) and calculate a mean period.

Key Takeaways

Use $N\geq 10$ to reduce percentage uncertainty.
Use a consistent reference point and direction.
Repeat readings and average.

Common Mistakes

Timing one oscillation only (large reaction-time percentage uncertainty).
Miscounting oscillations (counting half-oscillations).
Starting/stopping at different points in the motion.

Things to Be Careful About

Keep the amplitude small and similar for each run.
Ensure the string does not snag on the cylinder/stand during motion.
Quote $T$ with sensible precision (typically to $0.01\ \text{s}$ if timing many oscillations).

Techniques used

time multiple oscillations and divide to obtain the periodrepeat timing measurements and take a meanuse consistent start/stop reference points for oscillations

(c)

Change $L$ by attaching a different point on the string to the cylinder and determine $T$ . Repeat until you have six sets of values of $L$ and $T$ .

Record your results in a table. Include values of $T^3$ and $L^2$ in your table.

DifficultyMedium

Worked solution

Record at least six sets of $L$ and $T$ and calculate $T^3$ and $L^2$ .

Example of a suitable table format (values shown are representative):

$L\ /\ \text{m}$	$T\ /\ \text{s}$	$T^3\ /\ \text{s}^3$	$L^2\ /\ \text{m}^2$
0.350	1.16	1.56	0.123
0.400	1.26	2.00	0.160
0.450	1.36	2.52	0.203
0.500	1.46	3.11	0.250
0.550	1.55	3.72	0.303
0.600	1.64	4.41	0.360

Final answer

See working (student-dependent table with L, T, T^3, L^2)

Detailed explanation

Background Concept

A good practical data table must:

include all raw and derived quantities in one clear table,
have column headings with quantity and unit (e.g. $T/\text{s}$ ),
use consistent decimal places/significant figures within each column,
include enough data points across a suitable range to reveal a trend.

Derived quantities here are $T^3$ and $L^2$ , so you must calculate:

T^3 = T \times T \times T, \qquad L^2 = L \times L

Understanding the Question

You must change $L$ (by attaching a different point of the string to the cylinder) and measure the corresponding period $T$ . You need six pairs $(L, T)$ , then you must add two extra calculated columns $T^3$ and $L^2$ .

Approach

Choose a suitable range of $L$ values (not all close together), and take six readings.
For each $L$ , measure $T$ using the “time $N$ oscillations then divide by $N$ ” method.
Enter $L$ and $T$ in a table with units.
Calculate $T^3$ and $L^2$ for each row, recording them to sensible significant figures.

Step-by-Step Reasoning

Decide on six values of $L$ spanning a reasonable interval (e.g. roughly $0.35$ to $0.60\ \text{m}$ ). A wider range generally produces a clearer graph and a more reliable gradient.
For each length:
- measure and record $L$ (same instrument/precision each time),
- time $N$ oscillations (e.g. $10$ ) at least twice and take the mean period,
- record $T$ .
Compute $T^3$ from your recorded $T$ . For example if $T=1.36\ \text{s}$ :

T^3 = (1.36)^3 = 2.52\ \text{s}^3

Compute $L^2$ from your recorded $L$ . For example if $L=0.450\ \text{m}$ :

L^2 = (0.450)^2 = 0.203\ \text{m}^2

Keep the number of significant figures in calculated columns consistent and not exceeding that justified by the raw data.

Key Takeaways

Six readings with a good spread improve the reliability of the graph.
Correct table headings must include units.
Derived columns must be calculated and recorded properly.

Common Mistakes

Missing units in headings (e.g. writing just $T^3$ ).
Inconsistent precision down a column (e.g. mixing $0.4$ and $0.450$ in the same $L$ column).
Rounding too early and losing accuracy in $T^3$ .
Using fewer than six sets of readings.

Things to Be Careful About

Decide early whether you will use $L$ in $\text{m}$ or $\text{cm}$ and be consistent (this affects the units of $L^2$ and the gradient).
When cubing $T$ , keep extra digits in your calculator then round at the end.
If repeats of $T$ vary significantly, you should repeat again: large scatter will reduce the quality of the best-fit line.

Techniques used

vary the independent variable across a suitable range with multiple readingsrecord raw measurements with units and consistent precisioncalculate derived quantities and record them to appropriate significant figuresorganise results into a single table with correct headings

(d)

(i)

Plot a graph of $T^3$ on the $y$ -axis against $L^2$ on the $x$ -axis.

DifficultyMedium

Worked solution

Plot $T^3$ on the $y$ -axis against $L^2$ on the $x$ -axis.

Label axes with units (e.g. $T^3/\text{s}^3$ and $L^2/\text{m}^2$ ), use suitable scales (at least half the grid), and plot all six points accurately.

Final answer

Graph of T^3 vs L^2 (plotted)

Detailed explanation

Background Concept

A graph is used to display how one variable depends on another. For credit in Cambridge practical papers, you must:

put the correct variable on each axis,
label each axis with quantity and unit,
use a sensible scale (not cramped, not overly coarse),
plot points accurately with small, neat crosses.

Understanding the Question

You have calculated $T^3$ and $L^2$ in your table. You are asked to plot a graph of $T^3$ (vertical axis) against $L^2$ (horizontal axis). In other words, $T^3$ is the dependent variable and $L^2$ is the independent variable.

Approach

Decide the min and max values of $L^2$ and $T^3$ from your table.
Choose axis scales so that the plotted region uses most of the available grid.
Label axes correctly and plot each pair $(L^2, T^3)$ .

Step-by-Step Reasoning

From your table, find the range of $L^2$ and the range of $T^3$ .
On the horizontal axis write something like $L^2/\text{m}^2$ (or $L^2/\text{cm}^2$ if you used cm consistently).
On the vertical axis write $T^3/\text{s}^3$ .
Choose a scale such that:
- each large square corresponds to a convenient increment (e.g. $0.02\ \text{m}^2$ or $0.5\ \text{s}^3$ ),
- your smallest and largest data points are well within the grid.
Plot all points using small crosses; accuracy matters (use a ruler to help read coordinates).

Key Takeaways

Correct axes and units are essential.
Good scale choice makes gradient/intercept more accurate.
Plot all points clearly and accurately.

Common Mistakes

Swapping axes (plotting $L^2$ on $y$ ).
Missing units on one or both axes.
Using an awkward scale (e.g. 3 squares = 0.01) that increases reading errors.
Plotting blobs instead of fine crosses.

Things to Be Careful About

If you used $L$ in $\text{cm}$ in the table, then $L^2$ is in $\text{cm}^2$ ; do not label it $\text{m}^2$ .
Do not force the axes to start at zero if it wastes most of the grid; start at a convenient value that still shows the trend clearly.

Techniques used

choose appropriate axes and label with quantities and unitsselect a scale that uses at least half the graph grid in each directionplot points accurately from a results table

(ii)

Draw the straight line of best fit.

DifficultyMedium-Easy

Worked solution

Draw a single straight line of best fit (not dot-to-dot), balanced so that the points are roughly evenly distributed about the line.

Final answer

Straight best-fit line drawn

Detailed explanation

Background Concept

If the data are expected to follow a linear relationship, you represent the trend with a straight line of best fit. The best-fit line is not necessarily drawn through every point; it should reflect the overall trend considering experimental scatter.

Understanding the Question

After plotting $T^3$ against $L^2$ , you must draw the straight line of best fit on the graph.

Approach

Use a ruler.
Position the ruler so that the line follows the trend and leaves roughly equal scatter above and below.
Draw one clear straight line.

Step-by-Step Reasoning

Look at the plotted points and identify whether they show an approximately linear trend.
Place a ruler so the line passes through the middle of the scatter.
Aim for a “balanced” line: the number (and size) of deviations above the line should be similar to those below.
Draw the line across the full range of your data (not just between two central points).

Key Takeaways

A best-fit line represents the trend, not perfect passage through every point.
A longer line across the data range helps later gradient/intercept determination.

Common Mistakes

Joining points dot-to-dot.
Drawing a line that is forced through an outlier.
Drawing a line only across a short section of the graph.

Things to Be Careful About

Use a sharp pencil so the line is thin (thick lines reduce reading accuracy).
Do not assume the line must pass through the origin unless the data show that clearly and the relationship requires it.

Techniques used

judge the overall trend of plotted pointsdraw a balanced straight line of best fit through the scatteravoid joining dot-to-dot when a best-fit line is required

(iii)

Determine the gradient and $y$ -intercept of this line.

gradient = ______

$y$ -intercept = ______

DifficultyMedium

Worked solution

Using two widely separated points on the best-fit line (example):

(L^2,\ T^3) = (0.12\ \text{m}^2,\ 1.56\ \text{s}^3)\ \text{and}\ (0.36\ \text{m}^2,\ 4.44\ \text{s}^3)

\text{gradient} = \frac{\Delta T^3}{\Delta L^2} = \frac{4.44-1.56}{0.36-0.12} = \frac{2.88}{0.24} = 12.0\ \text{s}^3\ \text{m}^{-2}

Read at $L^2=0$ :

y\text{-intercept} = 0.10\ \text{s}^3

Final answer

Example: gradient = 12.0 s^3 m^-2, y-intercept = 0.10 s^3

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ , the gradient and intercept come from:

y = mx + c

Gradient: $m = \Delta y/\Delta x$ (use two points on the best-fit line, far apart, to reduce percentage reading error).
$y$ -intercept: $c$ is the value of $y$ when $x=0$ .

Units:

[m] = \frac{\text{units of }y}{\text{units of }x}

Understanding the Question

Your graph has $y=T^3$ and $x=L^2$ . You must find the gradient of the best-fit line and the $y$ -intercept.

Approach

Choose two points on the drawn best-fit line (not necessarily data points), widely separated.
Calculate $\Delta T^3$ and $\Delta L^2$ and divide to get the gradient.
Extend/inspect the best-fit line to find where it crosses the $y$ -axis ( $L^2=0$ ) to obtain the intercept.

Step-by-Step Reasoning

Pick two clear points where the best-fit line crosses grid intersections to improve reading accuracy.
Suppose the points are $(0.12\ \text{m}^2,\ 1.56\ \text{s}^3)$ and $(0.36\ \text{m}^2,\ 4.44\ \text{s}^3)$ .
Compute changes:

\Delta T^3 = 4.44 - 1.56 = 2.88\ \text{s}^3

\Delta L^2 = 0.36 - 0.12 = 0.24\ \text{m}^2

Then:

\text{gradient} = \frac{2.88}{0.24} = 12.0\ \text{s}^3\ \text{m}^{-2}

To find the $y$ -intercept, look where the line meets the $T^3$ axis at $L^2=0$ . Read off that value (with unit $\text{s}^3$ ).

Key Takeaways

Use two points on the line, far apart.
Gradient is always $\Delta y/\Delta x$ .
Intercept is the $y$ value at $x=0$ .

Common Mistakes

Using two plotted data points instead of points on the best-fit line.
Using a small triangle (large percentage uncertainty).
Calculating $\Delta x/\Delta y$ by mistake.
Forgetting units for gradient or intercept.

Things to Be Careful About

Ensure you read values from the line with the correct axis scale.
If your $x$ -axis does not start at zero, you may need to extend the line to reach $x=0$ to estimate the intercept.
Keep enough significant figures (typically 2–3) consistent with graph-reading precision.

Techniques used

use two well-separated points on the best-fit line to calculate a gradientcalculate the gradient as $\Delta y / \Delta x$ with unitsread the y-intercept from the graph at $x=0$

(e)

It is suggested that the quantities $T$ and $L$ are related by the equation

T^3 = EL^2 + F

where $E$ and $F$ are constants.

Using your answers in (d)(iii), determine the values of $E$ and $F$ . Give appropriate units.

$E$ = ______

$F$ = ______

DifficultyMedium-Easy

Worked solution

Compare

T^3 = EL^2 + F

with $y=mx+c$ for a graph of $T^3$ (y-axis) against $L^2$ (x-axis).

E = \text{gradient}, \qquad F = y\text{-intercept}

Using (d)(iii) (example values):

E = 12.0\ \text{s}^3\ \text{m}^{-2}

F = 0.10\ \text{s}^3

Final answer

Example: E = 12.0 s^3 m^-2, F = 0.10 s^3

Detailed explanation

Background Concept

When you plot $y$ against $x$ and get a straight line, the relationship is

y = mx + c

$m$ is the gradient (slope).
$c$ is the $y$ -intercept.

You can identify constants in an experimental equation by matching it term-by-term to $y=mx+c$ .

Units come from dimensional reasoning:

[m] = \frac{[y]}{[x]}, \qquad [c] = [y]

Understanding the Question

You are given the suggested relationship

T^3 = EL^2 + F

and you have already found the gradient and intercept from a graph of $T^3$ (vertical) against $L^2$ (horizontal). You must use those to determine $E$ and $F$ , including appropriate units.

Approach

Identify $T^3$ as $y$ and $L^2$ as $x$ .
Match $T^3 = EL^2 + F$ to $y = mx + c$ .
Set $E=m$ and $F=c$ .
Work out units from the axes units.

Step-by-Step Reasoning

On your graph:
- $y$ -axis quantity is $T^3$ with unit $\text{s}^3$ .
- $x$ -axis quantity is $L^2$ with unit $\text{m}^2$ (or $\text{cm}^2$ if you used cm).
Comparing:

T^3 = EL^2 + F \quad \leftrightarrow \quad y = mx + c

gives:

E = m, \qquad F = c

Units:

[E] = \frac{\text{s}^3}{\text{m}^2} = \text{s}^3\ \text{m}^{-2}

and

[F] = \text{s}^3

Substitute your measured gradient and intercept values directly.

Key Takeaways

Straight-line identification: gradient corresponds to the coefficient of $x$ , intercept is the constant term.
Units of gradient are always (units of $y$ )/(units of $x$ ).

Common Mistakes

Swapping $E$ and $F$ .
Giving $E$ the wrong units by forgetting the squared length on the $x$ -axis.
Quoting $F$ with length units (it should match $T^3$ units).

Things to Be Careful About

If you used $L$ in cm, then $L^2$ is in $\text{cm}^2$ and $E$ would be in $\text{s}^3\ \text{cm}^{-2}$ .
Do not round excessively; use a sensible number of significant figures consistent with graph-reading uncertainty.

Techniques used

match an experimental straight-line graph to the form y = mx + cidentify constants from the gradient and interceptdeduce correct units for constants from the plotted quantities

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