Physics 9702/33 — February/March 2020

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 oscillations of a rod.

(a)

• Assemble the apparatus as shown in Fig. 1.1.

• Adjust the apparatus until the two springs are approximately $15\text{ cm}$ apart. Each spring should be vertical and the same distance from the middle of the rod. The rod should be parallel to the bench.

Fig. 1.1

• The distance between the two springs where they support the rod is $x$ , as shown in Fig. 1.1. Measure and record $x$ .

$x$ = ______ $\text{cm}$

DifficultyEasy

Worked solution

Answer

Measure the separation $x$ between the two spring support points using a ruler and record to the nearest $1\ \text{mm}$ (e.g. $x = 15.0\ \text{cm}$ ).

Final answer

x ≈ 15.0 cm (example)

Detailed explanation

Background Concept

In practical work, a length measurement is only valid if (i) you measure the correct physical separation, (ii) you avoid parallax (eye not perpendicular to scale), and (iii) you record the reading to a precision consistent with the instrument (e.g. metre rule: typically to the nearest $1\ \text{mm} = 0.1\ \text{cm}$ ).

Understanding the Question

You have a rod supported by two vertical springs. The variable $x$ is the distance between the two springs at the points where they support the rod (i.e. the separation of the support points along the rod). You must measure and record $x$ in $\text{cm}$ .

Approach

Identify the two support points on the rod directly above each spring.
Use a ruler/metre rule aligned along the rod to measure the separation.
Read the scale with your eye directly above the mark to avoid parallax.
Record the value with unit and suitable precision.

Step-by-Step Reasoning

Place the ruler so that its edge lies parallel to the rod, and its scale runs along the line joining the two support points.
If possible, line up one support point with a clear scale mark (e.g. $0.0\ \text{cm}$ ). If not, take two readings (one at each support) and subtract.
Read to the nearest $0.1\ \text{cm}$ for a mm-scale ruler.
Record as, for example, $x = 15.0\ \text{cm}$ .

Key Takeaways

Measure between the correct physical points.
Avoid parallax.
Quote a realistic precision and always include units.

Common Mistakes

Measuring between the stands rather than between the spring contact/support points.
Recording an over-precise value (e.g. $15.03\ \text{cm}$ from a mm ruler).
Omitting the unit.

Things to Be Careful About

Ensure the springs are vertical and the rod is parallel to the bench before measuring, as instructed.
If you subtract two readings, keep consistent precision (e.g. both to $0.1\ \text{cm}$ ).

Techniques used

identify the correct points between which the distance is measureduse a ruler/metre rule with appropriate alignment to avoid parallaxrecord a length to an appropriate precision with unit

(b)

• Lift one end of the rod a short distance and push the other end of the rod down a short distance. Release the rod so that it oscillates with a rocking motion, as shown in Fig. 1.2.

Fig. 1.2

• Take measurements to determine the period $T$ of the oscillation.

$T$ = ______ $\text{s}$

DifficultyMedium-Easy

Worked solution

Working

Time $N$ complete oscillations (e.g. $N = 10$ ) using a stopwatch and compute

T = \frac{t}{N}

Repeat and average $T$ .

Example: if $t = 12.4\ \text{s}$ for $10$ oscillations,

T = \frac{12.4}{10} = 1.24\ \text{s}

Answer

$T \approx 1.24\ \text{s}$ (example, from timing 10 oscillations).

Final answer

T ≈ 1.24 s (example)

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. Using a stopwatch, human reaction time (starting/stopping) creates a significant uncertainty if you time just one oscillation. Timing many oscillations reduces the percentage uncertainty because the reaction time is spread over a larger total time.

Understanding the Question

The rod rocks up and down at its ends. You must measure the period $T$ of this rocking oscillation. The question expects a practical method (not a theoretical calculation).

Approach

Choose a clear reference position to define one complete oscillation (e.g. left end at its highest point).
Time $N$ oscillations where $N$ is reasonably large (often $10$ to $20$ ).
Calculate $T = t/N$ .
Repeat the timing at least once and average the values of $T$ .

Step-by-Step Reasoning

Start the oscillation with a small amplitude (so the motion stays regular and does not slip).
Decide what counts as “one oscillation”: for example, left end goes from highest point back to highest point.
Use a stopwatch to measure the total time $t$ for $N$ complete oscillations.
Compute

T = \frac{t}{N}

Repeat: obtain $t_1, t_2, \dots$ and calculate $T_1, T_2, \dots$ then average.

Key Takeaways

Always time multiple oscillations.
Define a consistent counting point.
Repeats and averaging improve reliability.

Common Mistakes

Timing only one oscillation (large percentage uncertainty).
Miscounting oscillations (e.g. counting half-oscillations).
Letting amplitude decay too much or the rod slip, making timing inconsistent.

Things to Be Careful About

Use the same amplitude range each time (small, but visible).
Count complete cycles only.
Record times to the stopwatch resolution (e.g. $0.01\ \text{s}$ ) but do not claim unrealistic precision in $T$ if your timings are inconsistent.

Techniques used

time multiple oscillations and divide to obtain the periodrepeat timings and average to reduce random erroruse a consistent reference point to count full oscillations

(c)

• Change $x$ by moving the stands. Adjust the apparatus until the springs are vertical and the rod is parallel to the bench. Measure $x$ and $T$ .

• Repeat until you have six sets of values of $x$ and $T$ .

• Record your results in a table. Include values of $\frac{1}{x}$ in your table.

DifficultyMedium

Worked solution

Answer

Record six sets of $x$ and $T$ and include a calculated column for $1/x$ .

Example of a correctly formatted table (values illustrative):

$x / \text{cm}$	$T / \text{s}$	$1/x / \text{cm}^{-1}$
12.0	1.47	0.083
14.0	1.37	0.071
16.0	1.30	0.062
18.0	1.24	0.056
20.0	1.20	0.050
22.0	1.16	0.045

(Consistent decimal places in each column; $1/x$ calculated for each $x$ .)

Final answer

Table of six values of x, T, and 1/x (student-dependent)

Detailed explanation

Background Concept

A good results table in Paper 3 must:

contain all readings in one clear table,
have column headings as “quantity / unit”,
use consistent precision (same decimal places) down a column,
include any calculated quantities requested (here $1/x$ ).

Also, to test a relationship reliably, you need enough data points over a sensible range of the independent variable.

Understanding the Question

You must change the spring separation $x$ and measure the period $T$ for each value. You need six pairs $(x, T)$ , and you must calculate and include $1/x$ in the table.

Approach

Choose a range of $x$ values (not all very close together), while keeping the springs vertical and symmetric about the rod’s midpoint.
For each $x$ , measure $T$ using the “time $N$ oscillations then divide” method.
Calculate $1/x$ for each row.
Present in a single table with correct headings and consistent precision.

Step-by-Step Reasoning

Decide values of $x$ (e.g. from about $12\ \text{cm}$ to $22\ \text{cm}$ , depending on what the apparatus allows). The key is that $1/x$ then changes noticeably.
For each $x$ $x$ :
- Ensure the rod is level and springs vertical before starting oscillation.
- Measure $T$ (preferably from timing $10$ oscillations, repeated and averaged).
Compute $1/x$ :

\frac{1}{x}

If $x$ is recorded in $\text{cm}$ , then $1/x$ is in $\text{cm}^{-1}$ . Choose a sensible rounding for $1/x$ (often 2–3 significant figures) and keep it consistent.

Construct the table with three columns: $x$ , $T$ , and $1/x$ .

Key Takeaways

Six data points give a meaningful graph.
A derived column must be calculated correctly and clearly labelled with units.
Consistent precision and clear headings are essential for marks.

Common Mistakes

Missing units in column headings.
Inconsistent decimal places within a column.
Only a narrow range of $x$ values (graph becomes unreliable).
Calculating $1/x$ using mixed units (e.g. sometimes using $x$ in m, sometimes in cm).

Things to Be Careful About

If you record $x$ in $\text{cm}$ , keep it that way throughout so $1/x$ is $\text{cm}^{-1}$ .
Re-check alignment each time you move the stands (springs vertical, rod level), otherwise $T$ may change for reasons unrelated to $x$ .
Don’t round $x$ too aggressively before calculating $1/x$ ; calculate from your measured $x$ then round $1/x$ for the table.

Techniques used

vary the independent variable over a suitable range and take multiple readingscalculate a derived quantity for each row of datarecord data in a table with correct headings, units, and consistent precision

(d)

(i)

Plot a graph of $T$ on the $y$ -axis against $\frac{1}{x}$ on the $x$ -axis.

DifficultyMedium

Worked solution

Answer

Plot a graph with:

$y$ -axis: $T / \text{s}$
$x$ -axis: $1/x / \text{cm}^{-1}$

Use a sensible scale (at least half the grid in each direction) and plot all six points accurately.

Final answer

Graph of T (y) against 1/x (x) with correct labels/units

Detailed explanation

Background Concept

A graph is used to show how one quantity depends on another. In Paper 3, marks are awarded for correct axis labels (quantity and unit), sensible scales, accurate plotting, and (in later parts) a correct best-fit line.

Understanding the Question

You have calculated $1/x$ and measured $T$ for six different values of $x$ . You are told to plot $T$ (vertical axis) against $1/x$ (horizontal axis). This means each point is $(1/x,\ T)$ .

Approach

Put the independent variable on the $x$ -axis: $1/x$ .
Put the dependent variable on the $y$ -axis: $T$ .
Label each axis as “quantity / unit”.
Pick scales that spread the data out.
Plot each point carefully.

Step-by-Step Reasoning

From your table, take each pair $\left(\frac{1}{x},\ T\right)$ .
Draw axes with enough room for labels.
Label axes clearly, e.g. $T / \text{s}$ and $1/x / \text{cm}^{-1}$ .
Choose scales:
- avoid awkward scales like 3 squares = 0.07,
- ensure points cover a large area (improves gradient accuracy).
Plot points with small, neat crosses (or dots in small circles), not blobs.

Key Takeaways

Always plot the variables specified in the question.
Correct units in axis labels matter.
Good scale choice improves accuracy of gradient/intercept.

Common Mistakes

Swapping axes (plotting $1/x$ on $y$ and $T$ on $x$ ).
Missing units, or writing only the unit without the quantity.
Using a scale that compresses the points into a small region.

Things to Be Careful About

If $x$ is in $\text{cm}$ , then $1/x$ is in $\text{cm}^{-1}$ (not $\text{m}^{-1}$ ).
Use consistent rounding: plot using the values recorded in your table.

Techniques used

choose appropriate axes and labels including unitsselect suitable scales to use most of the gridplot points accurately from a results table

(ii)

Draw the straight line of best fit.

DifficultyMedium-Easy

Worked solution

Answer

Draw a straight line of best fit through the plotted points (balanced with approximately equal scatter above and below the line).

Final answer

Straight line of best fit drawn

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data, not a point-to-point join. For a linear relationship, you draw a straight line that best represents all points, considering experimental scatter.

Understanding the Question

After plotting $T$ against $1/x$ , you must draw the straight line of best fit. The next part uses this line to calculate gradient and intercept, so the line must be carefully drawn.

Approach

Use a ruler.
Position the ruler so the line passes centrally through the cluster of points.
Ensure the number of points above and below is roughly balanced.

Step-by-Step Reasoning

Do not join the points.
If one point is clearly off the trend (an anomaly), the best-fit line should still represent the majority of points.
Draw the line thinly and extend it enough to read the intercept (where it crosses the $y$ -axis) clearly.

Key Takeaways

Best-fit means “overall trend”, not “through every point”.
A thin, well-placed line improves the accuracy of gradient and intercept.

Common Mistakes

Forcing the line through the origin when not required.
Drawing a line through the first and last point regardless of scatter.
Drawing a thick line, making readings of intercept/gradient inaccurate.

Things to Be Careful About

Use a long ruler and draw a single continuous straight line.
Extend the line across the full range of the plotted data for a more reliable gradient triangle.

Techniques used

draw a straight line that balances points above and belowignore anomalous points when justified by scatterextend the best-fit line to read intercepts reliably

(iii)

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

gradient = ______

$y$ -intercept = ______

DifficultyMedium

Worked solution

Working

Choose two well-separated points on the best-fit line and calculate

\text{gradient } m = \frac{\Delta T}{\Delta (1/x)}

Example (from line): using points $\left(0.0714\ \text{cm}^{-1},\ 1.37\ \text{s}\right)$ and $\left(0.0500\ \text{cm}^{-1},\ 1.20\ \text{s}\right)$ ,

m = \frac{1.37 - 1.20}{0.0714 - 0.0500} = \frac{0.17}{0.0214} \approx 7.9\ \text{s cm}

Read the $y$ -intercept $c$ where the line crosses the $T$ -axis (or calculate $c = T - m(1/x)$ ). Example:

c \approx 0.80\ \text{s}

Answer

Gradient $\approx 7.9\ \text{s cm}$

$y$ -intercept $\approx 0.80\ \text{s}$ (example values).

Final answer

gradient ≈ 7.9 s cm, y-intercept ≈ 0.80 s (example)

Detailed explanation

Background Concept

For a straight-line graph of the form

y = mx + c

the gradient is $m = \Delta y / \Delta x$ ,
the $y$ -intercept is $c$ (value of $y$ when $x = 0$ ).

On a $T$ vs $1/x$ graph, the gradient unit is

\frac{\text{s}}{\text{cm}^{-1}} = \text{s cm}

and the intercept has unit $\text{s}$ .

Understanding the Question

You must find the gradient and the $y$ -intercept of your best-fit line on the graph of $T$ (vertical) against $1/x$ (horizontal). These will be used later to find constants in a suggested equation.

Approach

Use a large triangle on the best-fit line (choose points far apart) to reduce reading error.
Compute the gradient with $\Delta T / \Delta(1/x)$ .
Find the intercept by reading where the line crosses the $T$ -axis, or calculate it using $c = T - m(1/x)$ with a point on the line.

Step-by-Step Reasoning

Pick two points on the drawn line (not necessarily actual data points), ideally near the ends of the plotted range.
Read their coordinates carefully:
- $x$ -coordinate is $1/x$ in $\text{cm}^{-1}$ ,
- $y$ -coordinate is $T$ in $\text{s}$ .
Calculate

m = \frac{T_2 - T_1}{(1/x)_2 - (1/x)_1}

Quote the gradient with unit $\text{s cm}$ .
For the intercept $c$ $c$ :
- either read directly at $1/x = 0$ (where the line crosses the $T$ axis),
- or compute using one point on the line:

c = T - m(1/x)

and quote $c$ in seconds.

Key Takeaways

Gradient is always “change in $y$ over change in $x$ ”.
Use a large triangle and points on the best-fit line.
Units of gradient come from the axis units.

Common Mistakes

Using $\Delta x / \Delta y$ instead of $\Delta y / \Delta x$ .
Using two nearby points (gives large percentage uncertainty in the gradient).
Forgetting that the $x$ -axis variable is $1/x$ , not $x$ .
Missing or incorrect gradient units.

Things to Be Careful About

Read coordinates from the line, not from the printed grid without checking scale.
Keep enough significant figures in intermediate steps so rounding does not distort the intercept.
If you calculate $c$ , use a point on the best-fit line (not a scattered experimental point).

Techniques used

determine the gradient using a large triangle on the best-fit lineread or calculate the y-intercept from the best-fit lineuse units consistently when calculating gradient and intercept

(e)

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

T = \frac{a}{x} + b

where $a$ and $b$ are constants.

Use your answers in (d)(iii) to determine the values of $a$ and $b$ . Give appropriate units.

$a$ = ______

$b$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

T = \frac{a}{x} + b

and plotting $T$ against $1/x$ gives

T = a\left(\frac{1}{x}\right) + b

So $a$ is the gradient and $b$ is the $y$ -intercept.

Using (d)(iii) example values: gradient $\approx 7.9\ \text{s cm}$ and intercept $\approx 0.80\ \text{s}$ .

Answer

a \approx 7.9\ \text{s cm}

b \approx 0.80\ \text{s}

Final answer

a = gradient (s cm), b = y-intercept (s)

Detailed explanation

Background Concept

If a graph is plotted with $y$ against $x$ and it is a straight line, it can be compared to

y = mx + c

where $m$ is the gradient and $c$ is the $y$ -intercept.

Here the suggested relationship is

T = \frac{a}{x} + b

which can be rewritten as

T = a\left(\frac{1}{x}\right) + b

So a plot of $T$ (vertical) against $1/x$ (horizontal) should be linear, with gradient $a$ and intercept $b$ .

Understanding the Question

You have already found the gradient and $y$ -intercept of the graph of $T$ against $1/x$ . You must now use those values to determine the constants $a$ and $b$ , including correct units.

Approach

Rearrange the given equation to match $y = mx + c$ .
Identify which constant corresponds to the gradient and which corresponds to the intercept.
Copy the numerical values from (d)(iii).
Assign units based on the graph axes (or from dimensional reasoning).

Step-by-Step Reasoning

Compare

T = a\left(\frac{1}{x}\right) + b

with

T = m\left(\frac{1}{x}\right) + c

Thus:

a = m \quad \text{and} \quad b = c

Units:
- $T$ is in $\text{s}$ .
- $1/x$ is in $\text{cm}^{-1}$ .
- Therefore

[a] = \frac{\text{s}}{\text{cm}^{-1}} = \text{s cm}

and

[b] = \text{s}

Substitute your measured gradient and intercept to obtain $a$ and $b$ .

Key Takeaways

Linearising the equation tells you exactly what the gradient and intercept represent.
Units come directly from axis units.

Common Mistakes

Stating $a$ has units of $\text{s}$ (it does not).
Using $x$ instead of $1/x$ when matching to $y = mx + c$ .
Mixing $\text{m}$ and $\text{cm}$ units between table, graph, and final constants.

Things to Be Careful About

If you plotted $1/x$ in $\text{cm}^{-1}$ , then $a$ must be in $\text{s cm}$ .
Quote $a$ and $b$ to a sensible number of significant figures consistent with the precision of your graph reading (often 2–3 s.f.).

Techniques used

match a linear equation to the form y = mx + cidentify constants from gradient and interceptdeduce units of constants from the equation and axis units

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Q2Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations20M

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