In this experiment, you will investigate the equilibrium of a wooden rod. (i) ● Assemble the apparatus as shown in Fig. 2.1. ● Adjust the apparatus so that the wooden rod is parallel to the bench and the spring is vertical. ● The distance between the ends of the spring is $L_0$, as shown in Fig. 2.1. Measure and record $L_0$. $L_0$ = \_\_\_\_\_\_ $\text{m}$ (ii) ● Pull the mass hanger down a short distance and then release it. The mass hanger will oscillate. ● Take…

(a)(i) L0 ≈ 0.120 m (example; student-dependent) · (a)(ii) T ≈ 0.630 s (example; student-dependent) · (a)(iii) k ≈ 19.9 N m^-1 (example; student-dependent) · (b)(i) L ≈ 0.135 m, θ ≈ 20° (example; student-dependent) · (b)(ii) 5% · (b)(iii) L ≈ 0.160 m, θ ≈ 45° (example; student-dependent) · (c) D ≈ 2.16 N and 1.98 N (example; student-dependent) · (d) Supports the relationship (values agree within 10%). · (e)(i) See working (four sources of uncertainty/limitations listed). ·…

Physics 9702/32 — May/June 2024

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

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate an electrical circuit.

(a)

● Connect the circuit shown in Fig. 1.1.

Fig. 1.1

● Ensure that the polarities of the power supply, component C and the voltmeter are as shown in Fig. 1.1.

● Close switch S for a short time and then open it.

● Watch the voltmeter reading as it reduces.

When the voltmeter reading passes a value $V_S$ of $8.00\text{ V}$ , start the stop-watch.

When the voltmeter reading passes a value of $7.00\text{ V}$ , stop the stop-watch.

● Record the starting value $V_S$ and the time $T$ for the voltmeter reading to fall by $1.00\text{ V}$ .

$V_S$ = ______
$T$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$V_S = 8.00\ \text{V}$

$T = 13.7\ \text{s}$ (example, to nearest $0.1\ \text{s}$ )

Final answer

Example: VS = 8.00 V, T = 13.7 s

Detailed explanation

Background Concept

A capacitor that has been charged and then allowed to discharge through a resistor will show a decreasing potential difference (p.d.) across it. In this circuit, the voltmeter measures the p.d. across the capacitor (and the resistor branch, since they are in parallel).

In a practical exam, you are not being tested on getting a particular numerical value (because your component values differ), but on:

using the apparatus correctly,
following the timing instruction exactly (start/stop at specified voltages),
recording values with appropriate precision.

Understanding the Question

You are told to:

Connect the circuit as in Fig. 1.1 with the stated polarities.
Close switch $S$ briefly (to charge the capacitor), then open it (so the capacitor discharges through $R$ ).
Watch the voltmeter as it falls.
Start timing when the reading passes $8.00\ \text{V}$ and stop timing when it passes $7.00\ \text{V}$ .

You must record:

the start voltage $V_S$ (which should be $8.00\ \text{V}$ ),
the time $T$ for a drop of $1.00\ \text{V}$ .

Approach

Ensure the capacitor is charged close to the supply p.d. by closing $S$ only briefly, then opening it.
Keep your eye on the voltmeter display and operate the stopwatch at the instants when the display passes $8.00\ \text{V}$ and $7.00\ \text{V}$ .
Record $V_S$ exactly as stated (to $0.01\ \text{V}$ because the question uses $8.00\ \text{V}$ ) and record $T$ to the stopwatch resolution (typically $0.1\ \text{s}$ ).

Step-by-Step Reasoning

diagram-rc-discharge

Build the circuit exactly as shown, including polarity markings (this matters because electrolytic capacitors are polarised and must not be reversed).
Close $S$ for a short time so that the capacitor charges.
Open $S$ . Now the capacitor discharges through the resistor branch, so the voltmeter reading decreases.
When the voltmeter reading passes $8.00\ \text{V}$ , start the stopwatch.
When it passes $7.00\ \text{V}$ , stop the stopwatch.
Record $V_S = 8.00\ \text{V}$ and the measured time $T$ (example shown in the solution).

Key Takeaways

In Paper 3, marks often come from correct method and correct recording precision.
“Passes a value” means you start/stop at the moment the reading crosses that value, not when it is merely close.

Common Mistakes

Starting at $8\ \text{V}$ instead of $8.00\ \text{V}$ (loss of required precision).
Timing from $8.00\ \text{V}$ down to $7.00\ \text{V}$ but accidentally stopping at $7.10\ \text{V}$ or $6.90\ \text{V}$ .
Leaving the switch closed too long (may heat the resistor or cause unstable readings) or not charging enough so the reading never reaches $8.00\ \text{V}$ .

Things to Be Careful About

Record $T$ with a sensible resolution (typically $0.1\ \text{s}$ ); do not overstate precision.
Make sure the capacitor polarity is correct; reversed polarity can damage the capacitor and change results.
Reacting time affects the measurement: try to be consistent in your start/stop technique.

Techniques used

set up the circuit with correct polaritiesuse a stopwatch to time a specified change in voltmeter readingrecord readings to an appropriate resolution

(b)

Choose another starting value $V_S$ . Close S for a short time and then open it. Measure the time $T$ for the voltmeter reading to fall by $1.00\text{ V}$ from the starting value $V_S$ .

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

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

10M Similar

DifficultyMedium

Worked solution

Answer

Record six sets of $V_S$ and $T$ and calculate $1/T$ .

Example table (headings with units, consistent dp):

$V_S\ /\ \text{V}$	$T\ /\ \text{s}$	$1/T\ /\ \text{s}^{-1}$
6.00	16.4	0.0610
7.00	14.9	0.0671
8.00	13.7	0.0730
9.00	12.7	0.0787
10.00	11.8	0.0847
11.00	11.0	0.0909

Final answer

See working (table of VS, T and 1/T)

Detailed explanation

Background Concept

To test a relationship experimentally you must:

vary an independent quantity (here the chosen starting voltage $V_S$ ),
measure the dependent quantity (here the time $T$ for a $1.00\ \text{V}$ drop),
take enough readings over a good range to reveal any trend,
calculate any required derived values (here $1/T$ ) and present everything clearly.

A good results table is a major source of marks in Paper 3.

Understanding the Question

You must choose different starting values $V_S$ and, for each one, measure the time $T$ for the voltmeter reading to fall by $1.00\ \text{V}$ (from $V_S$ to $V_S - 1.00\ \text{V}$ ). You need six pairs $(V_S, T)$ .

You must then present the results in a single table and include a third column for $1/T$ .

Approach

Choose a sensible range of $V_S$ values that are achievable during the discharge (e.g. several values between about $6\ \text{V}$ and $11\ \text{V}$ ).
For each $V_S$ $V_{S}$ :
- charge the capacitor by closing $S$ briefly,
- open $S$ and time the drop from $V_S$ to $V_S - 1.00\ \text{V}$ .
Keep technique consistent (same observer position, same way of starting/stopping the stopwatch).
Calculate $1/T$ for each row and record it to an appropriate number of significant figures.

Step-by-Step Reasoning

Selecting $V_S$ : If you choose values too close together, the graph trend is hard to see; if you choose values too near the supply limit or too low, you may not be able to start/stop cleanly at the required voltages.
Recording precision:
- Because the question specifies $8.00\ \text{V}$ , the voltmeter is effectively read to $0.01\ \text{V}$ , so record all $V_S$ values to $0.01\ \text{V}$ .
- A handheld stopwatch is typically read to $0.1\ \text{s}$ , so record all $T$ values to $0.1\ \text{s}$ .
Derived column:
- For each row compute $1/T$ .
- The unit of $1/T$ is $\text{s}^{-1}$ .
- Quote $1/T$ to typically 3 significant figures (or consistent dp) so the column is uniform.

A clear table must have:

headings with quantity and unit, e.g. $T / \text{s}$ ,
consistent decimal places within each column,
all six rows present.

Key Takeaways

Six readings over a sensible range is necessary to support a straight-line graph.
A correct table has clear headings, units, consistent precision, and correctly calculated derived values.

Common Mistakes

Omitting units in headings (e.g. writing just $T$ instead of $T / \text{s}$ ).
Mixing decimal places in the same column (e.g. $11.0\ \text{s}$ , $12.73\ \text{s}$ , $13\ \text{s}$ ).
Calculating $1/T$ but forgetting the unit $\text{s}^{-1}$ .
Not taking enough readings (fewer than six rows).

Things to Be Careful About

Make sure each measurement of $T$ corresponds to exactly a $1.00\ \text{V}$ drop.
Do not round $T$ too early before calculating $1/T$ ; calculate using your recorded $T$ then round $1/T$ appropriately.
Keep $V_S$ values within the region where the capacitor actually passes through them during discharge (so you can start/stop cleanly).

Techniques used

vary the independent variable over a suitable rangerepeat the timing procedure consistently for multiple valuesrecord results in a single table with correct headings and unitscalculate a derived quantity for each rowuse consistent significant figures within each column

(c)

(i)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $\dfrac{1}{T}$ (in $\text{s}^{-1}$ ) on the $y$ -axis against $V_S$ (in $\text{V}$ ) on the $x$ -axis using a sensible scale and all six data points.

Final answer

Graph of 1/T against VS plotted

Detailed explanation

Background Concept

When you are asked to plot a graph of one quantity against another, you must:

put the stated quantity on the stated axis,
label each axis with both the symbol and the unit,
choose a scale that spreads the points out (usually using at least half the grid in each direction),
plot each point accurately.

These are routine Paper 3 marks.

Understanding the Question

You have a table containing $V_S$ and $T$ and you have calculated $1/T$ .

You must plot a graph with:

horizontal axis: $V_S$ (unit V),
vertical axis: $1/T$ (unit $\text{s}^{-1}$ ).

Approach

Decide suitable axis ranges based on your smallest and largest $V_S$ and $1/T$ .
Use simple scales (e.g. 1 big square = 0.5 V or 1 V; 1 big square = 0.005 $\text{s}^{-1}$ , etc.) that make plotting easy.
Plot all six points with small, neat crosses.

Step-by-Step Reasoning

diagram-graph-layout

Draw axes and label them:
- $x$ -axis: $V_S / \text{V}$
- $y$ -axis: $1/T / \text{s}^{-1}$
Choose scales:
- Avoid awkward scales like 3 squares = 1 unit.
- Ensure data fills a large part of the grid.
Plot each point $(V_S, 1/T)$ $(V_{S}, 1/ T)$ :
- Read off carefully using the grid.
- Use a sharp pencil and fine crosses; do not use large dots.

Key Takeaways

Correct axis labels and sensible scales are as important as the points themselves.
Plot exactly what is requested: $1/T$ vs $V_S$ , not $T$ vs $V_S$ .

Common Mistakes

Swapping axes (plotting $V_S$ on $y$ instead of $x$ ).
Missing units in axis labels.
Using a tiny part of the grid (wastes resolution and reduces accuracy).
Plotting $T$ instead of $1/T$ .

Things to Be Careful About

If $1/T$ values are close together, choose a $y$ -axis scale that spreads them.
Start axes at zero only if it helps; it is acceptable to use a non-zero origin if clearly marked and it improves spread.

Techniques used

choose suitable axis scales that use most of the graph gridlabel axes with quantities and unitsplot points accurately from a results table

(ii)

Draw the straight line of best fit.

1M Similar

(iii)

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

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

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

(V_{S1}, y_1) = (6.00\ \text{V},\ 0.0610\ \text{s}^{-1}),\quad (V_{S2}, y_2) = (11.00\ \text{V},\ 0.0909\ \text{s}^{-1})

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{0.0909 - 0.0610}{11.00 - 6.00} = 5.98 \times 10^{-3}\ \text{s}^{-1}\text{V}^{-1}

$y$ -intercept (example) from $y = ax + b$ :

0.0610 = (5.98 \times 10^{-3})(6.00) + b \Rightarrow b = 2.51 \times 10^{-2}\ \text{s}^{-1}

Answer

gradient $= 6.0 \times 10^{-3}\ \text{s}^{-1}\text{V}^{-1}$

$y$ -intercept $= 2.5 \times 10^{-2}\ \text{s}^{-1}$

Final answer

gradient = 6.0×10^-3 s^-1 V^-1, y-intercept = 2.5×10^-2 s^-1 (example)

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ , the equation is

y = mx + c

where:

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

From a plotted best-fit line:

gradient is found using two points on the line and

m = \frac{\Delta y}{\Delta x}

$y$ -intercept is the value of $y$ when $x = 0$ (where the line crosses the $y$ -axis), or can be calculated using $c = y - mx$ .

Understanding the Question

You have a graph of $1/T$ (vertical) against $V_S$ (horizontal) with a best-fit straight line. You must determine:

the gradient of the line,
the $y$ -intercept.

These values will be used again in part (d).

Approach

diagram-gradient-triangle

Choose two points on the best-fit line (not necessarily measured points) that are far apart to reduce percentage uncertainty.
Read their coordinates accurately.
Compute gradient using $\Delta(1/T)/\Delta V_S$ .
Find the intercept either by reading off at $V_S = 0$ or by substituting one point into $y = mx + c$ .

Step-by-Step Reasoning

Pick two widely separated points on the line, e.g. at the left and right side of the plotted region.
Read $V_S$ values to the same precision as your axis scale allows, and read $1/T$ values likewise.
Compute changes:
- $\Delta y = \Delta(1/T)$ , unit $\text{s}^{-1}$
- $\Delta x = \Delta V_S$ , unit V
Gradient:

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

So the unit is:

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

Intercept:
- If the axis includes $V_S = 0$ , read where the line crosses the $y$ -axis.
- If not, calculate using $b = y - a x$ with one point.

The numerical example in the solution shows one valid method.

Key Takeaways

Use a large triangle on the best-fit line to reduce uncertainty.
Gradient must be $\Delta y / \Delta x$ , not $\Delta x / \Delta y$ .
Always include units for gradient and intercept.

Common Mistakes

Using two nearby points (gives a large percentage uncertainty in gradient).
Using data points that are not on the line (instead of points on the drawn line).
Calculating $\Delta x / \Delta y$ by mistake.
Forgetting that the intercept has units of $\text{s}^{-1}$ .

Things to Be Careful About

Read coordinates from the line with care; small reading errors can strongly affect the gradient.
Keep sufficient significant figures during the calculation, then round sensibly at the end.
Make sure your line is thin; a thick line increases reading uncertainty.

Techniques used

use a large triangle on the best-fit line to determine the gradientcalculate gradient as $\Delta y / \Delta x$determine the y-intercept from where the line crosses the y-axis

(d)

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

\frac{1}{T} = aV_S + b

where $a$ and $b$ are constants.

Using your answers in (c)(iii), determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______
$b$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{T} = aV_S + b

Comparing with $y = mx + c$ for graph of $y = 1/T$ against $x = V_S$ :

a = \text{gradient},\quad b = y\text{-intercept}

Units:

[a] = \frac{\text{s}^{-1}}{\text{V}} = \text{s}^{-1}\text{V}^{-1},\quad [b] = \text{s}^{-1}

Answer

$a = 6.0 \times 10^{-3}\ \text{s}^{-1}\text{V}^{-1}$

$b = 2.5 \times 10^{-2}\ \text{s}^{-1}$

Final answer

a = gradient (s^-1 V^-1), b = y-intercept (s^-1)

Detailed explanation

Background Concept

A straight-line graph is described by

y = mx + c

If you plot $y$ against $x$ , then:

gradient $m$ is the coefficient of $x$ ,
intercept $c$ is the value of $y$ when $x=0$ .

Here the suggested relationship is

\frac{1}{T} = aV_S + b

So it is already in straight-line form.

Understanding the Question

You have already obtained:

gradient of the graph of $1/T$ vs $V_S$ ,
$y$ -intercept of the same graph.

You must use these to state numerical values for $a$ and $b$ with appropriate units.

Approach

Identify $y$ and $x$ for your plotted graph.
Compare term-by-term with the given equation.
Copy the gradient as $a$ and the intercept as $b$ .
Work out units from the units of $1/T$ and $V_S$ .

Step-by-Step Reasoning

Your graph uses:
- $y = 1/T$ with unit $\text{s}^{-1}$ ,
- $x = V_S$ with unit V.
Compare

\frac{1}{T} = aV_S + b

with

y = mx + c

Therefore:

$a$ corresponds to the gradient $m$ .
$b$ corresponds to the intercept $c$ .

Units:

Since $aV_S$ must have the same unit as $1/T$ ,

[a] = \frac{[1/T]}{[V_S]} = \frac{\text{s}^{-1}}{\text{V}} = \text{s}^{-1}\text{V}^{-1}

Since $b$ is added directly to $1/T$ ,

[b] = \text{s}^{-1}

Substitute your measured gradient and intercept values (example shown in the solution).

Key Takeaways

For a plot of $y$ against $x$ , gradient gives the coefficient of $x$ and intercept gives the constant term.
Units of gradient come from $\text{(unit of }y\text{)}/\text{(unit of }x\text{)}$ .

Common Mistakes

Giving $a$ the wrong unit (e.g. $\text{V s}^{-1}$ instead of $\text{s}^{-1}\text{V}^{-1}$ ).
Using the gradient of the wrong graph (e.g. if axes were swapped earlier).
Confusing $b$ with the $x$ -intercept (it is the $y$ -intercept).

Things to Be Careful About

Ensure you are using the gradient and intercept from the best-fit line, not from two raw data points.
Quote $a$ and $b$ to sensible significant figures consistent with your graph-reading precision.

Techniques used

match a linear equation to the form $y = mx + c$identify constants from the gradient and intercept of a graphdeduce units of constants from the plotted variables

Q2MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate the equilibrium of a wooden rod.

(a)

(i)

● Assemble the apparatus as shown in Fig. 2.1.

● Adjust the apparatus so that the wooden rod is parallel to the bench and the spring is vertical.

Fig. 2.1

● The distance between the ends of the spring is $L_0$ , as shown in Fig. 2.1.

Measure and record $L_0$ .

$L_0$ = ______ $\text{m}$

1M Similar

DifficultyEasy

Worked solution

Answer

Example measurement (student-dependent):

$L_0 = 0.120\ \text{m}$

Final answer

L0 ≈ 0.120 m (example; student-dependent)

Detailed explanation

Background Concept

A length measurement must be taken between two clearly defined reference points, using an instrument of known resolution (e.g. a metre rule). The recorded value should match the instrument precision (typically to the nearest mm, i.e. $0.001\ \text{m}$ ).

Understanding the Question

You have a spring supporting one end of a rod. $L_0$ is the distance between the two ends of the spring when the spring is vertical and the rod is horizontal. You must measure and record $L_0$ in metres.

Approach

Make sure the rod is parallel to the bench and the spring is vertical (as instructed).
Use a metre rule to measure the separation between the top and bottom ends of the spring.
Record in $\text{m}$ to a suitable number of decimal places.

Step-by-Step Reasoning

Place the metre rule alongside the spring.
Align the zero (or a clear scale mark) with one end of the spring, and read the position of the other end.
Avoid parallax by viewing the scale perpendicularly.
Convert to metres if necessary and record, e.g. $L_0 = 0.120\ \text{m}$ .

Key Takeaways

Measure between correct reference points.
Record with a unit and appropriate precision.

Common Mistakes

Measuring the spring along a slanted line (when it is not vertical) or using the wrong endpoints.
Not giving a unit, or recording to an unrealistic precision (e.g. too many decimal places).

Things to Be Careful About

Ensure the spring is vertical before measuring.
Keep the metre rule parallel to the spring to reduce systematic error.
Read at eye level to minimise parallax.

Techniques used

identify the correct two reference points for a length measurementuse a metre rule to take a single reading to appropriate precisionrecord a measurement with a unit and consistent decimal places

(ii)

● Pull the mass hanger down a short distance and then release it. The mass hanger will oscillate.

● Take measurements to find the period $T$ of the oscillations.

$T$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Time $10$ oscillations (repeat and average):

$t_1 = 6.32\ \text{s}$ , $t_2 = 6.28\ \text{s}$

Mean time for $10$ oscillations:

\bar{t} = \frac{6.32 + 6.28}{2} = 6.30\ \text{s}

Period:

T = \frac{\bar{t}}{10} = \frac{6.30}{10} = 0.630\ \text{s}

Answer

$T = 0.630\ \text{s}$

Final answer

T ≈ 0.630 s (example; student-dependent)

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. Using a stopwatch, reaction time makes timing a single oscillation inaccurate, so you time $N$ oscillations (with $N\ge 5$ or $10$ ) and divide by $N$ :

T = \frac{t}{N}

Repeating timings and averaging reduces random uncertainty.

Understanding the Question

You pull the mass hanger down slightly and release it so it oscillates. You must obtain $T$ from measurements (not from theory), so the mark is for good technique: timing several oscillations, repeating, and calculating $T$ .

Approach

Choose a clear reference point (e.g. the lowest point of motion).
Time $10$ oscillations with a stopwatch.
Repeat at least once.
Average the times and divide by $10$ to get $T$ .

Step-by-Step Reasoning

Start timing as the mass passes the reference point in a chosen direction.
Count $10$ complete cycles and stop when the mass returns to the same point moving in the same direction.
Example readings: $t_1 = 6.32\ \text{s}$ and $t_2 = 6.28\ \text{s}$ .
Average: $\bar{t} = 6.30\ \text{s}$ .
Divide by $10$ :

T = \frac{6.30}{10} = 0.630\ \text{s}

Key Takeaways

Time many oscillations, not one.
Repeat and average to improve reliability.

Common Mistakes

Timing only one oscillation (large percentage reaction-time error).
Not counting complete oscillations correctly.
Stopping at a different point in the cycle (e.g. not matching direction).

Things to Be Careful About

Keep the amplitude small so the motion stays close to simple harmonic motion.
Avoid pushing the mass sideways (which can introduce extra motion and affect timing).
Use consistent significant figures for the stopwatch readings.

Techniques used

time multiple oscillations and divide to obtain the periodrepeat timing measurements and calculate a meanuse a stopwatch consistently with a clear timing reference point

(iii)

● Calculate the value of the spring constant $k$ using

k = \frac{\alpha\pi^2}{T^2}

where $\alpha = 0.800\text{ kg}$ .

$k$ = ______ $\text{N m}^{-1}$

● Justify the number of significant figures that you have given for your value of $k$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using

k = \frac{\alpha\pi^2}{T^2}

with $\alpha = 0.800\ \text{kg}$ and $T = 0.630\ \text{s}$ :

k = \frac{0.800\pi^2}{(0.630)^2} = 19.9\ \text{N m}^{-1}

Significant figures: $T$ is measured to $3\ \text{s.f.}$ (and $\alpha$ is $3\ \text{s.f.}$ ), so quote $k$ to $3\ \text{s.f.}$ .

Answer

$k = 19.9\ \text{N m}^{-1}$ (to $3\ \text{s.f.}$ )

Final answer

k ≈ 19.9 N m^-1 (example; student-dependent)

Detailed explanation

Background Concept

When a quantity is calculated from measurements, the result should be quoted to a sensible number of significant figures, usually based on the least precise measured quantity used in the calculation.

Here, $k$ is obtained from the given relationship:

k = \frac{\alpha\pi^2}{T^2}

where $\alpha$ is provided and $T$ is measured.

Understanding the Question

You must (1) calculate $k$ using your measured $T$ , and (2) justify the significant figures in your value of $k$ . The justification is about measurement precision, not about “how many digits look nice”.

Approach

Substitute $\alpha$ and your measured $T$ into the given formula.
Calculate $k$ and include units.
Decide the significant figures by looking at the precision of $T$ (and $\alpha$ ).

Step-by-Step Reasoning

With $\alpha = 0.800\ \text{kg}$ and (example) $T = 0.630\ \text{s}$ :

k = \frac{0.800\pi^2}{(0.630)^2}

Numerically, $\pi^2 \approx 9.87$ and $0.630^2 = 0.3969$ , so:

k \approx \frac{0.800 \times 9.87}{0.3969} = 19.9\ \text{N m}^{-1}

Significant figures: if $T$ is recorded as $0.630\ \text{s}$ then $T$ is $3\ \text{s.f.}$ , and $\alpha = 0.800$ is also $3\ \text{s.f.}$ , so quoting $k$ to $3\ \text{s.f.}$ is consistent.

Key Takeaways

Use the provided equation exactly.
Always include units.
Significant figures should reflect measurement precision.

Common Mistakes

Giving too many significant figures (implying unrealistic precision).
Forgetting units for $k$ .
Using $T$ for one oscillation timed poorly, leading to an unreliable $k$ .

Things to Be Careful About

Because $k \propto 1/T^2$ , any percentage uncertainty in $T$ roughly doubles in $k$ ; this is another reason not to over-quote digits.
Ensure $T$ is in seconds before substitution.

Techniques used

substitute measured values into a given equationcalculate a derived quantity with correct unitsjustify significant figures based on the least precise measurement

(b)

(i)

● Move the stand supporting the spring away from the other stand and add the plumb line, as shown in Fig. 2.2.

Fig. 2.2

● Adjust the apparatus so that the angle $\theta$ between the spring and the vertical is approximately $20^\circ$ and the wooden rod is parallel to the bench, as shown in Fig. 2.2.

● The new distance between the ends of the spring is $L$ , as shown in Fig. 2.2.

Measure and record $L$ .

$L$ = ______ $\text{m}$

● Measure and record $\theta$ .

$\theta$ = ______ $^{\circ}$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example measurements (student-dependent):

$L = 0.135\ \text{m}$

$\theta = 20^\circ$

Final answer

L ≈ 0.135 m, θ ≈ 20° (example; student-dependent)

Detailed explanation

Background Concept

A plumb line gives a true vertical reference. The angle $\theta$ is the angle between the spring and the vertical, so it should be measured between the spring and the plumb line (not between the spring and the rod).

The spring length $L$ should be measured along the spring between its endpoints.

Understanding the Question

You move the stand so the spring is no longer vertical, then set $\theta \approx 20^\circ$ while keeping the rod parallel to the bench. You must record:

$L$ (new spring length)
$\theta$ (angle between spring and vertical)

Approach

Adjust stand position until the spring makes about $20^\circ$ to the vertical (use the plumb line).
Re-adjust until the rod is horizontal.
Measure $L$ along the spring, and measure $\theta$ with a protractor against the vertical reference.

Step-by-Step Reasoning

Hang the plumb line from the top support so it hangs freely and is clearly visible.
Place a protractor so its centre is at (or close to) the top end of the spring and align the $0^\circ$ line with the plumb line.
Read the angle to the line of the spring (estimate to nearest degree).
Measure $L$ with a metre rule by aligning along the spring and reading endpoint-to-endpoint.
Example: $L = 0.135\ \text{m}$ , $\theta = 20^\circ$ .

Key Takeaways

Use the plumb line as the vertical reference.
Measure $L$ along the spring, not as a horizontal or vertical separation.

Common Mistakes

Measuring $\theta$ from the horizontal rod instead of the vertical.
Measuring $L$ as a vertical drop rather than the spring’s actual length.

Things to Be Careful About

Ensure the rod is parallel to the bench before taking readings.
Reduce parallax when reading a protractor and ruler.
Let the system settle (no oscillations) before measuring.

Techniques used

adjust the apparatus to achieve a target angle and a horizontal rodmeasure the spring length with a ruler along the springmeasure an angle to the vertical using a plumb line as reference

(ii)

Estimate the percentage uncertainty in your value of $\theta$ . Show your working.

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Take angle reading uncertainty as $\pm 1^\circ$ .

For $\theta = 20^\circ$ :

\%\ \text{uncertainty} = \frac{1}{20} \times 100 = 5\%

Answer

percentage uncertainty $= 5\%$

Final answer

Detailed explanation

Background Concept

Percentage uncertainty is

\%\ \text{uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100

For an angle measured with a typical protractor, a common estimate for absolute uncertainty is about $\pm 1^\circ$ (you may justify this from the smallest scale division and alignment limits).

Understanding the Question

You must estimate the percentage uncertainty in your measured $\theta$ (from part (b)(i)) and show working. The experiment uses a plumb line and protractor, so the limiting factor is usually reading/alignment to the nearest degree.

Approach

Choose a sensible absolute uncertainty for $\theta$ (e.g. $\pm 1^\circ$ ).
Substitute into the percentage-uncertainty formula using your measured $\theta$ .

Step-by-Step Reasoning

Suppose the protractor reading is to the nearest degree, so absolute uncertainty is $\pm 1^\circ$ .
If $\theta = 20^\circ$ :

\%\ \text{uncertainty} = \frac{1}{20} \times 100 = 5\%

Key Takeaways

Percentage uncertainty scales inversely with the size of the measured angle.
Always show the formula and substitution.

Common Mistakes

Using $\pm 1\%$ instead of $\pm 1^\circ$ .
Dividing by the wrong value (e.g. using $45^\circ$ when the angle was $20^\circ$ ).

Things to Be Careful About

Your chosen absolute uncertainty should match your instrument and reading method.
Quote the final percentage uncertainty sensibly (often to 1 s.f. is acceptable).

Techniques used

estimate an absolute uncertainty from instrument resolutioncalculate percentage uncertainty from absolute uncertaintyreport uncertainty to an appropriate number of significant figures

(iii)

Repeat (b)(i) using an angle $\theta$ of approximately $45^\circ$ .

$L$ = ______ $\text{m}$
$\theta$ = ______ $^{\circ}$

3M Similar

(c)

It is suggested that the relationship between $L$ and $\theta$ is

k(L - L_0) + B = \frac{D}{\cos\theta}

where $B = 2.0\text{ N}$ and $D$ is a constant.

Using your data, calculate two values of $D$ .

first value of $D$ = ______
second value of $D$ = ______

1M Similar

DifficultyMedium

Worked solution

Working

From

k(L-L_0)+B = \frac{D}{\cos\theta}

D = \left[k(L-L_0)+B\right]\cos\theta

Using example data: $k=19.9\ \text{N m}^{-1}$ , $L_0=0.120\ \text{m}$ , $B=2.0\ \text{N}$ .

For $\theta = 20^\circ$ , $L=0.135\ \text{m}$ :

D_1 = \left[19.9(0.135-0.120)+2.0\right]\cos 20^\circ = 2.16\ \text{N}

For $\theta = 45^\circ$ , $L=0.160\ \text{m}$ :

D_2 = \left[19.9(0.160-0.120)+2.0\right]\cos 45^\circ = 1.98\ \text{N}

Answer

first value of $D = 2.16\ \text{N}$

second value of $D = 1.98\ \text{N}$

Final answer

D ≈ 2.16 N and 1.98 N (example; student-dependent)

Detailed explanation

Background Concept

When a relationship contains a constant (here $D$ ), you can test it by calculating $D$ from different sets of measurements. If the relationship is correct, the calculated values of $D$ should agree within experimental uncertainty.

Given:

k(L - L_0) + B = \frac{D}{\cos\theta}

You rearrange to make $D$ the subject:

D = \left[k(L-L_0)+B\right]\cos\theta

Understanding the Question

You have measured $L$ and $\theta$ for two angles (about $20^\circ$ and $45^\circ$ ), and you already have $L_0$ and $k$ from part (a). You must use these to compute two values of the constant $D$ .

Approach

Rearrange the given equation to $D = (k(L-L_0)+B)\cos\theta$ .
For each data set (each $\theta$ ), substitute $k$ , $L$ , $L_0$ , $B$ , and $\theta$ .
Calculate $D$ twice.

Step-by-Step Reasoning

Start from

k(L - L_0) + B = \frac{D}{\cos\theta}

Multiply both sides by $\cos\theta$ :

\left[k(L-L_0)+B\right]\cos\theta = D

Now use your measured values.
Example using earlier sample numbers:
- $k = 19.9\ \text{N m}^{-1}$ , $L_0 = 0.120\ \text{m}$ , $B = 2.0\ \text{N}$ .
- For $\theta=20^\circ$ , $L=0.135\ \text{m}$ :

L-L_0 = 0.015\ \text{m}

k(L-L_0) = 19.9 \times 0.015 = 0.299\ \text{N}

D_1 = (0.299+2.0)\cos 20^\circ = 2.16\ \text{N}

For $\theta=45^\circ$ , $L=0.160\ \text{m}$ :

L-L_0 = 0.040\ \text{m}, \quad k(L-L_0)= 19.9\times 0.040 = 0.796\ \text{N}

D_2 = (0.796+2.0)\cos 45^\circ = 1.98\ \text{N}

(Your own values will differ because your measurements differ.)

Key Takeaways

Rearranging correctly is crucial before substituting data.
Comparing the computed constant from two trials is a basic consistency check.

Common Mistakes

Using $D = \frac{k(L-L_0)+B}{\cos\theta}$ (incorrect rearrangement).
Forgetting that $\cos\theta$ uses $\theta$ in degrees on a calculator set to degree mode.
Mixing units (e.g. using $L$ in cm while $k$ is in $\text{N m}^{-1}$ ).

Things to Be Careful About

Convert all lengths to $\text{m}$ before using $k$ in $\text{N m}^{-1}$ .
Keep consistent significant figures (your result cannot be more precise than your measured $L$ and $\theta$ ).

Techniques used

rearrange a given relationship to make the unknown the subjectsubstitute experimental measurements into the rearranged equationcalculate and compare a constant from two sets of readings

(d)

It is suggested that the percentage uncertainty in the values of $D$ is $10\%$ .

Using this uncertainty, explain whether your results support the relationship in (c).

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Example values: $D_1 = 2.16\ \text{N}$ and $D_2 = 1.98\ \text{N}$ .

Percentage difference:

\frac{|2.16-1.98|}{\frac{2.16+1.98}{2}}\times 100 = 8.7\%

Since $8.7\% < 10\%$ , the two values agree within the stated uncertainty.

Answer

Yes. The calculated values of $D$ are consistent within $10\%$ , so the results support the relationship.

Final answer

Supports the relationship (values agree within 10%).

Detailed explanation

Background Concept

To decide whether two experimental values agree, you compare their difference with the uncertainty. If the stated percentage uncertainty is $10\%$ , then results that differ by less than about $10\%$ are usually considered consistent (given the level of the uncertainty statement).

A convenient check is the percentage difference:

\%\ \text{difference} = \frac{|D_1-D_2|}{\text{mean of }D_1\text{ and }D_2}\times 100

Understanding the Question

You have two values of $D$ from part (c). The question tells you the percentage uncertainty in the values of $D$ is $10\%$ . You must use this to state whether your results support the suggested relationship.

Approach

Find how far apart $D_1$ and $D_2$ are (absolute difference).
Convert that to a percentage difference (relative to a representative value, often the mean).
If the percentage difference is $\le 10\%$ , conclude the results support the relationship; otherwise they do not.

Step-by-Step Reasoning

Using example values $D_1=2.16\ \text{N}$ and $D_2=1.98\ \text{N}$ :

|D_1-D_2| = 0.18\ \text{N}

Mean value:

\bar{D} = \frac{2.16+1.98}{2} = 2.07\ \text{N}

Percentage difference:

\frac{0.18}{2.07}\times 100 = 8.7\%

Since $8.7\%$ is less than the stated $10\%$ , the difference is within the uncertainty, so the data are consistent with the relationship.

Key Takeaways

A model is supported if values calculated from different trials agree within experimental uncertainty.

Common Mistakes

Comparing the absolute difference in $\text{N}$ directly to $10\%$ (different types of quantity).
Using $\frac{|D_1-D_2|}{D_1}$ or $\frac{|D_1-D_2|}{D_2}$ without stating what you are doing; mean is the clearest.

Things to Be Careful About

The $10\%$ uncertainty is itself an estimate; your conclusion should be phrased appropriately (e.g. “support” / “consistent with”, not “proves”).

Techniques used

compare two values using a stated percentage uncertaintycalculate a percentage difference between experimental valuesmake a consistency judgement based on overlap within uncertainty

(e)

(i)

Describe four sources of uncertainty or limitations of the procedure for this experiment.

For any uncertainties in measurement that you describe, you should state the quantity being measured and a reason for the uncertainty.

4M Similar

DifficultyMedium

Worked solution

Answer

Any four, e.g.

$L$ (and $L_0$ ) reading uncertainty: ends of the spring are not sharply defined / coils obscure exact endpoints, so length reading is uncertain.
Parallax in $L$ (and $L_0$ ): metre rule and spring may not be in the same plane; eye not normal to scale gives a reading error.
$\theta$ measurement uncertainty: difficult to align protractor with the plumb line and the spring; thickness of spring and plumb line makes the angle ambiguous.
Timing $T$ : human reaction time starting/stopping stopwatch and deciding the exact same point in the oscillation causes random uncertainty (especially if damping occurs).

(Other valid limitations: rod not perfectly horizontal each time; pivot friction / movement at supports; oscillations not purely vertical.)

Final answer

See working (four sources of uncertainty/limitations listed).

Detailed explanation

Background Concept

In practical exams, “uncertainty/limitations” marks are awarded for specific, physics-based issues that affect measurements. For measurement uncertainties you should:

name the quantity (e.g. $L$ , $\theta$ , $T$ )
state why it is uncertain (e.g. parallax, poor definition of reference points, reaction time)

A limitation is a feature of the method that restricts accuracy/validity (e.g. only two angles, difficulty maintaining equilibrium conditions).

Understanding the Question

You must give four sources of uncertainty or limitations in this procedure. For any measurement uncertainty, you must state the measured quantity and the reason.

Approach

Think through each measured quantity in the experiment:

$L_0$ and $L$ (length measurements)
$\theta$ (angle measurement)
$T$ (timing)
Then add any procedural limitations (keeping rod horizontal, pivot friction, movement of supports).

Step-by-Step Reasoning

Examples of four good points:

Uncertainty in $L$ and $L_0$ (endpoints): the “end of the spring” may be hard to define because the hooks/coils are curved and the attachment points are bulky, so the measured length depends on judgement.
Parallax in $L$ and $L_0$ : if the metre rule is not in the same plane as the spring, or your eye is not perpendicular to the scale, the reading shifts.
Uncertainty in $\theta$ : aligning the protractor with the plumb line and the axis of the spring is difficult; both have thickness so there is not a single precise line to measure from.
Uncertainty in $T$ (stopwatch timing): reaction time and deciding the exact instant a cycle completes introduce random errors; damping can make the “same point” harder to judge.

Other acceptable limitations (if needed): keeping the rod exactly parallel to the bench for each reading is difficult; supports may slip; pivot friction may change the equilibrium slightly.

Key Takeaways

State the quantity and the cause.
Be specific (what is hard to see/align/judge).

Common Mistakes

Writing vague phrases like “human error” without stating what measurement it affects.
Repeating the same point in different words (e.g. parallax twice as two separate uncertainties without a genuinely different context).

Things to Be Careful About

Each of the four points should be distinct.
Focus on uncertainties that genuinely affect the measured values used later ( $L_0$ , $L$ , $\theta$ , $T$ , hence $k$ and $D$ ).

Techniques used

identify measurement uncertainties and link each to a causedescribe procedural limitations that affect validity or reliabilitystate uncertainties with the quantity affected and a clear reason

(ii)

Describe four improvements that could be made to this experiment. You may suggest the use of other apparatus or different procedures.

4M Similar

DifficultyMedium

Worked solution

Answer

Any four, e.g.

Improve timing of $T$ : use a light gate/data logger (or video analysis) to obtain the period automatically and remove reaction-time error.
Improve $L$ and $L_0$ measurement: attach a pointer at each spring end and read against a fixed vertical scale (or use a set square to transfer positions to a ruler) to reduce parallax and define endpoints.
Improve $\theta$ measurement: use a larger/digital protractor or angle sensor fixed at the top attachment so the angle to the vertical is read more precisely.
Improve reliability/validity: take more than two values of $\theta$ (e.g. $10^\circ$ to $60^\circ$ ) with repeats, so $D$ can be checked over a wider range rather than just two points.

(Other valid improvements: use a spirit level to ensure rod horizontal; clamp supports more rigidly to prevent slipping.)

Final answer

See working (four improvements listed).

Detailed explanation

Background Concept

An improvement should directly reduce a named uncertainty/limitation or increase the reliability of the conclusion. Strong improvements either:

change the instrument (better resolution, remove subjective judgement), or
change the method (more repeats, wider range, better control of variables).

Understanding the Question

You must describe four improvements. These can be new apparatus or altered procedures. The best answers clearly link each improvement to an uncertainty/limitation from (e)(i).

Approach

Match improvements to the main problem areas:

timing $T$
measuring lengths $L, L_0$
measuring angle $\theta$
limited data / poor control of equilibrium conditions

Step-by-Step Reasoning

Four good improvements (with why they help):

Light gate / data logger for $T$ : records times precisely and consistently, removing human reaction time and judgement of cycle completion.
Fixed scale with pointers for $L$ and $L_0$ : if pointers mark the endpoints and a fixed scale is read, endpoints are clearer and parallax is reduced compared with holding a rule next to a spring.
Better angle measurement: a large/digital protractor, or an angle sensor mounted at the top support, gives a clearer reference and finer resolution than estimating with a small handheld protractor.
More angles and repeats: using several values of $\theta$ and repeating each measurement allows you to check whether $D$ is constant over a range and to average out random scatter.

Additional sensible improvements include using a spirit level to ensure the rod is horizontal and improving clamping to prevent slipping.

Key Takeaways

Improvements should be practical and targeted.
Link each to a specific uncertainty/limitation.

Common Mistakes

Suggesting vague improvements like “be more careful” without stating what to change.
Proposing unrealistic apparatus that does not measure the required quantity.

Things to Be Careful About

Do not repeat the same improvement in different words.
Ensure each improvement is feasible within a school laboratory context and clearly affects $L_0$ , $L$ , $\theta$ , or $T$ (and hence $k$ and $D$ ).

Techniques used

propose procedural improvements linked to stated limitationsselect more appropriate apparatus to reduce measurement uncertaintyincrease data quantity or reliability through repeats and range