Part 1 --- The Physics Behind the Practical

When a tensile force $F$ is applied to a wire of original length $L$ and cross-sectional area $A$, the wire extends by $\Delta L$. Two ratios capture what is happening inside the material independently of the wire's dimensions.

For the same pulling force, a smaller cross-sectional area gives a larger stress.

Stress tells you how concentrated the pull is. A 10 N force on a thick cable produces much less stress than the same 10 N force on a very thin wire.

Strain tells you how much the material has deformed relative to its original size. A 1 mm extension on a 2 m wire is a much smaller strain than a 1 mm extension on a 10 cm wire.

The original length appears in the denominator, which is why strain has no units.

The Young modulus links the two.

This equation is usually provided on the data sheet, but you must be able to rearrange and use it fluently. $E$ is a material property --- it does not depend on the wire's length or thickness, only on the substance it is made from. Steel has $E \approx 200\;\text{GPa}$; copper has $E \approx 130\;\text{GPa}$.

Linearisation for graphical analysis

Rearranging the Young modulus equation to isolate the measured quantities:

This has the form $y = mx$ where:

• $y = \Delta L$ (extension, plotted on the $y$-axis)
• $x = F = mg$ (applied force, plotted on the $x$-axis)
• gradient $= \dfrac{L}{EA}$

Therefore:

Alternatively, if you plot $F$ on the $y$-axis against $\Delta L$ on the $x$-axis (which some mark schemes prefer), the gradient $= \dfrac{EA}{L}$ and so $E = \dfrac{\text{gradient} \times L}{A}$. Either convention is acceptable provided you are consistent.

Short exam link: stress--strain graphs and energy

This practical is usually analysed with an extension--force graph, but sometimes the same data are converted into stress and strain. In that case, the straight-line region still represents Hooke's law behaviour.

This is because $\text{energy} = \frac{1}{2}F\,\Delta L$, and dividing both sides by the volume $AL$ gives $\frac{1}{2}\,\frac{F}{A}\,\frac{\Delta L}{L} = \frac{1}{2}\sigma\varepsilon$. The key exam check is to read the axis labels carefully: area under a force--extension graph gives energy in joules, whereas area under a stress--strain graph gives energy per unit volume in $\text{J m}^{-3}$.

Part 2 --- Equipment, Setup, and Method

Equipment list

Apparatus diagram

The figure below shows the standard two-wire setup; notice how the comparison wire and the scale-vernier pair let you measure the test wire's true extension while cancelling beam movement and thermal effects.

[DIAGRAM: asset_name: RP 04 - Determination of the Young Modulus by a Simple Method - Diagram 1; asset_slug: RP 04 - Determination of the Young Modulus by a Simple Method - Diagram 1; recommended_method: retained_png; description: Two long wires hang vertically side-by-side from a rigid ceiling beam. The left wire is labelled "comparison wire" and the right wire is labelled "test wire". Both are clamped to the beam at the top. A horizontal millimetre scale is clamped to the comparison wire at the bottom; a vernier slider is clamped to the test wire at the same height and slides against the scale. The comparison wire supports a single 1 kg hanger (to keep it taut). The test wire supports a hanger with a stack of slotted masses. A sand tray sits on the floor beneath the masses. Safety goggles are shown beside the setup. Labels: ceiling beam, clamps, comparison wire, test wire, mm scale, vernier scale, mass hanger, slotted masses, sand tray.]

Step-by-step method

1. Set up the two-wire system. Clamp both wires to the ceiling beam so they hang vertically, side by side, with the scale-and-vernier arrangement connecting them at the bottom.

   • Why two wires? The comparison wire compensates for any sagging of the beam under load and for thermal expansion. Without it, a warm room could cause the beam to sag or the wire to lengthen, producing a false "extension". The comparison wire experiences the same environmental changes, so the vernier reading only reflects the true extension caused by the added mass.

2. Hang a 1 kg mass hanger from each wire. This pre-tensions both wires, removing any kinks and ensuring they hang straight.

   • Why pre-tension? A kinked wire will "extend" as kinks straighten out, giving false readings that are not true elastic extension. Pre-tensioning ensures you are only measuring genuine elastic strain from the start.

3. Measure the original length $L$ of the test wire from the clamp at the beam down to the vernier clamp, using the metre ruler. Record $L$ in metres.

   • Why measure to the vernier clamp? This is the length of wire that is actually stretching. Any wire above the beam clamp is not under the test load.

4. Record the initial vernier scale reading $r_0$. This is your zero reference.

5. Add a 1 kg slotted mass to the test wire hanger. Wait a few seconds for the wire to stop oscillating, then read the new vernier position $r_1$. The extension is $\Delta L_1 = r_1 - r_0$.

   • Why wait? The mass causes the wire to bounce briefly. Reading while oscillating introduces random error.

6. Continue adding masses in equal increments (e.g. 1 kg each time) up to about 7--8 kg total on the test wire, recording the vernier reading after each addition.

7. Unload the wire by removing masses one at a time, recording the vernier reading at each step.

   • Why unload? Comparing loading and unloading readings checks whether the elastic limit has been exceeded. If unloading readings are larger than loading readings for the same mass, the wire has been permanently deformed and those data points must be discarded. If they match, the wire was within its elastic limit and you can average loading/unloading values to reduce random error.

8. Measure the wire diameter using the micrometer screw gauge at a minimum of six different positions along the wire (and at different orientations at each position in case the cross-section is not perfectly circular). Record all readings and calculate the mean diameter $d$.

   • Why multiple positions and orientations? The wire may not be perfectly uniform. Averaging reduces the effect of any localised variation.
   • Why a micrometer? A ruler cannot resolve $0.01\;\text{mm}$. Since the diameter is small (~0.45 mm), even a $0.01\;\text{mm}$ error represents a ~2% uncertainty in diameter, which doubles to ~4% in the area calculation.

9. Calculate the cross-sectional area $A = \frac{\pi d^2}{4}$.

10. Plot extension $\Delta L$ against the added load $F = mg$ from the slotted masses and draw an unconstrained line of best fit. The theoretical relationship passes through the origin, but a measured best-fit line should not be forced through it; a significant intercept can reveal a zero offset or other systematic effect.

If your zero reading $r_0$ was taken with the hanger already attached, the hanger provides the baseline tension and the changing force on the graph is the force from the added slotted masses only.

Use the value of $g$ stated in the question or data sheet. If none is specified, $9.81\;\text{N kg}^{-1}$ is a sensible value to use for school practical calculations.

Why use a long, thin wire?

This is a common focus in practical discussions because a long, thin wire gives a larger extension for a given load, making the extension large enough to measure accurately.

Quantitatively, $\Delta L = \frac{FL}{EA}$. For a fixed force and material:

• Doubling $L$ doubles $\Delta L$ --- a longer wire gives a larger, more measurable extension.
• Halving the diameter makes $A$ one quarter as large (since $A \propto d^2$), which quadruples $\Delta L$.

If the wire were short and thick, extensions would be fractions of a millimetre --- far too small to measure reliably even with a vernier, leading to enormous percentage uncertainties.

Alternative methods

A common alternative is a horizontal bench method where the wire is clamped at one end, run across the bench over a pulley, and loaded with hanging masses. A travelling microscope or marker-and-ruler system measures the extension. This method is simpler to set up but less precise because friction at the pulley can reduce the effective tension, and the wire sags under its own weight. The vertical two-wire arrangement remains the standard method.

Part 3 --- Variables, Controls, and Experimental Design

Repeats: Load and unload the wire at least once (giving two sets of extension readings for each mass). If time allows, repeat the entire loading--unloading cycle a second time. Averaging loading and unloading values reduces random error in the extension readings and confirms the wire has not exceeded its elastic limit.

Suitable range: Use enough mass increments (at least 6--8 different loads) to produce a clear spread of data points on the graph. The maximum load should be chosen so that the wire does not exceed its elastic limit --- check by comparing loading and unloading values.

Part 4 --- Expected Results, Graphs, and Interpretation

Sample results table

The standard school setup for this required practical is the vertical two-wire apparatus described above. The sample numbers below come from a horizontal copper-wire version of the same experiment and are included purely to demonstrate the graph and calculation steps; once you have measured force and extension, the Young modulus analysis is the same.

Using sample data for a copper wire ($L = 2.4\;\text{m}$, $d = 0.274\;\text{mm}$):

Graph

The graph below shows the expected straight-line trend. For this illustrative horizontal setup, the vertical error bars are $\pm 0.25\;\text{mm}$, representing the estimated uncertainty from the resolution and repeat-to-repeat spread of the extension readings. The unconstrained best-fit line is shown with independently chosen steepest and shallowest acceptable lines that intersect every error bar.

[DIAGRAM: asset_name: RP 04 - Determination of the Young Modulus by a Simple Method - Diagram 2; asset_slug: RP 04 - Determination of the Young Modulus by a Simple Method - Diagram 2; recommended_method: deterministic_chart; description: Monochrome graph of extension $\Delta L$ / mm against force $F$ / N for eight sample measurements. Every point has a visible vertical uncertainty bar of $\pm 0.25$ mm. An unconstrained best-fit line has gradient 0.320 mm N$^{-1}$ and a negative intercept. Independently calculated dashed steepest and dotted shallowest acceptable lines intersect every error bar; neither is automatically constrained through the origin. Axes, units, line identities, and gradients are clearly labelled.]

Interpretation

The graph should be linear over the elastic range. The sample best-fit line is not forced through the origin and has a negative intercept, which could indicate a zero offset or another systematic effect. Its gradient still equals $\frac{L}{EA}$, from which the Young modulus is extracted.

If the line curves upward at high loads, the wire has exceeded its limit of proportionality. Data points in the curved region should be excluded from the gradient calculation.

The physical reason for the straight-line relationship is that, within the elastic limit, atomic bonds in the metal lattice behave like tiny springs. The macroscopic extension is the sum of billions of these tiny atomic-level stretches, and since each atom--atom "spring" obeys Hooke's law, so does the entire wire.

Stress--strain graph interpretation

If you convert the same measurements to stress and strain, the gradient of the straight-line region is $E$ directly. For this practical, the important point is that only the linear region should be used to determine Young modulus.

The area under the linear portion of the stress--strain graph, a triangle of base $\varepsilon$ and height $\sigma$, gives $\frac{1}{2}\sigma\varepsilon$ --- the elastic strain energy stored per unit volume.

Part 5 --- Worked Example with Full Calculation

Using the sample data: $L = 2.4\;\text{m}$, $d = 0.274\;\text{mm}$.

Step 1: Calculate the cross-sectional area

Step 2: Determine the gradient

From the graph, select two widely separated points on the best-fit line (not data points, but points on the line itself).

Taking approximately $(2.00,\;0.36)$ and $(15.00,\;4.53)$ from the unconstrained best-fit line:

Step 3: Calculate the Young modulus

Step 4: Compare with the reference value

The reference Young modulus for copper is approximately $130\;\text{GPa}$.

This is well within typical experimental uncertainty, indicating that the method is sound and the wire was within its elastic limit throughout.

Step 5: Energy stored in the wire

Suppose the wire is loaded to $F = 15.70\;\text{N}$ with extension $\Delta L = 5.00\;\text{mm}$.

This energy is stored as elastic potential energy in the stretched atomic bonds. If the wire is unloaded within the elastic limit, all of this energy is recovered. On a force--extension graph this energy equals the triangular area under the line.

For the same copper wire, the stress at 9.81 N is $\sigma = \frac{F}{A} = \frac{9.81}{5.9 \times 10^{-8}} = 1.66 \times 10^{8}\;\text{Pa}$, and the strain energy stored at an extension of 2.7 mm is $\frac{1}{2}F\Delta L = \frac{1}{2} \times 9.81 \times 2.7 \times 10^{-3} = 1.3 \times 10^{-2}\;\text{J}$. This is a good reminder that the same dataset can be used to extract stiffness, stress, and stored energy.

Part 6 --- Uncertainty and Error Analysis

Systematic errors

Random errors

Full uncertainty calculation (worked through with numbers)

Uncertainty in diameter $d$:

Suppose six micrometer readings (in mm) are: 0.273, 0.275, 0.274, 0.272, 0.276, 0.274.

Mean $d = 0.274\;\text{mm}$.

The range is $0.276 - 0.272 = 0.004\;\text{mm}$, so the uncertainty from scatter is $\frac{\text{range}}{2} = 0.002\;\text{mm}$.

The micrometer's resolution is $0.01\;\text{mm}$, giving an instrument uncertainty of $\pm 0.005\;\text{mm}$ (half the smallest division).

The scatter-based uncertainty ($0.002\;\text{mm}$) is smaller than the instrument uncertainty ($0.005\;\text{mm}$), so we use the larger value: $\Delta d = \pm\,0.005\;\text{mm}$.

Uncertainty in area $A$:

Since $A = \frac{\pi d^2}{4}$, and $A \propto d^2$, the percentage uncertainty in $A$ is double the percentage uncertainty in $d$:

The diameter is squared in the area formula, so its percentage uncertainty is doubled when you calculate the area. That makes diameter an important contributor, but you should still compare the actual percentage uncertainties in the data rather than assume it is always the largest term.

Uncertainty in length $L$:

$L = 2.400\;\text{m}$, measured with a metre ruler ($\pm\,1\;\text{mm}$).

This is negligible compared to the uncertainties in $d$ and $\Delta L$.

Uncertainty in gradient (from the graph):

Draw the steepest and shallowest acceptable straight lines that intersect the uncertainty bars. Do not force either line through the origin.

For the plotted sample, the best-fit gradient is $3.203 \times 10^{-4}\;\text{m\,N}^{-1}$, the steepest acceptable gradient is $3.398 \times 10^{-4}\;\text{m\,N}^{-1}$, and the shallowest acceptable gradient is $3.057 \times 10^{-4}\;\text{m\,N}^{-1}$.

Using half the range as a symmetric estimate of the gradient uncertainty:

Combining uncertainties for $E$:

Since $E = \frac{L}{A \times \text{gradient}}$, and $L$, $A$, and the gradient are all multiplied or divided:

So if $E = 127\;\text{GPa}$:

The reference value of $130\;\text{GPa}$ lies within this range, confirming the result is consistent with the known value.

As a quick check on the diameter calculation, readings of 0.44, 0.46, 0.45, 0.44, 0.45, and 0.46 mm give a mean diameter of $0.450\;\text{mm}$. The half-range is $0.01\;\text{mm}$, so the percentage uncertainty in $d$ is $2.2\%$, and the percentage uncertainty in the cross-sectional area is therefore $4.4\%$ because the diameter is squared.

Sources of error and improvements table

Part 7 --- Pulling the Practical Together

Most questions on this practical come back to the same chain of reasoning: measure the wire's geometry carefully, apply a range of known forces, measure the extensions as precisely as possible, and then use the graph gradient to separate the material property from the dimensions of the sample.

The common checks are whether you can justify the comparison wire, explain why the wire should be long and thin, recognise when the elastic limit has been exceeded, and decide which measured quantity is contributing most to the uncertainty in the final value of $E$.

If you can explain why each piece of apparatus is there, you are much less likely to get stuck when the practical is described in an unfamiliar way.

Use that same idea when answering uncertainty questions: do the arithmetic first, then decide which source is largest from the numbers rather than from a rule of thumb.

A full method question then asks you to connect the measurements, the graph, and the error-reduction steps into one coherent answer.

The same practical logic is used well beyond the school lab whenever engineers need a reliable value for stiffness before choosing a material for a structure.