The key idea
Hooke's law grows up: springs become wires, force-extension becomes stress-strain, and the Young modulus experiment becomes one of the most reliably examined practicals at AS.
Understand it · 01
Separate specimen properties from material properties
Two wires made from the same material can extend by different amounts under the same force. A longer wire extends more, while a thicker wire extends less. Stress, F/A, and strain, ΔL/L, remove these simple geometrical effects so that different specimens can be compared. Stress has units of pascals and strain has no unit because it is a ratio of lengths. Young modulus E = stress/strain describes the linear elastic response of a material.
A large Young modulus means that a relatively large stress produces a small strain: the material is stiff. This is different from strength, which concerns the stress a material can withstand before failure, and different from toughness, which concerns energy absorbed before fracture. A stiff material can be brittle. Keep these distinctions visible when interpreting material choices rather than treating every favourable mechanical property as another word for strong.
Understand it · 02
Distinguish proportional behaviour from elastic recovery
The limit of proportionality marks the end of a straight-line force–extension or stress–strain relationship. The elastic limit marks the boundary beyond which unloading leaves a permanent deformation. These limits are related but need not coincide. Elastic behaviour means recovery of the original dimensions after unloading; it does not automatically mean that the force is proportional to extension. Hooke’s law is restricted to the linear region where that proportionality holds.
The area under a force–extension graph gives work done in stretching. For a straight line from the origin, the area is a triangle and the stored energy is ½FΔL. For a nonlinear curve, estimate the area instead of applying the triangle formula blindly. The area under a stress–strain graph represents energy per unit volume. If loading and unloading follow different paths, the enclosed area corresponds to energy dissipated in a cycle.
Understand it · 03
Measure the small extension carefully
A Young modulus experiment needs the original wire length, its diameter, the applied force and the resulting extension. Measure the diameter at several positions and orientations using a micrometer; calculate cross-sectional area as πd²/4. A small diameter error is amplified when the diameter is squared. Measure extension relative to a reference and avoid including support movement as if it were wire extension. A long thin wire makes the extension easier to resolve.
Add loads gradually, keep the wire below the range where permanent deformation occurs, and protect against falling masses or a snapping wire. Plot force against extension and use the gradient k from the straight region; then E = kL/A. A graph uses multiple measurements and can reveal an offset or the onset of nonlinear behaviour. Do not force a curved set of results into a straight-line fit just to obtain a modulus.
Key definitions
- Stress (σ)
- The force applied per unit cross-sectional area: σ = F ÷ A. Units: pascals.
- Strain (ε)
- The extension per unit original length: ε = x ÷ L. Dimensionless.
- Young modulus (E)
- The ratio of stress to strain in the region where they are proportional: E = σ ÷ ε.
- Elastic deformation
- Deformation in which the object returns to its original length when the load is removed; plastic deformation is permanent.
Equations, units & conditions
Stress
σ = F ÷ A
Units: Pa
Strain
ε = x ÷ L
Units: no unit
Young modulus
E = σ ÷ ε = FL ÷ Ax
Units: Pa
Strain energy
U = ½Fx = ½kx²
Units: J
Linear elastic behaviour from the unstretched state; otherwise use the actual force–extension area.
Worked examples, step by step
Example 1
A steel wire of length 2.0 m and diameter 0.50 mm extends by 1.2 mm under a load of 30 N. Calculate the Young modulus.
Follow the method
- A = πd²/4 = π × (0.50 × 10⁻³)² ÷ 4 = 1.96 × 10⁻⁷ m².
- Stress = F/A = 30 ÷ (1.96 × 10⁻⁷) = 1.53 × 10⁸ Pa.
- Strain = x/L = 1.2 × 10⁻³ ÷ 2.0 = 6.0 × 10⁻⁴.
- E = stress ÷ strain = (1.53 × 10⁸) ÷ (6.0 × 10⁻⁴).
E ≈ 2.6 × 10¹¹ Pa (260 GPa)
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Calculate Young modulus
A wire of original length 2.0 m and area 1.0 × 10⁻⁶ m² extends by 1.0 mm under a force of 100 N within its linear region. Find Young modulus.
Show solution and self-check: question 1
- Strain = 0.0010/2.0 = 5.0 × 10⁻⁴.
- Stress = 100/(1.0 × 10⁻⁶) = 1.0 × 10⁸ Pa.
- E = stress/strain.
2.0 × 10¹¹ Pa.
Self-check · one mark per point
- Convert extension to metres.
- Calculate stress and strain.
- Divide to obtain 2.0 × 10¹¹ Pa.
Build · 3 suggested marks
2. Elastic energy
An ideal spring has stiffness 240 N m⁻¹ and is extended by 0.075 m from its natural length. Find the force and elastic energy stored.
Show solution and self-check: question 2
- F = kx = 240 × 0.075 = 18 N.
- E = ½kx² = 0.5 × 240 × 0.075².
18 N and 0.675 J.
Self-check · one mark per point
- Use F = kx.
- Use the triangular force–extension area or ½kx².
- Obtain 0.675 J.
Common mistakes & how to avoid them
- Diameter for area. The wire's micrometer reading is a diameter: A = πd²/4, and the d² doubles its percentage uncertainty. Both facts are marks.
- Strain energy read as Fx instead of the area ½Fx. The area under the force-extension line is the energy — for the linear region, a triangle.
- Confusing the elastic limit with the limit of proportionality. The proportional limit ends the straight-line relationship; the elastic limit marks the onset of permanent deformation. They can be distinct, with nonlinear but recoverable deformation between them.
- In the Young modulus practical, measuring extension without a reference wire or ignoring temperature — the standard evaluation marks ask for exactly these.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
What is the difference between the limit of proportionality and the elastic limit?
The limit of proportionality is where force stops being proportional to extension. The elastic limit is where unloading begins to leave permanent deformation. A material can recover its original shape even after its graph has become nonlinear.
Elastic describes recovery, while proportional describes the mathematical relationship during loading. These are different tests. To locate a proportional limit, inspect where a force–extension or stress–strain graph departs from its initial straight line. To investigate an elastic limit, unload and measure whether the specimen returns to its original dimensions. A loading curve alone may not establish recovery without further information. Avoid assuming every material has the same separation between these points: the exact response depends on the material and treatment. The term elastic therefore does not automatically justify using Hooke’s law.
How is strain energy calculated?
Work done in stretching is the area under the force–extension graph. In a linear elastic region starting at zero extension, this is ½Fx = ½kx². For nonlinear loading, use the actual area rather than the triangular shortcut.
The force is initially smaller than its final value, so multiplying final force by the whole extension overestimates the area for a linear spring. Distinguish work supplied during loading from energy recovered on unloading. If the paths differ, their enclosed area represents energy dissipated; not all input returns as useful mechanical energy. A stress–strain graph has a related but different area: energy per unit volume. Multiplying that area by the specimen volume gives energy. Axis labels therefore determine whether the result is joules or joules per cubic metre.
Linear spring energy
A spring reaches 24 N at an extension of 0.060 m within its linear elastic range. Find stiffness and stored energy.
- k = F/x = 24/0.060 = 400 N m⁻¹.
- Stored energy = ½Fx = 0.5 × 24 × 0.060.
Stiffness 400 N m⁻¹; energy 0.72 J.
Why does measuring the diameter dominate the uncertainty in a Young modulus experiment?
Diameter can dominate because the cross-sectional area is proportional to its square, so a Unit 6 maximum estimate doubles its percentage uncertainty. It does not always dominate: compare its contribution with force, length and extension uncertainties.
A thin wire may have a diameter only a small number of micrometer divisions across, making a small absolute reading uncertainty relatively important. Measure several positions and orientations to detect thickness variation and an imperfectly circular cross-section. Repetition can estimate variation but cannot repair a zero offset, so check the instrument first. In E = 4FL/(πd²x), the diameter-squared term supplies the factor of two. These combined-percentage calculations are an A2 Unit 6 extension, while the AS practical still benefits from careful diameter measurement and a clear evaluation of the method.
Unit 6: diameter contribution
A wire diameter is 0.50 ± 0.01 mm. Estimate the diameter contribution to percentage uncertainty in Young modulus.
- Percentage diameter uncertainty = 100 × 0.01/0.50 = 2.0%.
- Young modulus contains d⁻², so multiply by |−2|.
Approximately 4.0%, before adding other measurement contributions.
How do wire length and thickness affect extension?
For the same material within its linear elastic range, extension is FL/(AE). Doubling the original length doubles extension at fixed force and area; doubling cross-sectional area halves it at fixed force and length.
Stiffness belongs to the whole specimen, while Young modulus describes its material response. A longer specimen places more deforming material in sequence, whereas a larger cross-section shares the force across more material. Stress and strain remove those geometrical differences so comparisons between specimens become meaningful. Be careful when a question doubles diameter rather than area: circular area then increases by a factor of four. These proportional arguments assume temperature and material condition remain comparable and that the force does not push either specimen beyond the range where the stated modulus applies.
What is the difference between stiffness, strength and toughness?
Stiffness describes resistance to elastic deformation, strength concerns the stress sustained before failure, and toughness concerns energy absorbed before fracture. A material can be stiff but brittle, so these properties should not be used interchangeably.
Young modulus compares stiffness in the linear stress–strain region. Breaking stress describes one aspect of strength, while the total area under a stress–strain curve up to fracture relates to toughness. A brittle specimen may reach a high stress but fail after little strain, giving a different energy absorption from a ductile specimen that deforms considerably. When choosing a material, connect the desired property to its job: a protective structure may need controlled deformation to absorb energy, while a measuring frame may need small elastic deflections. The graph helps distinguish those demands.
How do I calculate Young modulus from a force–extension graph?
For a linear wire, the gradient of force against extension is specimen stiffness k. Young modulus follows from E = kL/A; the gradient alone is not Young modulus because its units are N m⁻¹ rather than pascals.
The conversion accounts for the specimen geometry. Measure original length L and cross-sectional area A separately, then use a gradient fitted to the straight region of the force–extension data. If the graph has extension vertically and force horizontally, its gradient is the reciprocal stiffness, so invert it before applying the relation. A graph can reveal an initial slack region or a zero offset that a single reading hides. Do not force the fit through the origin without considering the experiment, and do not include a visibly nonlinear region when estimating a constant modulus.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Distinguish elastic and plastic deformation; identify the elastic limit
- Define stress, strain and the Young modulus
- Describe the Young modulus experiment for a wire
- Interpret force-extension and stress-strain graphs
- Calculate work from the force-extension area; use U = ½Fx = ½kx² only in the linear elastic region
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Deformation of solids. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/deformation-of-solids
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.