OCR A Level Physics A: Springs, stress, strain and the Young modulus (H556)

This guide covers sections 3.4.1 (Springs) and 3.4.2 (Mechanical properties of matter) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017.

See the OCR A Level Physics hub and the printable checklist. Pair this page with the revision notes and the practice questions. Check your starting point with the free diagnostics. Uncertainties build on Foundations of Physics.

What you must be able to do

Statement	Content
3.4.1(a)-(c)	Tensile and compressive deformation; Hooke’s law; F = kx
3.4.1(d)	Force-extension graphs; investigating springs, rubber bands, polythene strips
3.4.2(a)-(b)	Work done is the area under the graph; elastic potential energy
3.4.2(c)-(d)	Stress, strain, ultimate tensile strength; Young modulus and its measurement
3.4.2(e)-(f)	Stress-strain graphs for ductile, brittle and polymeric materials; elastic and plastic deformation

3.4.1(a) Tensile and compressive deformation

Forces that pull on opposite ends of an object are tensile and make it longer: the change in length is the extension. Forces that push on opposite ends are compressive and make it shorter: the change is the compression. A spring can be loaded either way; a wire or rubber band goes slack when pushed.

In every case x is the change in length from the natural length, not the total length.

3.4.1(b)-(c) Hooke’s law and the force constant

Hooke’s law: while the limit of proportionality is not passed, a spring or wire changes length by an amount (extension or compression) in direct proportion to the force applied.

F = kx

Here k is the force constant (also called the spring constant), measured in N m⁻¹. A large k means a stiff spring. The value of k belongs to one particular spring or wire.

Worked example 1. A force of 7.2 N stretches a spring by 45 mm. Find k, then the force that compresses the same spring by 30 mm.

x = 45 mm = 0.045 m. k = F/x = 7.2 / 0.045 = 160 N m⁻¹. F = kx = 160 × 0.030 = 4.8 N.

3.4.1(d) Force-extension graphs

Plot extension on the horizontal axis and force on the vertical axis, or the other way round if the question asks for it. For a spring obeying Hooke’s law the graph is a straight line through the origin. If force is on the vertical axis, the gradient is k. If extension is on the vertical axis, the gradient is 1/k.

A straight line through the origin shows Hooke’s law is obeyed. The point where the line starts to curve is the limit of proportionality. A steeper line means a stiffer spring. Compression can be plotted as negative extension with negative force; the line continues straight through the origin.

Rubber bands. A rubber band does not obey Hooke’s law. Its graph is a curve, and the unloading curve lies below the loading curve. The band returns to its original length, and the area between the curves is energy dissipated as thermal energy.

Polythene strips. A polythene strip extends roughly in proportion at first, then yields and stretches a long way for little extra force. It does not return to its original length, so it has deformed plastically.

Investigating force-extension (3.4.1(d)(ii))

This links to practical activity group 2. A sound method for a spring:

1. Hang the spring from a rigid clamp beside a vertical metre rule, with a pointer on the lowest coil. Read the position with your eye level with it, to avoid parallax.
2. Record the unloaded position. Add a 100 g slotted mass (weight 0.98 N) and record the new position, and so on in equal steps.
3. Unload in the same steps and record again. Agreement shows no permanent change.
4. Extension = position - unloaded position. Plot force against extension and find the gradient.

For a polythene strip or rubber band, wear eye protection and keep feet clear of falling masses. Use smaller steps where the graph starts to curve, to locate the limit of proportionality.

3.4.2(a)-(b) Work done and elastic potential energy

The work done by a force is force times distance moved in the direction of the force. When the force changes with extension, the work done in stretching is the area under the force-extension graph. For a straight line through the origin the area is a triangle:

E = ½Fx

Using F = kx gives

E = ½kx²

This is elastic potential energy, recovered when the spring returns to its natural length. Doubling the extension multiplies the stored energy by 4. The symbol E is also used for the Young modulus, so read the context.

Worked example 2. The spring in example 1 is stretched by 0.045 m. Find the stored energy two ways.

E = ½Fx = ½ × 7.2 × 0.045 = 0.162 J. E = ½kx² = ½ × 160 × 0.045² = 0.162 J.

Worked example 3. A toy launcher has k = 360 N m⁻¹ and is compressed by 50 mm. It fires a 45 g ball horizontally along a smooth surface. Find the ball’s speed, assuming all the stored energy becomes kinetic energy.

E = ½ × 360 × 0.050² = 0.45 J. ½mv² = 0.45, so v² = 2 × 0.45 / 0.045 = 20. v = 4.5 m s⁻¹ (2 s.f.).

When the graph is not a straight line

If the graph is not a single straight line, split the area into triangles and trapeziums, or count squares and multiply by the energy value of one square.

Worked example 4. A metal strip is loaded so that force rises linearly from 0 to 20 N as extension goes from 0 to 0.040 m, then linearly to 36 N at 0.100 m. Find the work done in stretching it.

Triangle: ½ × 20 × 0.040 = 0.40 J. Trapezium: ½ × (20 + 36) × (0.100 - 0.040) = 28 × 0.060 = 1.68 J. Total work = 0.40 + 1.68 = 2.08 J.

3.4.2(f) Elastic and plastic deformation

Elastic deformation: once the forces are taken off, the object goes back to exactly its starting dimensions. Plastic deformation: some of the change is permanent, so the object stays longer or bent. Plastic behaviour begins beyond the elastic limit. The elastic limit and the limit of proportionality are different ideas: one is about returning, the other about proportionality.

When a metal is unloaded after plastic deformation, the unloading line is approximately parallel to the initial loading line. The graph does not return to the origin, and the horizontal gap at zero force is the permanent extension.

In example 4 the initial slope is 20 / 0.040 = 500 N m⁻¹. Unloading from 36 N at 0.100 m reaches zero force at 0.100 - 36/500 = 0.028 m, the permanent extension.

Energy recovered = ½ × 36 × (0.100 - 0.028) = 1.30 J. The rest of the 2.08 J, 0.78 J, was dissipated as thermal energy in the metal.

3.4.2(c) Stress, strain and ultimate tensile strength

Force constant and extension depend on the size of the sample. Stress and strain allow materials to be compared.

• Tensile stress σ = F / A, where F is the force and A the cross-sectional area. Unit: pascal, Pa (N m⁻²).
• Tensile strain ε = x / L, where x is the extension and L the original length. Strain is a ratio and has no unit.
• Ultimate tensile strength (UTS): the maximum tensile stress a material can withstand before it breaks. Use the original cross-sectional area unless told otherwise.

For a wire, A = πd²/4.

Worked example 5. A wire of diameter 0.60 mm breaks when the tension reaches 480 N. Find the UTS.

A = π × (0.30 × 10⁻³)² = 2.83 × 10⁻⁷ m². UTS = 480 / 2.83 × 10⁻⁷ = 1.7 × 10⁹ Pa.

3.4.2(d)(i) The Young modulus

Within the limit of proportionality, stress is proportional to strain. The ratio is the Young modulus E:

E = tensile stress / tensile strain = (F/A) / (x/L) = FL / (Ax)

The unit is the pascal. The Young modulus belongs to the material, not the sample: two wires of the same steel have the same E but different k. Rearranging gives k = EA/L.

Worked example 6. A steel wire of original length 2.40 m and diameter 0.50 mm carries a load of 62 N and extends by 3.6 mm. Find E, and the energy stored.

A = π × (0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m². Stress = 62 / 1.96 × 10⁻⁷ = 3.16 × 10⁸ Pa. Strain = 3.6 × 10⁻³ / 2.40 = 1.5 × 10⁻³. E = 3.16 × 10⁸ / 1.5 × 10⁻³ = 2.1 × 10¹¹ Pa. Energy stored = ½Fx = ½ × 62 × 3.6 × 10⁻³ = 0.11 J.

3.4.2(d)(ii) Measuring the Young modulus of a metal

This is the practical named for practical activity group 2.

1. Use a long thin wire (about 2 m), clamped firmly at the top, with a small fixed weight at the lower end to keep it straight. Long and thin gives a measurable extension.
2. Measure the original length L from the clamp to the reference marker with a metre rule or tape.
3. Use a micrometer to find the diameter at several positions along the wire, in two perpendicular directions each time, and take the mean. Calculate A = πd²/4.
4. Add equal masses in steps, recording the marker position against a fixed scale each time.
5. Unload and repeat to check the wire stays elastic. Keep extensions small.
6. Plot extension (y) against force (x). The gradient is x/F = L/(EA), so E = L / (A × gradient).

For example, with L = 2.40 m, d = 0.50 mm and a gradient of 5.8 × 10⁻⁵ m N⁻¹, E = 2.40 / (1.96 × 10⁻⁷ × 5.8 × 10⁻⁵) = 2.1 × 10¹¹ Pa.

Uncertainty. A depends on d², so the percentage uncertainty in A is twice that in d. A diameter of 0.50 ± 0.01 mm is a 2% uncertainty in d and 4% in A. This is why the diameter is measured at many points. The extension is small, so its percentage uncertainty is also large.

Safety. Wear eye protection and put sand or foam below the masses.

3.4.2(e) Stress-strain graphs

Plot stress on the vertical axis and strain on the horizontal axis. The gradient of the straight part is the Young modulus, because gradient = stress/strain.

Ductile material (for example copper). A straight line through the origin to the limit of proportionality, then a curve. A yield point may appear, where the material extends a lot for little extra stress. The curve reaches a peak, the UTS, then the wire narrows (necks) and breaks.

Brittle material (for example glass). A straight line almost to the break, then sudden fracture with little or no plastic deformation.

Polymeric material. Rubber gives a curved graph; the loading and unloading curves differ, but it returns to its original length. A polythene strip extends a long way once it yields and does not return to its original length.

Worked example 7. The straight part of a metal’s stress-strain graph passes through (strain 0.0012, stress 240 MPa). Find E.

E = 240 × 10⁶ / 0.0012 = 2.0 × 10¹¹ Pa.

Common errors

• Using total length instead of extension for x.
• Leaving diameters in millimetres in A = πd²/4, or using the diameter as the radius.
• Saying stress is proportional to strain for all loads; it holds only within the limit of proportionality.
• Saying the Young modulus depends on the wire’s length or thickness.
• Using ½Fx on a curved graph.

Try the practice questions next.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 3.4.1 Springs and 3.4.2 Mechanical properties of matter.