— Physics · AS Level (7408) · Hooke's Law, Stress/Strain, Young Modulus —
★ Exam Tip: Stress and strain are ratios, not rates — read
graphs directly at a point, never from a gradient. When a question says "use Figure X to...", always use that
graph/area method, even if you know a direct formula shortcut — marks are tied to the specified technique.
1. Hooke's Law & Spring Constant
Deformation in solids: Change in shape (AS content: extension/compression only — any
other deformation type, e.g. twisting, is outside scope).
Hooke's law: "The tensile force applied on an object under stress is directly
proportional to the change in length of the object until the limit of proportionality is not exceeded."
F ∝ ΔL ⟹ F = KΔL
Spring constant: Ratio of the tensile force to the change in length of the spring. K =
F/ΔL. Constant for only ONE spring — not springs in general, not even two similar-looking
springs.
F-ΔL graph: straight to P (limit of proportionality), curving to E (elastic limit). Shaded area = EPE.
Intuition — K∝1/ΔL: same weight W on a high-K spring vs a low-K
spring: high K ⟹ small ΔL (stiffer, extends less); low K ⟹ large ΔL. "Springs with a higher spring constant extend
less for the same force."
2. Elastic Potential Energy & Key Points
Elastic potential energy: The energy stored in an object by virtue of its deformation.
E.P.E = ½ × ΔL × F | E.P.E = ½KΔL² (only valid till the limit of
proportionality)
Limit of proportionality
After this point, Hooke's law is not obeyed — an increment in force no longer causes a proportional
increment in change of length.
Elastic
Once deformed, object returns to its original shape/length.
Elastic limit
Point after which object does NOT return to its original shape/length — gets permanently (plastically)
deformed.
Plastic
Opposite of elastic — after deformation, does not return; e.g. play-doh. (Rubber band / ballpoint-pen spring
= elastic examples.)
Breaking point
The point after which the object ruptures — the absolute end of the graph.
★ Full logic chain: if loading stays BELOW the elastic limit (even
past the limit of proportionality, where the curve bends), the unloading curve EXACTLY RETRACES the loading curve
back to zero — exceeding the limit of proportionality only breaks Hooke's law's straight-line behaviour, NOT the
return to original length. Only exceeding the ELASTIC limit causes permanent deformation.
within proportionality: no losspast P, before E: still no losspast E: permanent set, loop = loss
★ Hysteresis loss = (area under loading curve) − (area under unloading
curve) = the enclosed loop area. Two destinations: 1) lost as heat (stretch a spring vigorously — it warms
slightly), 2) used to permanently deform the object (rearranging its structure costs energy).
4. Stress, Strain & Young Modulus
Stress: Ratio of tensile force applied to the cross-sectional area for a deformed solid.
Scalar. Stress = Force/Area (units: kg m⁻¹s⁻² = Pa)
Strain: Ratio of change in length to the original length of a deformed solid. Not a physical quantity (no units — m/m cancels). Strain = ΔL/L₀
Young Modulus: Ratio of the stress to strain of an object. Scalar.
Derivation:
E = Stress/Strain = (F/A)÷(ΔL/L₀) = (F/A)×(L₀/ΔL)
⟹ E = (F×L₀)/(A×ΔL)
★ E = (F × L₀)/(A × ΔL) (units: Pa)
★ Constant for a TYPE of material: two copper springs, different L₀
and A — E is the SAME (same material) but K will be DIFFERENT (K depends on geometry too). Same material does
NOT mean same spring constant — only same E; K=EA/L₀ still needs L₀ and A to match too.
The Significance of Young Modulus
E = (F×L₀)/(A×ΔL) ⟹ E∝F/ΔL (since L₀, A ≈ constant) | also K=F/ΔL ⟹ E∝K
"The greater the Young modulus, the less the extension compared to an object with a lower Young modulus."
★ Unifying insight: the ratio from a force-extension graph gives K;
the ratio from a stress-strain graph gives E. Since E=K×(L₀/A), these are "basically synonymous" — E is just K
rescaled by geometry to depend on the MATERIAL alone. A stress-strain graph looks qualitatively identical in shape
to a force-extension graph for the same material — same P and E points, only the axis scales differ.
Stress-Strain Graph → Strain Energy per Unit Volume
Area under stress-strain graph = strain energy PER UNIT VOLUME
★ Key distinction: Area under a force-extension graph = plain EPE.
Area under a STRESS-STRAIN graph = EPE PER UNIT VOLUME — always append "per unit volume," a commonly-lost
mark. (Derivation: ½×stress×strain = ½×(F/A)×(ΔL/L₀) = [½FΔL]/[A×L₀] = EPE/Volume.)
Sanity check: on a stress-strain graph, strain=0 automatically means ΔL=0
(zero extension), since strain=ΔL/L₀.
5. Searle's Experiment — Measuring Young Modulus
Purpose: To measure the Young Modulus.
★ Why a control wire? It compensates for changes NOT due to the load
— e.g. temperature changes cause thermal expansion in BOTH wires equally. Measuring the test wire's extension
RELATIVE to the identical, unloaded control wire cancels out thermal drift, leaving only the load-caused
extension.
Apparatus
Purpose
Test wire
Young modulus of this is required
Control wire
Identical to test wire (material, length, cross-sectional area) — cancels thermal effects
Slotted masses
For loading (precise, known increments)
Wooden blocks & clamps
To hold the wires in place
Main scale + Vernier/marker
To work out ΔL
Metre rule
To measure L₀ of wires (avoid parallax error)
Micrometer screw gauge
Measure wire diameter ⟹ A=πd²/4
Method:
Get a test wire and control wire — must be identical in every sense (E same, L₀ same, A same).
Measure L₀ with a metre rule (avoid parallax error).
Measure diameter with a micrometer screw gauge at ≥3 different points along the wire; average it. A=πr² or
πd²/4.
Set up the experiment as shown.
Attach small masses initially on BOTH wires to make them taut (removing coiling).
Add slotted masses on the test wire only. Allow it to settle, take the reading using marker and main scale
(avoid parallax errors).
Make a table.
E_avg = [(σ₁/s₁)+(σ₂/s₂)+(σ₃/s₃)] / 3 (average E from multiple stress-strain data points)
★ Alternative: use the F-ΔL graph's GRADIENT directly: E=(F×L₀)/(A×ΔL) ⟹
F=(EA/L₀)×ΔL (y=mx form) ⟹ gradient m = EA/L₀.
6. Springs in Series & Parallel
Series: same W acts on all; ΔL_net=ΔL₁+ΔL₂+ΔL₃Parallel: all springs extend by the SAME amount
Series derivation:
Each spring: W=K₁ΔL₁, W=K₂ΔL₂, W=K₃ΔL₃ ⟹ 1/K₁=ΔL₁/W etc.
"The same W acts on all 3 springs" ⟹ ΔL_net=ΔL₁+ΔL₂+ΔL₃
W=K_net×ΔL_net ⟹ 1/K_net = ΔL_net/W = ΔL₁/W+ΔL₂/W+ΔL₃/W
★ Series: 1/K_net = 1/K₁ + 1/K₂ + 1/K₃
Parallel derivation:
"All springs extend by the same amount" ⟹ W=K_net×ΔL_net; W=F₁+F₂+F₃
K_net×ΔL_net = (K₁×ΔL_net)+(K₂×ΔL_net)+(K₃×ΔL_net) — ΔL_net cancels from every term
★ Parallel: K_net = K₁ + K₂ + K₃
7. Worked MCQ Practice
Bolt with tapered cross-sectionMCQ · stress ratio
Bolt, circular cross-section, diameter 2d at X, diameter d at Y. Ratio (stress at Y)/(stress at X)?
Trick — tension is the same throughout: pulling a wire/cord from
both ends means every internal cross-section is in equilibrium — the tensile force is IDENTICAL everywhere along
it (F_X = F_Y), so it cancels out of the ratio entirely.
Rubber band stretched and relaxedMCQ · hysteresis areas
Force-extension graph: path OPQ (loading) to extension e, then QRO (unloading). Area X between the curves; Area
Y under QRO. Which statement is correct?
Area under the LOADING curve (X+Y combined) = total energy put in = minimum
energy required to stretch to e
Answer: B — (Area X + Area Y) is the minimum energy required to stretch the band to extension e
Suspension bridge cable — increase in strain energyMCQ · trapezium
area
Cable stretches 47mm under 19.3×10⁵N; a lorry increases the force to 23.3×10⁵N, extension to 57mm. Increase in
strain energy?
Trapezium area between the two points: ½×[(19.3+23.3)×10⁵]×(57−47)×10⁻³ ⟹
21,360 ≈ 21 kJ
Answer: A (21 kJ)
Which material is purely elastic?MCQ · 4 stress-strain graphs
4 loading/unloading stress-strain graphs. Which shows PURELY ELASTIC behaviour?
Purely elastic = loading and unloading curves are IDENTICAL (retrace exactly — no hysteresis
loop, no permanent deformation).
Answer: B (loading = unloading curve exactly)
Elastic cord between two pins, angled loadCalculation · Young
modulus
Cord unstretched L₀=16.0cm, A=2.0×10⁻⁶m², held by 2 smooth pins 8.0cm apart. Load 0.40kg hung centrally, angle
between the two sides=60°. Find E.
60° angle ⟹ equilateral triangle ⟹ each side = 8.0cm ⟹ stretched total =
8+8+8=24cm ⟹ ΔL=24−16=8cmW=2Tcos30°: (0.40)(9.81)=2×T×cos30 ⟹ T≈2.27NE=(F×L₀)/(A×ΔL) = (2.27×16×10⁻²)/(2×10⁻⁶×8×10⁻²) ⟹ ≈2.3×10⁶ Pa
Answer: C (2.3×10⁶ Pa)
Three springs, series + parallel combinedCalculation · combined
K
Springs P, Q identical (constant k) in parallel; combo in series with spring R (constant 3k). Increase in
overall length when force W applied?
Scale sensitive to small masses, less to largeMCQ · spring
arrangement design
Two springs, one low k, one high k. Which arrangement gives high sensitivity for SMALL masses but low
sensitivity for LARGE masses?
Sensitivity = "the smallest change you can measure." Small masses need e↑ ⟹ low K (responds with
large, readable extension). Large masses need e↓ ⟹ high K (doesn't over-extend/max out the scale).
Design logic: arrange so the LOW-k spring acts first (small loads
only stretch it, high sensitivity), and the HIGH-k spring only engages once a rigid box's internal slack is
taken up (for large loads).
Answer: A (outer low-k spring on a rigid box, inner high-k spring engages only for larger loads)
8. Structured Problems
Spring launches a toy car around a vertical loopStructured Q —
EPE→KE→GPE chain
Spring compressed 16.0cm, F-x graph shows EPE=0.64J at that compression. Toy car (0.076kg) leaves spring at X
with KE=initial EPE, travels a track including a 0.12m-radius vertical loop, to point Y.
(i) Speed at X: KE=EPE: 0.64=½(0.076)v² ⟹ v≈4.1 m/s(ii) KE at top of loop (A, height 0.24m): E_X=E_A: KE_X=GPE_A+KE_A ⟹
0.64=(0.076)(9.81)(0.24)+KE_A ⟹ KE_A≈0.46 J(iii) Speed at Y (same height as X, no losses assumed): same as X ≈4.1 m/s
(d) In practice, KE at Y is 0.23J LESS than at X (resistive force). Find average resistive force
X→Y.
Block at constant speed hits a springStructured Q — equilibrium
trap
Block 0.40kg slides at constant 0.30 m/s, hits a spring, decelerates to zero when compressed 8.0cm.
(i) Initial KE=½mv²=½(0.40)(0.30)² ⟹ 0.018 J(ii) F-x graph given, straight to F_MAX at x=8.0cm. KE=EPE: 0.018=½×F_MAX×(8×10⁻²) ⟹
F_MAX≈0.45 N(iii) Max deceleration: F=ma: 0.45=0.40×a ⟹ a≈1.1 ms⁻²
★ Common trap: momentarily having ZERO velocity does NOT mean
equilibrium — equilibrium requires zero RESULTANT FORCE, not zero speed. At the instant of max compression, the
spring's built-up tension is still acting with nothing to balance it.
(iv)1. Before hitting the spring: "Yes, as it is travelling with a constant speed." 2. When speed
becomes zero: "Not in equilibrium, as the tension developed provides a net force horizontally on the block."
Wire E, energy stored, comparison to a 2L/2d wireStructured Q —
comparison question
(i) Cross-sectional area measurement: "Measure the diameter of the wire at least 3 different points
using a micrometer screw gauge. Average the diameters and calculate the area using A=πd²/4."
(ii) E=(F×L₀)/(A×ΔL) = (4×2.5)/[(9.4×10⁻⁸)×(0.8×10⁻³)] ⟹ ≈1.3×10¹¹
Pa(iii) Increase in energy 2.0N→4.0N (trapezium): ½×(2+4)×(0.4×10⁻³) ⟹ 1.2×10⁻³ J
(c) New wire, SAME material, TWICE the length (2L₀), TWICE the diameter (2d). Compare the force
needed for the same extension.
★ Comparison-question method: write E for both wires, make F the
subject in each, substitute. A₂=π(2d)²/4=4A₁.
The new wire needs TWICE the force for the same extension — its F-x line is twice as steep.
Block hanging from a spring, pulled and releasedStructured Q — full
energy chain
Spring unstretched L₀=0.080m. Block hangs in equilibrium at X (spring length 0.095m). Pulled down to Y (spring
length 0.120m), released, shoots upward.
★ Exam-technique reminder: when a question says "use Figure X
to...", always use that graph/area method (e.g. the trapezium), even if you know ½KΔL² would give the same
number — marks are tied to demonstrating the specified technique.
✓ Quick Recap — Full Chapter Cheat-Sheet
Hooke's law: F=KΔL, valid only up to the limit of proportionality. K constant for ONE spring only.
E.P.E = ½FΔL = ½KΔL² (valid to limit of proportionality). Stress=F/A (Pa). Strain=ΔL/L₀ (no units — a pure
ratio).
E = (F×L₀)/(A×ΔL) — constant for a TYPE of material. Same material ⟹ same E, but NOT necessarily same
K (K also needs same L₀, A).
E∝K (same underlying meaning, different scale). Area under F-x graph = EPE. Area under stress-strain graph =
EPE per unit volume.
Loading past the elastic limit ⟹ hysteresis loop ⟹ energy lost as heat + permanent deformation. Within the
elastic limit (even past P): no loss, full retrace.
Searle's experiment: control wire cancels thermal expansion effects, not just "for comparison."