Hooke's Law | 9231 Further Mechanics

1. Hooke's Law — The Core Formula

When an elastic string or spring is stretched beyond its natural length, the tension in it is directly proportional to the extension. This is Hooke's Law.

⭐ Hooke's Law

T = λx / l

where λ = modulus of elasticity (N), x = extension (m), l = natural length (m)

Key Terminology

lNatural lengthunstretched length (m)

xExtensionamount stretched beyond l (m)

λModulus of elasticitymeasured in Newtons

TTension in string/springalways along the string

Physical Meaning of λ

λ is a property of the material and the string. It equals the force needed to double the string's length (extend it by l). A stiffer string has a larger λ.

💡 Alternative form

Sometimes written as T = kx where k = λ/l is the spring stiffness (N/m). Cambridge uses the λ/l form.

2. Strings vs Springs — Critical Difference

Left: slack elastic string (T=0 when not stretched). Right: compressed spring exerts thrust.

⚠ String vs Spring — Must Know

Elastic string: T = λx/l only when stretched (x > 0). If x ≤ 0, the string is slack and T = 0. A string cannot push.
Spring: can be both stretched (tension, T = λx/l) and compressed (thrust, T = λx/l with x = compression). A spring can push and pull.
Always check whether the string is taut before applying Hooke's Law.

3. Elastic Potential Energy (EPE)

When a string or spring is stretched or compressed, elastic potential energy is stored in it. This is the work done against the elastic force during extension.

⭐ Elastic Potential Energy Formula

EPE = λx² / (2l)

Also written as ½kx² where k = λ/l. Units: Joules (J)

Derivation (for understanding)

The tension T = λx/l varies with extension x. The work done stretching from 0 to x is:

EPE= ∫₀ˣ T dx = ∫₀ˣ (λx/l) dx = λx²/(2l)integration

This derivation is not required by Cambridge, but understanding it solidifies why the formula has the x² term — it is the area under the force-extension graph (a triangle).

4. Energy Conservation with Elastic Strings

The total mechanical energy of a system involving an elastic string is conserved (assuming no friction or air resistance):

🔑 Conservation of Energy — Extended Form

KE + GPE + EPE= constant

½mv²+ mgh + λx²/2l= constant

This is the most powerful tool in Hooke's Law problems. Apply it between two positions where enough information is known.

Important: When is EPE Zero?

EPE = 0 when the string/spring is at its natural length or slack (x = 0). When the particle passes through the natural length position, all EPE has been converted to KE and GPE.

Equilibrium Position with a Hanging Elastic String

A mass m hanging on an elastic string of natural length l and modulus λ reaches equilibrium at extension e:

T = mgat equilibrium

λe/l= mg

e= mgl/λequilibrium extension

5. Two Springs / Strings Connected in Series

When two springs are joined end-to-end and both are stretched, they have the same tension throughout (light springs). The extensions add up.

Springs in series: same tension T, extensions x₁ and x₂

T= λ₁x₁/l₁ = λ₂x₂/l₂equal tension

Total extension= x₁ + x₂

📐 Exam Note — String Becomes Slack

For a string between two fixed points: it becomes slack when the particle moves past the natural length position. At that moment T drops to zero and EPE = 0.
After the string goes slack, the particle moves under gravity alone (projectile or vertical free fall) until the string becomes taut again.
This creates a two-phase motion — elastic phase and free phase — very common in 9231 questions.

Example 1 Equilibrium Extension — Finding λ ★☆☆ Standard

A particle of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.8 m. The other end is fixed to a ceiling. The particle hangs in equilibrium with the string extended by 0.3 m. Find (a) the modulus of elasticity, (b) the total length of the string in equilibrium. (g = 9.8 m/s²)

1

Equilibrium condition

At equilibrium T = mg. Using Hooke's Law T = λx/l:

λ × 0.3 / 0.8= 0.5 × 9.8 = 4.9 N

λ= 4.9 × 0.8 / 0.3 = 13.07 N

2

Part (b) — Total length

Total length= l + x = 0.8 + 0.3 = 1.1 m

Example 2 Energy Method — Maximum Extension ★★☆ Challenging

A particle of mass 0.4 kg is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 24 N. The other end is fixed to a point O on a ceiling. The particle is held at O and released from rest. Find (a) the position where the particle first comes to instantaneous rest, (b) the speed of the particle as it passes through the natural length position. (g = 9.8 m/s²)

1

Set up — define positive direction and reference levels

Take O as origin, downward positive. GPE reference: O. At O: KE = 0, GPE = 0, EPE = 0 (string natural length = 0 extension).

2

Part (a) — Maximum extension (instantaneous rest)

String becomes taut when particle is 1.2 m below O. After that, let extension = x. At maximum extension, v = 0.

Distance below O = 1.2 + x, so GPE lost = mg(1.2 + x):

Energy conserved:KE + EPE = GPE gained from O

0 + λx²/(2l)= mg(1.2 + x)

24x²/(2×1.2)= 0.4 × 9.8 × (1.2 + x)

10x²= 3.92(1.2 + x) = 4.704 + 3.92x

10x² − 3.92x − 4.704= 0

3

Solve the quadratic

x= [3.92 ± √(15.37 + 188.16)] / 20 = [3.92 ± √203.53] / 20

x= [3.92 ± 14.27] / 20

x= 0.910 m (taking positive root)

Particle first rests at 1.2 + 0.910 = 2.11 m below O.

4

Part (b) — Speed at natural length position (x = 0, 1.2 m below O)

At natural length: EPE = 0. Energy from O to this point:

½mv²= mg × 1.2 = 0.4 × 9.8 × 1.2 = 4.704 J

v²= 2 × 4.704 / 0.4 = 23.52

v= 4.85 m/s

Example 3 Two Springs Between Fixed Points ★★☆ Challenging

A light spring PQ has natural length 1.6 m and modulus 20 N. A second spring QR has natural length 1.4 m and modulus 28 N. The ends P and R are fixed to two points 4 m apart on a horizontal surface. A particle is attached at Q. Find the extension in each spring when the system is in equilibrium.

1

Setup — equal tension throughout (horizontal, no gravity component)

Total natural length = 1.6 + 1.4 = 3.0 m. Distance between P and R = 4 m. Total extension = 1.0 m.

Let x₁ = extension in PQ, x₂ = extension in QR. Then x₁ + x₂ = 1.0 m.

2

Equal tension condition

T₁ = T₂⟹ 20x₁/1.6 = 28x₂/1.4

12.5x₁= 20x₂

x₁= 1.6x₂

3

Solve simultaneously

1.6x₂ + x₂= 1.0 ⟹ 2.6x₂ = 1.0

x₂= 0.385 m

x₁= 1.6 × 0.385 = 0.615 m

T= 12.5 × 0.615 = 7.69 N

Example 4 String Becomes Slack — Two-Phase Motion ★★★ A* Level

A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus 15 N, the other end fixed to point O. The particle is projected vertically downward from O with speed 2 m/s. Find (a) the maximum extension of the string, (b) the speed of the particle when it returns to O. (g = 10 m/s²)

1

Part (a) — Maximum extension

At O: KE = ½(0.3)(4) = 0.6 J, GPE = 0, EPE = 0. Let max extension = x, particle is at 0.5 + x below O:

0.6 + 0.3×10×(0.5+x)= λx²/(2l) = 15x²/(2×0.5) = 15x²

0.6 + 1.5 + 3x= 15x²

15x² − 3x − 2.1= 0

50x² − 10x − 7= 0

x= [10 ± √(100 + 1400)] / 100 = [10 ± √1500] / 100

x= (10 + 38.73)/100 = 0.487 m

2

Part (b) — Speed when returning to O

By conservation of energy (no energy loss): when particle returns to O, it has the same energy as at the start. KE at O (upward) = initial KE = 0.6 J.

½(0.3)v²= 0.6

v= 2 m/s (upward)

The motion is symmetric in energy — the particle returns to O with the same speed it left with. This is because the whole path from O back to O involves no net change in GPE and no net change in EPE (string taut only below natural length).

Interactive Spring Simulator

Explore Hooke's Law and energy storage visually. Switch between the static Hooke's Law explorer and the dynamic energy conservation simulator.

🔧 Spring & String Visualiser

Natural Length l (m) 1.0 m

Modulus λ (N) 20 N

Extension x (m) 0.4 m

—Tension T (N)

—EPE (J)

—Total Length (m)

—Stiffness k (N/m)

Elastic Potential Energy stored

Practice Questions

Question 1 — Hooke's Law Basics

An elastic spring has natural length 0.6 m and modulus of elasticity 30 N. Find (a) the tension when the spring is extended by 0.2 m, (b) the compression force when the spring is compressed by 0.1 m, (c) the extension when a force of 20 N is applied.

Apply T = λx/l directly. For compression, x is the compression amount and the formula gives the thrust (pushing force). For part (c), rearrange: x = Tl/λ.

✓ Solution

(a) T= 30 × 0.2 / 0.6 = 10 N

(b) T= 30 × 0.1 / 0.6 = 5 N (thrust/compression)

(c) x= 20 × 0.6 / 30 = 0.4 m

Question 2 — Equilibrium and EPE

A particle of mass 2 kg hangs in equilibrium on an elastic string of natural length 1.5 m and modulus 40 N. Find (a) the extension, (b) the total length of the string, (c) the elastic potential energy stored. (g = 9.8 m/s²)

At equilibrium T = mg. So λx/l = mg gives x. Then EPE = λx²/(2l).

✓ Solution

(a) x= mgl/λ = 2×9.8×1.5/40 = 29.4/40 = 0.735 m

(b) Total= 1.5 + 0.735 = 2.235 m

(c) EPE= 40×(0.735)²/(2×1.5) = 40×0.540/3 = 7.20 J

Question 3 — Energy Conservation

A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus 25 N. The other end is fixed to a ceiling. P is held at the ceiling and released from rest. Find (a) the speed of P as it passes through the natural length position, (b) the maximum extension of the string. (g = 10 m/s²)

For (a): from ceiling to natural length (0.8 m drop), EPE = 0, so ½mv² = mgh = 0.5×10×0.8. For (b): set up energy equation from ceiling to maximum extension x, where total drop = 0.8 + x.

✓ Solution

(a) ½(0.5)v²= 0.5×10×0.8 = 4 J → v = √16 = 4 m/s

For (b): at max extension x, v = 0:

25x²/(2×0.8)= 0.5×10×(0.8+x)

15.625x²= 4 + 5x

15.625x² − 5x − 4= 0

x= [5 ± √(25+250)]/31.25 = [5 ± 16.58]/31.25

x= 21.58/31.25 = 0.691 m

Question 4 — A* String and Projectile

A particle of mass 0.2 kg is attached to one end of a light elastic string of natural length 1 m and modulus 10 N. The other end is fixed to a point O. The particle is released from rest at O. Find (a) the speed when the string becomes taut again on the way up (i.e. when the particle returns through the natural length position moving upward), and (b) the greatest height above O reached by the particle after the string goes slack. (g = 10 m/s²)

The motion is symmetric. The particle returns to the natural length position with the same speed as it passed through it on the way down. Once the string goes slack (particle above natural length), it moves as a free projectile under gravity alone.

✓ Solution

(a) At natural length on way down (1 m below O): ½mv² = mgl = 0.2×10×1 = 2 J. v = √20 m/s. By symmetry, speed returning upward through natural length = √20 = 4.47 m/s (upward).

(b) Once string goes slack, particle is at 1 m below O with speed 4.47 m/s upward. Under gravity alone:

Additional height= v²/(2g) = 20/(2×10) = 1.0 m above natural length

Above O= 1.0 − 1.0 = 0 m

The particle just reaches O — a beautiful symmetric result. The maximum height equals the starting point O.

Formula Reference Sheet

Complete reference for Hooke's Law — Cambridge 9231 P3, Section 3.4.

Hooke's Law

T = λx / ltension / thrust

λmodulus of elasticity (N)

lnatural length (m)

xextension or compression (m)

k = λ/lspring stiffness (N/m)

Elastic Potential Energy

EPE = λx² / (2l)

= ½kx²alternative form

EPE = 0 when x = 0at natural length

Units: Joules (J)

Conservation of Energy

KE + GPE + EPE = const

½mv² + mgh + λx²/2l = const

Use between 2 known states

Take GPE = 0 at convenient level

Equilibrium Extension

T = mg at equilibrium

e = mgl / λextension at equilib.

Check x > 0 (string taut)

String vs Spring

String: T = 0 when slack

String: x ≥ 0 always

Spring: tension or thrust

Spring: x can be + or −

Two Springs in Series

T₁ = T₂ = T (same tension)

x₁ + x₂ = total extension

λ₁x₁/l₁ = λ₂x₂/l₂

📋 Cambridge Exam Strategy — Hooke's Law

Identify the state clearly — is the string taut or slack? If slack, T = 0 and EPE = 0.
For energy problems, always write out the full energy equation before substituting — examiners follow your method.
When the particle hangs in equilibrium, the equilibrium position is NOT the position of zero EPE — it is where T = mg.
For two-phase motion: solve the elastic phase first (string taut), then the free phase (string slack, no EPE) separately.
State GPE reference level explicitly: "Taking the initial position as zero GPE level..."
The maximum speed of a falling particle on a string occurs at the equilibrium position, not the natural length position.