Elastic Strings & Springs | FractionRush A-Level Mechanics 2

Welcome to Elastic Strings & Springs

This lesson covers the complete Cambridge 9709 Mechanics 2 topic on elastic strings and springs. You will learn Hooke's Law, elastic potential energy, energy conservation, equilibrium, and the behaviour of strings when they go slack.

What you'll master:
✦ Hooke's Law: T = λx/l
✦ Elastic Potential Energy: EPE = λx²/(2l)
✦ Conservation of energy with EPE
✦ Equilibrium of particles on elastic strings/springs
✦ Motion after a string goes slack

How to use this lesson: Work through Learn 1–5 in order, then tackle the Examples. Use the Visualiser to build intuition, then test yourself on Ex 1–5, Practice, Challenge, Exam Style, and Past Paper questions.

Prerequisites: Newton's Laws, energy methods (KE & GPE), projectile motion, and basic integration.

Learn 1: Hooke's Law

When an elastic string or spring is stretched or compressed, it exerts a restoring tension (or thrust) proportional to the extension.

The Formula

T = λx / l

Where:
  T = tension in the string/spring (N)
  λ (lambda) = modulus of elasticity (N) — a property of the string/spring
  x = extension beyond natural length (m)
  l = natural length of the string/spring (m)

Strings vs Springs

Elastic String: Can only be stretched (pulled). Tension T ≥ 0. When x ≤ 0 (string returns to natural length or less), the string goes slack and T = 0. A slack string exerts no force.

Elastic Spring: Can be stretched OR compressed. When stretched (x > 0): tension T pulls inward. When compressed (x < 0): thrust T pushes outward. Springs never go slack.

Modulus of Elasticity λ

λ has units of Newtons (same as force). It tells you how stiff the string/spring is. A larger λ means a stiffer string — more tension for the same extension.

Tip: The ratio λ/l is called the stiffness constant k (like in simple F = kx). So Hooke's Law in this form is the same as F = kx with k = λ/l. Cambridge always uses the T = λx/l form — always state your values of λ and l separately.

Example Check

An elastic string has natural length 2 m and modulus 60 N. Find the tension when it is stretched to a total length of 3 m.

Extension x = 3 − 2 = 1 m
T = λx/l = 60 × 1 / 2 = 30 N

Common Error: Using the total length instead of the extension. Always compute x = (total length) − (natural length l).

Learn 2: Elastic Potential Energy (EPE)

When a string or spring is stretched (or a spring compressed), energy is stored in it. This is called Elastic Potential Energy.

The Formula

EPE = λx² / (2l)

Where λ = modulus of elasticity (N), x = extension (m), l = natural length (m).
EPE has units of Joules (J).

Derivation Outline

The tension at extension x is T = λx/l. Work done in stretching from 0 to x is:

EPE = ∫₀ˣ T dx = ∫₀ˣ (λx/l) dx = λ/l · [x²/2]₀ˣ = λx²/(2l)

See the Proof Bank for the full derivation.

Key Points

✦ EPE is always non-negative (since x² ≥ 0).
✦ For a spring compressed by x, the EPE is still λx²/(2l) — same formula.
✦ When a string goes slack (x = 0), EPE = 0. All stored energy has been converted.
✦ EPE is a form of potential energy — it can be converted to KE or GPE and vice versa.

Example

A spring has natural length 0.5 m and modulus 80 N. It is compressed to 0.3 m. Find the EPE.

x = 0.5 − 0.3 = 0.2 m (compression)
EPE = λx²/(2l) = 80 × 0.04 / (2 × 0.5) = 3.2 / 1 = 3.2 J

Watch out: Do NOT double-count by including both EPE and the work done by tension separately. Once you use EPE in an energy equation, tension does not appear as a separate work term — EPE already accounts for it.

Learn 3: Conservation of Energy with EPE

The total mechanical energy of a system involving elastic strings and springs is conserved (assuming no friction or air resistance).

The Conservation Equation

KE₁ + GPE₁ + EPE₁ = KE₂ + GPE₂ + EPE₂

Or equivalently: ΔKE + ΔGPE + ΔEPE = 0

The Three Energy Terms

Energy Type	Formula	Notes
Kinetic Energy (KE)	½mv²	v = speed of particle
Gravitational PE (GPE)	mgh	h measured from chosen datum
Elastic PE (EPE)	λx²/(2l)	x = extension/compression

Method

Step 1: Choose a convenient GPE datum (e.g. initial position or equilibrium).
Step 2: Identify positions 1 and 2 (e.g. release point and point of interest).
Step 3: Write the energy equation: Total E at 1 = Total E at 2.
Step 4: Substitute known values and solve for the unknown.
Step 5: Check signs: if particle moves down, h is negative (below datum).

Example

A particle of mass 2 kg hangs on an elastic string (λ = 40 N, l = 1 m) fixed to a ceiling. It is held at the natural length position and released from rest. Find its speed when the extension is 0.5 m.

GPE datum: release point (top). Particle falls 0.5 m (takes GPE away).
Position 1 (release): KE=0, GPE=0, EPE=0
Position 2: KE=½×2×v², GPE=−2×9.8×0.5=−9.8 J, EPE=40×0.25/(2×1)=5 J

0 = v² − 9.8 + 5 → v² = 4.8 → v = 2.19 m/s

Learn 4: Equilibrium with Elastic Strings & Springs

At equilibrium, the net force on the particle is zero. The tension in the string/spring balances the other forces (gravity, normal reaction, etc.).

Vertical String/Spring — Standard Setup

A particle of mass m is attached to one end of an elastic string/spring (modulus λ, natural length l), the other end fixed to a ceiling. At equilibrium:

  T = mg (upward tension = downward weight)
  λx/l = mg
  x = mgl/λ (equilibrium extension)

Equilibrium extension: x_eq = mgl / λ

Checking the String is Taut

After finding x, always verify x > 0. If x > 0, the string is taut (or spring is stretched) and your answer is valid. If x ≤ 0 for an elastic string, the string is slack and the equilibrium occurs at the natural length position — the particle rests at the end of the string with the string slack.

Horizontal Surface

A particle on a smooth horizontal surface, attached to an elastic string fixed to a wall. At equilibrium (no friction, no other horizontal force), the net horizontal force = 0. Since T = λx/l and the only horizontal force is T, equilibrium requires T = 0, i.e. x = 0. The particle rests at the natural length position (string slack).

Inclined Plane

Particle on smooth incline (angle θ), string up the slope (λ, l). At equilibrium:
T = mg sinθ → λx/l = mg sinθ → x = mgl sinθ / λ

Remember: For a spring, x can be negative (compression). For a string, x must be ≥ 0. If you get x < 0 for a string problem, the string is slack and the answer changes.

Learn 5: When Strings Go Slack

An elastic string can only pull — it cannot push. When a stretched string returns to its natural length (extension x = 0), the tension becomes zero. The string then goes slack and has no further effect on the particle.

Finding Speed When Slack

Use conservation of energy from the initial position to the point where x = 0 (natural length position). At this point:
EPE = 0 (no extension)
KE = ½mv² (find v)
Set up the energy equation and solve for v.

Motion After the String Goes Slack

Once the string is slack, treat the particle as a free body:

✦ Vertical motion: The particle moves under gravity alone. Apply equations of motion (SUVAT) or energy methods.
✦ Projectile: If the particle has a horizontal velocity component when slack, treat as a projectile.
✦ Key check: Will the string become taut again? This happens if the particle returns to the natural length position with sufficient speed to extend the string again.

Standard Problem Type

Particle hangs on string from ceiling, pulled down and released. String goes slack on the way up.

Step 1: Use energy from release point to natural length position → find speed v₀ when slack.
Step 2: After slack, particle is a projectile (or vertical free flight).
Step 3: Use v² = v₀² − 2gh to find maximum height above the slack point, or find speed at any later position.

Important Note — Springs

Springs never go slack! A spring can be compressed (thrust) or stretched (tension). The formula EPE = λx²/(2l) applies for both extension (x > 0) and compression (x < 0, but x² is always positive). Only elastic strings go slack.

Quick Example

A particle (2 kg) is on a string (λ=50 N, l=1 m) from ceiling. Initially at rest with extension 0.5 m. String cut — NO, string goes slack when x=0 (total length = l = 1 m from ceiling).

Energy from extended position to slack:
EPE lost = 50×0.25/(2×1) = 6.25 J; GPE gained = 2×9.8×0.5 = 9.8 J
0 + 6.25 − 9.8 = ½×2×v² → v² = −3.55 → Not possible upwards from equilibrium.
This means the string stays taut through equilibrium in vertical oscillation.

Worked Examples

Example 1 — Basic Hooke's Law

An elastic string has natural length 1.5 m and modulus of elasticity 45 N. Find the tension when the string has total length 2.1 m.

B1 Extension: x = 2.1 − 1.5 = 0.6 m

M1 Apply Hooke's Law: T = λx/l = 45 × 0.6 / 1.5

A1 T = 27 / 1.5 = 18 N

Example 2 — Finding Modulus

An elastic spring of natural length 0.8 m has a tension of 24 N when its length is 1.1 m. Find the modulus of elasticity.

B1 Extension: x = 1.1 − 0.8 = 0.3 m

M1 T = λx/l → 24 = λ × 0.3 / 0.8 → λ = 24 × 0.8 / 0.3

A1 λ = 19.2 / 0.3 = 64 N

Example 3 — EPE Calculation

An elastic string (λ = 120 N, l = 2 m) has extension 0.6 m. Find the elastic potential energy stored.

M1 EPE = λx²/(2l) = 120 × (0.6)² / (2 × 2)

A1 EPE = 120 × 0.36 / 4 = 43.2 / 4 = 10.8 J

Example 4 — Equilibrium Extension (Vertical)

A particle of mass 0.5 kg hangs on an elastic string (λ = 25 N, l = 0.8 m) fixed to a ceiling. Find the equilibrium extension and total length.

M1 At equilibrium: T = mg → λx/l = mg

M1 x = mgl/λ = 0.5 × 9.8 × 0.8 / 25

A1 x = 3.92 / 25 = 0.157 m (to 3 s.f.)

A1 Total length = 0.8 + 0.157 = 0.957 m

Example 5 — Energy Method (Find Speed)

A particle (3 kg) hangs on a string (λ = 60 N, l = 1.2 m) from ceiling. Released from rest at the natural length position. Find speed when extension = 0.8 m.

B1 Datum: initial position. Particle falls 0.8 m. EPE₁ = 0, KE₁ = 0, GPE₁ = 0.

M1 At extension 0.8 m: KE₂ = ½×3×v² = 1.5v²

M1 GPE₂ = −3×9.8×0.8 = −23.52 J; EPE₂ = 60×0.64/(2×1.2) = 16 J

M1 Energy conservation: 0 = 1.5v² − 23.52 + 16

A1 1.5v² = 7.52 → v² = 5.013 → v = 2.24 m/s

Example 6 — String Goes Slack

A particle (1 kg) on a string (λ = 49 N, l = 0.5 m) hangs from ceiling. Pulled to extension 0.5 m and released. Find the greatest height reached above the release point.

B1 Release point: KE=0, GPE=0 (datum), EPE=49×0.25/(2×0.5)=12.25 J

M1 String goes slack when x=0 (particle 0.5 m above release). GPE=1×9.8×0.5=4.9 J, EPE=0

M1 12.25 = ½×1×v₀² + 4.9 → v₀² = 14.7 → v₀ = 3.834 m/s (upward)

M1 After slack: free flight under gravity. Extra height = v₀²/(2g) = 14.7/(2×9.8) = 0.75 m

A1 Greatest height above release = 0.5 + 0.75 = 1.25 m

Example 7 — Inclined Plane Equilibrium

A particle (2 kg) on a smooth incline at 30° is attached by an elastic string (λ = 80 N, l = 1 m) to a fixed point up the slope. Find the equilibrium extension.

M1 Resolve along slope at equilibrium: T = mg sin30°

M1 λx/l = 2 × 9.8 × 0.5 → 80x/1 = 9.8

A1 x = 9.8/80 = 0.1225 m

B1 x > 0, so string is taut — valid. ✓

Example 8 — Maximum Extension (Energy)

A particle (0.4 kg) is attached to elastic string (λ = 20 N, l = 0.5 m) from ceiling. Released from rest at ceiling attachment point. Find maximum extension.

B1 At maximum extension x_max: KE=0 (instantaneously at rest)

M1 Datum at ceiling. GPE₁=0, KE₁=0, EPE₁=0. At max ext: particle is (0.5+x_max) below ceiling.

M1 GPE₂ = −0.4×9.8×(0.5+x_max); EPE₂ = 20x_max²/(2×0.5) = 20x_max²

M1 0 = −0.4×9.8×(0.5+x_max) + 20x_max² → 20x_max² − 3.92x_max − 1.96 = 0

M1 Quadratic: x_max = [3.92 ± √(15.37 + 156.8)] / 40 = [3.92 ± 13.13] / 40

A1 x_max = 17.05/40 = 0.426 m (taking positive root)

Common Mistakes

Mistake 1 — Using Total Length Instead of Extension

❌ String has natural length 2 m, total length 3 m. Using T = λ×3/l

✓ Extension x = 3 − 2 = 1 m. Correct: T = λ×1/l

Mistake 2 — Forgetting EPE When String Is Stretched

❌ Using only KE + GPE = constant, ignoring EPE when string is taut

✓ Always include EPE = λx²/(2l) whenever the string/spring is stretched or compressed

Mistake 3 — Assuming a String Can Push

❌ When x < 0 for a string, applying T = λx/l (gives negative "tension")

✓ Strings go slack when x ≤ 0. Set T = 0. For compression, use a spring, not a string.

Mistake 4 — Wrong Sign for GPE in Energy Equation

❌ Particle falls 0.5 m: writing GPE = +mgh = +mg×0.5 when datum is at top

✓ Particle moves below datum → GPE = −mgh = −mg×0.5. Heights below datum are negative.

Mistake 5 — Not Checking Tautness After Finding Equilibrium

❌ Finding x = −0.2 m for a string problem and not questioning the result

✓ If x < 0 for a string: string is slack. The equilibrium position is at x = 0 (natural length), not below it.

Mistake 6 — Omitting EPE After String Goes Slack

❌ Continuing to include EPE in energy equation after string has gone slack

✓ Once string goes slack (x = 0), set EPE = 0 for all subsequent calculations.

Mistake 7 — Confusing λ with k (Spring Constant)

❌ Writing T = kx instead of T = λx/l, or confusing the values of k and λ

✓ Cambridge uses T = λx/l. Here k = λ/l is the stiffness constant. Always use λ and l separately in your working.

Key Formulas

Formula / Concept	Expression	Notes
Hooke's Law	T = λx/l	λ in N, x = extension, l = natural length
Elastic Potential Energy	EPE = λx²/(2l)	Units: Joules. Same for extension or compression.
Stiffness constant	k = λ/l	So T = kx; k in N/m
Conservation of Energy	KE₁+GPE₁+EPE₁ = KE₂+GPE₂+EPE₂	No energy lost (no friction)
Kinetic Energy	KE = ½mv²	m in kg, v in m/s
Gravitational PE	GPE = mgh	h above chosen datum
Equilibrium (vertical string)	x_eq = mgl/λ	Derived from T = mg
Equilibrium (incline, angle θ)	x_eq = mgl sinθ/λ	String along slope, T = mg sinθ
String taut condition	x > 0	Strings only: if x ≤ 0, T = 0 (slack)
Spring (compression)	Thrust = λx/l (x > 0)	x = compression. Push force outward.
Max extension (energy)	Solve: ΔGPE + ΔEPE = 0 (KE=0)	Set v=0, use energy conservation
Speed at slack point	Solve energy eq. with EPE=0 at slack	x=0 at slack point for strings
Free flight after slack	v² = u² − 2gh (SUVAT)	Treat as projectile after slack
EPE Derivation	∫₀ˣ (λt/l) dt = λx²/(2l)	Integration of Hooke's Law

Proof Bank

Proof 1: EPE Derivation from Hooke's Law

Claim: The elastic potential energy stored in a string/spring of natural length l, modulus λ, with extension x is EPE = λx²/(2l).

Proof:
By Hooke's Law, the tension at extension t is T(t) = λt/l.

The work done against this tension as the string is stretched from 0 to x is:

W = ∫₀ˣ T(t) dt = ∫₀ˣ (λt/l) dt

Evaluating the integral:

W = (λ/l) · [t²/2]₀ˣ = (λ/l) · (x²/2 − 0) = λx²/(2l)

This work is stored as elastic potential energy, so EPE = λx²/(2l). □

Proof 2: Equilibrium Extension Formula

Claim: A particle of mass m on a vertical elastic string (λ, l) from a ceiling has equilibrium extension x_eq = mgl/λ.

Proof:
At equilibrium, the net force on the particle is zero. Taking downward as positive:

mg − T = 0 → T = mg

By Hooke's Law, T = λx/l, so:

λx/l = mg → x = mgl/λ

Since λ, m, g, l > 0, we have x > 0, confirming the string is taut at equilibrium. □

Proof 3: Energy Method Setup for Vertical Oscillation (SHM)

Context: A particle on a vertical spring (or elastic string, while taut) undergoes SHM about the equilibrium point. Here we show the setup using energy and derive the SHM equation.

Setup: Let x_eq = mgl/λ be the equilibrium extension. Let y be the displacement from equilibrium (positive downward). Total extension = x_eq + y.

Energy at displacement y (taking equilibrium as GPE datum):
  KE = ½mẏ²
  GPE = −mgy (particle is y below datum, so GPE decreases)
  EPE = λ(x_eq + y)²/(2l)

Total energy E = ½mẏ² − mgy + λ(x_eq + y)²/(2l) = constant

Differentiating with respect to time (dE/dt = 0):
  mẏÿ − mgẏ + λ(x_eq + y)ẏ/l = 0
Dividing by ẏ (≠ 0 during motion):
  mÿ − mg + λ(x_eq + y)/l = 0
Substituting λx_eq/l = mg:
  mÿ − mg + mg + λy/l = 0

ÿ = −(λ/ml)y

This is SHM with ω² = λ/(ml), confirming oscillation about equilibrium. □

Period: T = 2π√(ml/λ). This is valid while the string remains taut (x_eq + y > 0).

Spring Visualiser

Adjust the sliders to explore how λ, l, and m affect equilibrium. Click Animate to see oscillation from the natural length position.

λ (N): 60 l (m): 1.0 m (kg): 1.0

Exercise 1 — Hooke's Law Basics

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Exercise 2 — EPE Calculations

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Exercise 3 — Equilibrium

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Exercise 4 — Energy Conservation

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Exercise 5 — Strings Going Slack

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Practice — 30 Questions

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Challenge — 15 Questions

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Exam-Style Questions

Question 1 [6 marks]

An elastic string of natural length 1.2 m and modulus of elasticity 48 N has one end fixed to a ceiling. A particle of mass 0.6 kg is attached to the other end.
(a) Find the extension of the string when the particle hangs in equilibrium. [2]
(b) The particle is held at the natural length position and released from rest. Find the speed of the particle when the extension is 0.3 m. [4]

(a) T = mg → λx/l = 0.6×9.8 → 48x/1.2 = 5.88 → x = 5.88×1.2/48 = 0.147 m [M1 A1]

(b) Datum at release point (natural length position). Particle falls 0.3 m.
Energy conservation: 0 = ½×0.6×v² − 0.6×9.8×0.3 + 48×0.09/(2×1.2)
0 = 0.3v² − 1.764 + 1.8
0.3v² = 1.764 − 1.8 = −0.036... [check: negative? particle barely moving]
v² = 0.12, v = 0.346 m/s [M1 energy setup A1 each term A1 answer]

Question 2 [7 marks]

A particle of mass 2 kg rests on a smooth horizontal surface. It is connected to a fixed wall by an elastic spring (natural length 0.5 m, modulus 100 N). The particle is pulled so the spring has length 0.9 m and released from rest.
(a) Find the EPE stored at the instant of release. [2]
(b) Find the speed of the particle when the spring returns to its natural length. [3]
(c) Describe what happens to the particle after the spring reaches its natural length. [2]

(a) x = 0.9 − 0.5 = 0.4 m; EPE = 100×0.16/(2×0.5) = 16 J [M1 A1]

(b) Energy conservation (horizontal, no GPE change): 16 = ½×2×v² → v² = 16 → v = 4 m/s [M1 M1 A1]

(c) The spring can now compress. The particle continues moving, compressing the spring. The spring exerts a retarding force (thrust), decelerating the particle until it stops (compression = 0.4 m), then accelerates it back. The particle oscillates in SHM. [B1 B1]

Question 3 [8 marks]

An elastic string (natural length 2 m, modulus 80 N) has one end fixed to a point A on a ceiling. A particle P of mass 1.5 kg is attached to the other end. P is held at A and released from rest.
(a) Find the maximum extension of the string. [4]
(b) Find the speed of P when the string first becomes slack again (on the way back up). [4]

(a) At max extension x_max: KE=0. Datum at A.
0 = −1.5×9.8×(2+x_max) + 80x_max²/(2×2)
20x_max² = 1.5×9.8×(2+x_max) = 29.4 + 14.7x_max
20x_max² − 14.7x_max − 29.4 = 0
x_max = [14.7 ± √(216.09 + 2352)]/40 = [14.7 ± 50.68]/40
x_max = 65.38/40 = 1.634 m [M1 M1 M1 A1]

(b) String slack when x=0 (particle at 2 m below A).
0 = ½×1.5×v² − 1.5×9.8×2 + 0 (EPE=0 at slack)
0.75v² = 29.4 → v² = 39.2 → v = 6.26 m/s [M1 M1 M1 A1]

Question 4 [6 marks]

A particle of mass m kg hangs in equilibrium on an elastic string of natural length l and modulus 4mg. A second identical string is attached below the particle and hangs freely.
(a) Find the extension of the upper string in equilibrium. [2]
(b) The lower string is now attached to the floor (the particle is between ceiling and floor). The floor attachment causes the lower string to also have extension l/8. Find the tension in the lower string. [2]
(c) Hence find the new extension in the upper string. [2]

(a) T_upper = mg → 4mg·x/l = mg → x = l/4 [M1 A1]

(b) T_lower = 4mg × (l/8) / l = 4mg/8 = mg/2 [M1 A1]

(c) Particle equilibrium: T_upper = mg + T_lower (upper pulls up, weight + lower tension pull down... wait: lower string attached to floor pulls particle DOWN)
T_upper = mg + T_lower = mg + mg/2 = 3mg/2
4mg·x_new/l = 3mg/2 → x_new = 3l/8 [M1 A1]

Question 5 [7 marks]

A bungee jumper of mass 70 kg falls from a bridge. The elastic rope has natural length 15 m and modulus 2000 N. Taking g = 9.8 m/s²:
(a) Find the equilibrium extension of the rope. [2]
(b) Find the maximum extension of the rope (starting from rest at bridge level). [3]
(c) Find the speed of the jumper at the equilibrium position. [2]

(a) x_eq = mgl/λ = 70×9.8×15/2000 = 10290/2000 = 5.145 m [M1 A1]

(b) Datum at bridge. At max extension x_max: KE=0.
0 = −70×9.8×(15+x_max) + 2000x_max²/(2×15)
(200/3)x_max² = 686(15+x_max) = 10290 + 686x_max
(200/3)x_max² − 686x_max − 10290 = 0
200x_max² − 2058x_max − 30870 = 0
x_max = [2058 ± √(4235364 + 24696000)]/400 = [2058 ± 5397]/400
x_max = 7455/400 = 18.6 m [M1 M1 A1]

(c) At equilibrium (x = 5.145 m below natural length end):
0 + 0 + 0 = ½×70×v² − 70×9.8×(15+5.145) + 2000×(5.145)²/(30)
35v² = 70×9.8×20.145 − 2000×26.47/30 = 13819.5 − 1764.7 = 12054.8... [recalculate]
35v² = 70×9.8×20.145 − (2000×26.47/30) → v = 18.5 m/s [M1 A1]

Question 6 [5 marks]

An elastic string has natural length l and modulus λ. One end is fixed at point O on a smooth horizontal table. A particle of mass m is attached to the other end and placed at distance 2l from O. It is released from rest. Find the speed of the particle when the string returns to its natural length.

Initial extension x = 2l − l = l.
EPE₁ = λl²/(2l) = λl/2. KE₁=0. (No GPE change — horizontal.)
At natural length: EPE=0, KE=½mv².
½mv² = λl/2 → v² = λl/m → v = √(λl/m) [M1 M1 M1 A1 A1]

Question 7 [8 marks]

A particle P of mass 0.3 kg is attached to one end of an elastic string of natural length 0.4 m and modulus 7.5 N. The other end is fixed to a point O. P hangs in equilibrium below O.
(a) Show that the equilibrium extension is 0.157 m (3 s.f.). [2]
(b) P is pulled down a further 0.1 m from equilibrium and released. Show that it performs SHM with period T = 2π√(ml/λ). [4]
(c) Find the amplitude of oscillation and speed at equilibrium. [2]

(a) x_eq = mgl/λ = 0.3×9.8×0.4/7.5 = 1.176/7.5 = 0.1568 ≈ 0.157 m ✓ [M1 A1]

(b) Let y = displacement from equilibrium (down positive). Extension = x_eq + y.
Newton's 2nd Law: mÿ = mg − T = mg − λ(x_eq+y)/l
= mg − λx_eq/l − λy/l = mg − mg − λy/l = −λy/l
ÿ = −(λ/ml)y → SHM ✓ with ω² = λ/(ml), T = 2π/ω = 2π√(ml/λ) [M1 M1 A1 A1]

(c) Amplitude = 0.1 m (released 0.1 m from equilibrium). Max speed at equilibrium = ωA = √(λ/ml) × 0.1 = √(7.5/0.3×0.4)×0.1 = √62.5×0.1 = 0.791 m/s [M1 A1]

Question 8 [6 marks]

Two elastic strings, each of natural length 0.5 m and modulus 20 N, are attached to a particle of mass 0.4 kg. The other ends are attached to fixed points A and B on a smooth horizontal surface, with AB = 1.5 m. Find the equilibrium position of the particle and the tension in each string.

Let particle be at distance d from A. Extension of string 1: x₁ = d − 0.5 (if d > 0.5). Extension of string 2: x₂ = (1.5 − d) − 0.5 = 1 − d (if 1−d > 0 i.e. d < 1).
Both taut when 0.5 < d < 1.
At equilibrium (horizontal, no gravity component along surface): T₁ = T₂
20(d−0.5)/0.5 = 20(1−d)/0.5
d − 0.5 = 1 − d → 2d = 1.5 → d = 0.75 m from A (midpoint) [M1 M1 A1]
x₁ = x₂ = 0.25 m. T = 20×0.25/0.5 = 10 N in each string. [M1 A1 A1]

Past Paper Questions (Adapted Cambridge 9709 M2)

Past Paper 1 — 9709/52/O/N/20 Q3 (Adapted)

An elastic string of natural length 0.6 m and modulus of elasticity 30 N has one end attached to a fixed point A. A particle of mass 0.45 kg is attached to the other end and hangs in equilibrium.
(i) Find the length of the string. [3]
(ii) The particle is now pulled down 0.12 m from its equilibrium position and released from rest. Find the speed of the particle when it returns to the equilibrium position. [4]

(i) x_eq = mgl/λ = 0.45×9.8×0.6/30 = 2.646/30 = 0.0882 m
Total length = 0.6 + 0.0882 = 0.688 m [B1 M1 A1]

(ii) Amplitude A = 0.12 m. ω² = λ/(ml) = 30/(0.45×0.6) = 111.1
Speed at equilibrium = ωA = √111.1 × 0.12 = 10.54 × 0.12 = 1.26 m/s [M1 M1 M1 A1]
(Also valid: energy method — EPE at lowest = KE at equilibrium + GPE difference)

Past Paper 2 — 9709/53/M/J/19 Q4 (Adapted)

A particle P of mass 2 kg is attached to one end of an elastic string of natural length 1 m and modulus 50 N. The other end is attached to a fixed point O on a ceiling. P is held at O and released from rest.
(i) Find the extension when P first comes to instantaneous rest. [4]
(ii) Find the speed of P at the point where the string becomes natural length on the way down. [3]

(i) Datum at O. At max extension x: KE=0.
0 = −2×9.8×(1+x) + 50x²/(2×1) [energy: GPE lost = EPE gained + KE]
25x² = 19.6(1+x) = 19.6 + 19.6x
25x² − 19.6x − 19.6 = 0
x = [19.6 ± √(384.16 + 1960)]/50 = [19.6 ± 48.42]/50
x = 68.02/50 = 1.36 m [M1 M1 M1 A1]

(ii) String natural length at 1 m below O (x=0). EPE=0.
Energy from O: 0 = ½×2×v² − 2×9.8×1 → v² = 19.6 → v = 4.43 m/s [M1 M1 A1]

Past Paper 3 — 9709/51/M/J/21 Q5 (Adapted)

A particle of mass 0.6 kg is connected to a point A on a smooth inclined plane (angle 35°) by an elastic string of natural length 0.9 m and modulus 24 N. The string lies along the line of greatest slope. The particle is held at A and released.
(i) Find the equilibrium extension. [2]
(ii) Show that the string goes slack. State where it goes slack. [3]
(iii) Find the speed of the particle when the string goes slack. [4]

(i) x_eq = mgl sinθ/λ = 0.6×9.8×0.9×sin35°/24 = 3.089×0.5736/24 = 0.0738 m [M1 A1]

(ii) The particle is released from A (top of string). Moving down slope, string extends. The string goes slack only if particle overshoots equilibrium enough. The string goes slack when x = 0, i.e. when particle is 0.9 m down slope from A (at natural length position). Since the particle is released from A and swings down, it reaches the natural length position (0.9 m down), then if it has enough KE it goes beyond — but wait, beyond 0.9 m the string is taut. The string goes slack if the particle reverses direction and moves ABOVE the natural length position. In SHM, amplitude = distance from equilibrium to start = x_eq (from A to equilibrium is x_eq down, from start to equilibrium is x_eq). Particle oscillates x_eq either side of equilibrium, so highest point = equilibrium − x_eq = 0. Particle returns to A, and the string goes slack when particle returns to A. String goes slack at point A. [B1 B1 B1]

(iii) By energy at A: EPE=0, KE=½×0.6×v². At A same GPE (=0), EPE=0. All energy is KE = 0 (released from rest at A). Wait — by SHM symmetry, speed at A = 0 (same point). Actually v=0 at A since it oscillates. Speed at slack = 0 m/s. [M1 M1 M1 A1]

Past Paper 4 — 9709/52/O/N/18 Q6 (Adapted)

Two points A and B are 2.4 m apart on a smooth horizontal surface, with B directly below a fixed point. A particle P of mass 0.3 kg is attached by an elastic string (natural length 0.8 m, modulus 15 N) to A and by another elastic string (natural length 1 m, modulus 12 N) to B. P rests in equilibrium on the surface.
(i) Find the distance AP. [4]
(ii) Find the tension in each string. [3]

(i) Let AP = d. Then PB = 2.4 − d.
Extension in string 1 (A): x₁ = d − 0.8 (need d > 0.8)
Extension in string 2 (B): x₂ = (2.4−d) − 1 = 1.4 − d (need d < 1.4)
Equilibrium: T₁ = T₂ (horizontal, smooth surface)
15(d−0.8)/0.8 = 12(1.4−d)/1
(15/0.8)(d−0.8) = 12(1.4−d)
18.75d − 15 = 16.8 − 12d
30.75d = 31.8 → d = 1.034 m ≈ 1.03 m [M1 M1 M1 A1]

(ii) x₁ = 1.034 − 0.8 = 0.234 m; T₁ = 15×0.234/0.8 = 4.39 N
x₂ = 1.4 − 1.034 = 0.366 m; T₂ = 12×0.366/1 = 4.39 N ✓ [M1 A1 A1]

Past Paper 5 — 9709/53/O/N/21 Q7 (Adapted)

A particle of mass 1.2 kg is attached to one end of an elastic string of natural length 2 m and modulus 60 N. The other end is fixed to point O on a ceiling. The particle is held 3.5 m below O and released from rest.
(i) Find the initial EPE and verify the string is extended. [2]
(ii) Find the speed of the particle when it has risen 1 m from the release point. [4]
(iii) Find the greatest height above the release point reached by the particle. [5]

(i) Initial extension x₀ = 3.5 − 2 = 1.5 m > 0 ✓ (string taut)
EPE₀ = 60×1.5²/(2×2) = 60×2.25/4 = 33.75 J [B1 B1]

(ii) After rising 1 m: particle is 2.5 m below O, extension = 0.5 m.
GPE gained = 1.2×9.8×1 = 11.76 J; EPE₂ = 60×0.25/(2×2) = 3.75 J
KE = 33.75 − 11.76 − 3.75 + 0 [original KE=0]
½×1.2×v² = 18.24 → v² = 30.4 → v = 5.51 m/s [M1 A1 M1 A1]

(iii) String goes slack when particle is 2 m below O (x=0). Particle has risen 1.5 m from release.
Energy at slack: 33.75 = ½×1.2×v₀² + 1.2×9.8×1.5
33.75 = 0.6v₀² + 17.64 → v₀² = 26.85 → v₀ = 5.18 m/s (upward)
After slack: free flight. Extra height = v₀²/(2g) = 26.85/19.6 = 1.37 m
Greatest height above release = 1.5 + 1.37 = 2.87 m [M1 M1 A1 M1 A1]

Elastic Strings & Springs M2

Welcome to Elastic Strings & Springs

Learn 1: Hooke's Law

The Formula

Strings vs Springs

Modulus of Elasticity λ

Example Check

Learn 2: Elastic Potential Energy (EPE)

The Formula

Derivation Outline

Key Points

Example

Learn 3: Conservation of Energy with EPE

The Conservation Equation

The Three Energy Terms

Method

Example

Learn 4: Equilibrium with Elastic Strings & Springs

Vertical String/Spring — Standard Setup

Checking the String is Taut

Horizontal Surface

Inclined Plane

Learn 5: When Strings Go Slack

Finding Speed When Slack

Motion After the String Goes Slack

Standard Problem Type

Important Note — Springs

Quick Example

Worked Examples

Example 1 — Basic Hooke's Law

Example 2 — Finding Modulus

Example 3 — EPE Calculation

Example 4 — Equilibrium Extension (Vertical)

Example 5 — Energy Method (Find Speed)

Example 6 — String Goes Slack

Example 7 — Inclined Plane Equilibrium

Example 8 — Maximum Extension (Energy)

Common Mistakes

Mistake 1 — Using Total Length Instead of Extension

Mistake 2 — Forgetting EPE When String Is Stretched

Mistake 3 — Assuming a String Can Push

Mistake 4 — Wrong Sign for GPE in Energy Equation

Mistake 5 — Not Checking Tautness After Finding Equilibrium

Mistake 6 — Omitting EPE After String Goes Slack

Mistake 7 — Confusing λ with k (Spring Constant)

Key Formulas

Proof Bank

Proof 1: EPE Derivation from Hooke's Law

Proof 2: Equilibrium Extension Formula

Proof 3: Energy Method Setup for Vertical Oscillation (SHM)

Spring Visualiser

Exercise 1 — Hooke's Law Basics

Exercise 2 — EPE Calculations

Exercise 3 — Equilibrium

Exercise 4 — Energy Conservation

Exercise 5 — Strings Going Slack

Practice — 30 Questions

Challenge — 15 Questions

Exam-Style Questions

Question 1 [6 marks]

Question 2 [7 marks]

Question 3 [8 marks]

Question 4 [6 marks]

Question 5 [7 marks]

Question 6 [5 marks]

Question 7 [8 marks]

Question 8 [6 marks]

Past Paper Questions (Adapted Cambridge 9709 M2)

Past Paper 1 — 9709/52/O/N/20 Q3 (Adapted)

Past Paper 2 — 9709/53/M/J/19 Q4 (Adapted)

Past Paper 3 — 9709/51/M/J/21 Q5 (Adapted)

Past Paper 4 — 9709/52/O/N/18 Q6 (Adapted)

Past Paper 5 — 9709/53/O/N/21 Q7 (Adapted)