Elasticity and simple harmonic motion

CIE A-Level Further Mathematics· 25 min read

1. Deriving SHM for Elastic Systems★★☆☆☆⏱ 15 min

When a mass attached to an elastic string or spring is displaced from equilibrium, the restoring force follows Hooke's Law, which satisfies the core requirement for SHM: $a = - ω^{2} x$ . We first analyze simple horizontal systems, where gravity does not affect motion along the oscillation axis.

📘 Definition

Restoring Force

$F = - k x$

A force acting opposite to displacement from equilibrium, pulling the system back to equilibrium, required for SHM

Example:

For a spring with spring constant $k$ , displacement $x$ , restoring force is $F = - k x$

📐 Worked Example

A mass $m$ is attached to a horizontal spring with spring constant $k$ , natural length $l_{0}$ . Show that the motion is SHM and find $ω$ .

1
Let displacement from equilibrium (natural length for horizontal systems) be $x$ .
2
$F = -kx$
3
By Newton's second law, $F = ma$ , so:
4
$ma = -kx \implies a = -\frac{k}{m}x$
5
This matches the SHM differential equation $a = - ω^{2} x$ , so motion is SHM with:
6
$\omega = \sqrt{\frac{k}{m}}$

Exam tip:

Always measure displacement from the equilibrium position, not the natural length of the string or spring.

2. Vertical Elastic SHM★★★☆☆⏱ 20 min

For vertical systems, gravity extends the string or spring to a new equilibrium position before oscillation begins. Gravity cancels out when deriving the SHM equation, leaving a simple expression for angular frequency.

📐 Worked Example

A light elastic string of natural length $l$ and modulus $λ$ is fixed at one end, with mass $m$ attached to the other. The mass is displaced slightly from equilibrium. Show motion is SHM and find the period.

1
First find equilibrium extension $e$ : at equilibrium, tension equals weight:
2
$\frac{\lambda e}{l} = mg \implies e = \frac{mgl}{\lambda}$
3
Let $x$ be displacement downwards from equilibrium. Total extension is $e + x$ .
4
Find net force downwards (weight minus tension):
5
$F = mg - \frac{\lambda (e + x)}{l}$
6
Substitute $m g = \frac{λ e}{l}$ from equilibrium:
7
$F = \frac{\lambda e}{l} - \frac{\lambda e}{l} - \frac{\lambda x}{l} = -\frac{\lambda}{l}x$
8
By Newton's second law $F = ma$ , so:
9
$a = -\frac{\lambda}{ml}x = -\omega^2 x \implies \omega = \sqrt{\frac{\lambda}{ml}}$
10
Period $T$ is:
11
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{ml}{\lambda}}$

3. Energy in Elastic SHM★★★☆☆⏱ 15 min

Elastic SHM systems have three forms of mechanical energy: kinetic energy of the mass, elastic potential energy of the string/spring, and gravitational potential energy (for vertical systems). Total mechanical energy is conserved for undamped motion.

At equilibrium: Kinetic energy is maximum, net potential energy is minimum
At maximum displacement: Kinetic energy is zero, total potential energy is maximum
Total energy: $E_{total} = \frac{1}{2} m ω^{2} A^{2}$ , where $A$ is amplitude, same as standard SHM

📐 Worked Example

A 1 kg mass undergoes SHM on a vertical elastic spring with $ω = 2 rad s^{- 1}$ and amplitude 0.5 m. Calculate the maximum kinetic energy of the mass.

1
Maximum kinetic energy equals the total energy of the system:
2
$E_{\text{max}} = \frac{1}{2}m\omega^2 A^2$
3
Substitute values $m = 1$ , $ω = 2$ , $A = 0.5$ :
4
$E_{\text{max}} = \frac{1}{2}(1)(2^2)(0.5^2) = 0.5 \text{ J}$

4. Problem Solving for Elastic SHM★★★★☆⏱ 20 min

Most exam questions require you to find period, amplitude, maximum speed, or maximum displacement for elastic SHM. The key first step is always to find the equilibrium position, then derive $ω$ , then apply standard SHM results.

✓ Quick check

Check your understanding of the core first step:

For a vertical elastic SHM problem, which point do you measure displacement from to get the simple SHM equation $a = - ω^{2} x$ ?
- Natural length of the string
- Equilibrium position
- Lowest point of the oscillation
Reveal answer
1 —
Correct! Displacement must always be measured from equilibrium for the simple SHM form, gravity cancels out around this point.

📐 Worked Example

A 2 kg mass is attached to an elastic string of natural length 1 m, modulus 40 N, hung vertically. The mass is pulled down 0.2 m from equilibrium and released from rest. Find the maximum speed of the mass.

1
Calculate $ω$ for the system:
2
$\omega = \sqrt{\frac{\lambda}{ml}} = \sqrt{\frac{40}{(2)(1)}} = 2\sqrt{5} \text{ rad s}^{-1}$
3
Amplitude $A = 0.2$ m (released from rest). Maximum speed $v_{max} = A ω$ :
4
$v_{\text{max}} = 0.2 \times 2\sqrt{5} = 0.4\sqrt{5} \approx 0.89 \text{ m s}^{-1}$

5. Common Pitfalls

Wrong move:

Measuring displacement from natural length instead of equilibrium for vertical SHM

Why:

This leaves a constant gravity term in the force equation, so you do not get the standard SHM form

Correct move:

First calculate equilibrium extension, then measure all displacements from this point; gravity cancels out

Wrong move:

Forgetting elastic strings cannot exert compressive force, only tension

Why:

If displacement goes above natural length, the string goes slack and motion is no longer SHM

Correct move:

Check if amplitude is large enough to make the string slack, split motion into SHM and free fall if needed

Wrong move:

Mixing up modulus of elasticity and spring constant in the period formula

Why:

Many learners substitute the wrong values when calculating $ω$ for elastic strings

Correct move:

Remember $k = \frac{λ}{l}$ , so $ω = \frac{λ}{m l}$ for elastic strings

Wrong move:

Double-counting gravitational potential energy for vertical SHM energy calculations

Why:

Gravitational potential energy change is already accounted for after shifting to equilibrium coordinates

Correct move:

Use the standard SHM total energy formula $E = \frac{1}{2} m ω^{2} A^{2}$ , it works for vertical systems

6. Quick Reference Cheatsheet

System	Angular Frequency $ω$	Period $T$
Horizontal spring ( $k$ )	$\frac{k}{m}$	$2 π \frac{m}{k}$
Vertical elastic string ( $λ, l$ )	$\frac{λ}{m l}$	$2 π \frac{m l}{λ}$
Vertical spring ( $k$ )	$\frac{k}{m}$	$2 π \frac{m}{k}$
Maximum speed	$A ω$
Total energy	$\frac{1}{2} m ω^{2} A^{2}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Vertical elastic spring SHM problem
2021 · 1
Horizontal elastic string SHM
2023 · 2
Energy elastic SHM question

Going deeper

syllabusCIE 9231 Further Mechanics Syllabus

What's Next

Elasticity and SHM forms a foundation for more advanced oscillatory motion topics in further mechanics, including damped and forced oscillations, and energy analysis of driven systems. It is also commonly combined with work-energy principles and connected systems problems in CIE 9231 exams, so mastery of this subtopic is critical for scoring high marks. The concepts here also underpin many university-level classical mechanics topics, so building a strong understanding now will support future study.