# Elasticity and simple harmonic motion

> CIE A-Level Further Mathematics · Further Mechanics Unit 3
> Source: https://www.owlsprep.com/study/cie-9231-u3-elasticity-and-simple-harmonic-motion/

This subtopic explores how elastic forces from strings and springs produce simple harmonic motion (SHM), covering equilibrium derivation, period calculation, and energy analysis for horizontal and vertical systems, common in CIE 9231 exams.

**Prerequisites:** [Basic simple harmonic motion principles](https://www.owlsprep.com/study/cie-9231-u3-basic-simple-harmonic-motion/); [Hooke's Law and elastic potential energy](https://www.owlsprep.com/study/cie-9231-u3-elasticity/)

## Learning objectives

- Derive that elastic systems satisfy the conditions for simple harmonic motion
- Calculate period, amplitude and maximum speed for horizontal and vertical elastic SHM
- Analyse energy conservation in elastic SHM systems
- Solve common CIE exam problems for elastic SHM

## Deriving SHM for Elastic Systems

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: $a = -\omega^2 x$. We first analyze simple horizontal systems, where gravity does not affect motion along the oscillation axis.

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

*Notation:* $F = -kx$

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

**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 $\omega$.

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 $a = -\omega^2 x$, 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.

## Vertical Elastic SHM

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 $\lambda$ 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 $mg = \frac{\lambda e}{l}$ 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. $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{ml}{\lambda}}$$

> **tip**
>
> Gravity only shifts the equilibrium position, it does not change the angular frequency or period of vertical elastic SHM.

## Energy in Elastic SHM

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: $E_{\text{total}} = \frac{1}{2}m\omega^2 A^2$, where $A$ is amplitude, same as standard SHM

**Worked example:** A 1 kg mass undergoes SHM on a vertical elastic spring with $\omega = 2 \text{ rad s}^{-1}$ 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 $m=1$, $\omega=2$, $A=0.5$:
4. $$E_{\text{max}} = \frac{1}{2}(1)(2^2)(0.5^2) = 0.5 \text{ J}$$

## Problem Solving for Elastic SHM

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 $\omega$, then apply standard SHM results.

**Check your understanding**

Check your understanding of the core first step:

1. For a vertical elastic SHM problem, which point do you measure displacement from to get the simple SHM equation $a = -\omega^2 x$?

   - Natural length of the string
   - Equilibrium position
   - Lowest point of the oscillation

   *Answer:* Equilibrium position

   *Why:* 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 $\omega$ 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 $v_{\text{max}} = A\omega$:
4. $$v_{\text{max}} = 0.2 \times 2\sqrt{5} = 0.4\sqrt{5} \approx 0.89 \text{ m s}^{-1}$$

## Common pitfalls

- **Wrong:** Measuring displacement from natural length instead of equilibrium for vertical SHM
  - Why it fails: This leaves a constant gravity term in the force equation, so you do not get the standard SHM form
  - Correct: First calculate equilibrium extension, then measure all displacements from this point; gravity cancels out
- **Wrong:** Forgetting elastic strings cannot exert compressive force, only tension
  - Why it fails: If displacement goes above natural length, the string goes slack and motion is no longer SHM
  - Correct: Check if amplitude is large enough to make the string slack, split motion into SHM and free fall if needed
- **Wrong:** Mixing up modulus of elasticity and spring constant in the period formula
  - Why it fails: Many learners substitute the wrong values when calculating $\omega$ for elastic strings
  - Correct: Remember $k = \frac{\lambda}{l}$, so $\omega = \sqrt{\frac{\lambda}{ml}}$ for elastic strings
- **Wrong:** Double-counting gravitational potential energy for vertical SHM energy calculations
  - Why it fails: Gravitational potential energy change is already accounted for after shifting to equilibrium coordinates
  - Correct: Use the standard SHM total energy formula $E = \frac{1}{2}m\omega^2 A^2$, it works for vertical systems

## Cheatsheet

| System | Angular Frequency $\omega$ | Period $T$ |
| --- | --- | --- |
| Horizontal spring ($k$) | $\sqrt{\frac{k}{m}}$ | $2\pi\sqrt{\frac{m}{k}}$ |
| Vertical elastic string ($\lambda, l$) | $\sqrt{\frac{\lambda}{ml}}$ | $2\pi\sqrt{\frac{ml}{\lambda}}$ |
| Vertical spring ($k$) | $\sqrt{\frac{k}{m}}$ | $2\pi\sqrt{\frac{m}{k}}$ |
| Maximum speed | $A\omega$ | - |
| Total energy | $\frac{1}{2}m\omega^2 A^2$ | - |

## 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.

- [Further Probability & Statistics](https://www.owlsprep.com/study/cie-9231-u4-overview/)
- [Discrete probability distributions](https://www.owlsprep.com/study/cie-9231-u4-discrete-probability-distributions/)
- [Continuous probability distributions](https://www.owlsprep.com/study/cie-9231-u4-continuous-probability-distributions/)

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