Ultimate Guide to Simple Harmonic Motion in HSC Mathematics Extension 2

Introduction

Simple Harmonic Motion (SHM) is one of the most elegant and universally applicable models in physics and mathematics. In HSC Mathematics Extension 2, you are required to rigorously define, prove, and model SHM using differential equations. Whether it's a pendulum swinging, a mass bouncing on a spring, or the rhythmic rising of the tides, SHM provides the mathematical language to describe periodic, oscillating motion perfectly.

Executive Summary

This comprehensive guide covers the core concepts of Simple Harmonic Motion:

Defining SHM: Understanding the core principle that acceleration is proportional and opposite to displacement ( $\ddot{x} = -n^2(x - c)$ ).
The Restoring Force: Recognizing how $F = -mn^2(x - c)$ acts to pull a particle back to the centre of motion.
Solving the Differential Equation: Verifying that $x(t) = a\cos(nt + \alpha) + c$ is a valid solution.
Key Characteristics: Calculating amplitude, period, center of motion, velocity, and acceleration.
Graphs and Modelling: Graphing SHM functions and determining equations from visual representations.

What is this about?

In the previous topic, we looked at general straight-line motion where forces varied. SHM is a very specific, highly predictable type of straight-line motion. It occurs whenever a particle is subjected to a "restoring force"—a force that gets stronger the further the particle moves away from its resting position, always pulling it back. Because the force (and therefore acceleration) scales linearly with displacement, the resulting position-time graph forms a perfect sine or cosine wave.

Main Content

1. Defining Simple Harmonic Motion

By definition, a particle undergoes Simple Harmonic Motion if its acceleration ( $\ddot{x}$ ) is directly proportional to its displacement from a central point ( $x-c$ ), and acts in the opposite direction.

Mathematically, this is expressed as: $\ddot{x} = -n^2(x - c)$ Where:

$x$ is the displacement of the particle.
$c$ is the centre of motion.
$n^2$ is a positive constant (we use $n^2$ to guarantee the constant is positive, ensuring the acceleration is always opposite to displacement).

Using Newton's Second Law ( $F = ma$ ), the force causing SHM is $F = -mn^2(x-c)$ . This is called a restoring force.

2. Solving the SHM Differential Equation

If we are given that $\ddot{x} = -n^2x$ (assuming the centre is at the origin, $c=0$ ), we can prove that equations of the form $x = a\cos(nt + \alpha)$ or $x = a\sin(nt + \alpha)$ are solutions.

Let's verify $x = a\cos(nt + \alpha)$ :

Differentiate to find velocity: $v = \dot{x} = -an\sin(nt + \alpha)$
Differentiate again to find acceleration: $a = \ddot{x} = -an^2\cos(nt + \alpha)$
Notice that $a\cos(nt + \alpha)$ is exactly our original $x$ .
Therefore, $\ddot{x} = -n^2x$ .

This confirms that the differential equation $\ddot{x} = -n^2x$ perfectly describes a cosine (or sine) wave!

3. Key Characteristics of SHM

For a particle moving according to $x(t) = a\cos(nt + \alpha) + c$ :

Centre of Motion: $c$ . The point where acceleration is zero and velocity is maximum.
Amplitude: $a$ . The maximum distance the particle travels from the centre. The particle oscillates between $x = c + a$ and $x = c - a$ .
Period: $T = \frac{2\pi}{n}$ . The time it takes to complete one full oscillation.
Velocity Equation: We can relate velocity and displacement without time using the formula derived from $a = v\frac{dv}{dx}$ . For SHM, this evaluates to: $v^2 = n^2(a^2 - (x-c)^2)$

4. Graphing and Modelling

When graphing SHM:

Displacement vs Time ( $x-t$ ): A standard sine or cosine wave.
Velocity vs Time ( $v-t$ ): Also a sinusoidal wave, phase-shifted by $90^\circ$ ( $\pi/2$ ) relative to displacement. Velocity is maximum when displacement is zero (at the centre), and velocity is zero when displacement is at a maximum (the endpoints).
Acceleration vs Time ( $a-t$ ): A sinusoidal wave completely out of phase ( $180^\circ$ or $\pi$ ) with displacement.

When modelling real-world problems (like tides), you will often be given the maximum and minimum values (e.g., high tide is 4m, low tide is 1m). From this, you can find:

Amplitude $a = \frac{\text{High} - \text{Low}}{2}$
Centre $c = \frac{\text{High} + \text{Low}}{2}$

Simple Worked Example

Question: A particle is moving in Simple Harmonic Motion. Its velocity $v$ (in m/s) at displacement $x$ (in metres) is given by the equation $v^2 = -9x^2 + 36x - 27$ . (a) Prove that the motion is Simple Harmonic. (b) Find the centre of motion, the amplitude, and the period.

Solution: (a) Prove it is SHM: To prove SHM, we must show that acceleration $\ddot{x}$ is in the form $-n^2(x-c)$ . We know $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ . First, halve the velocity equation: $\frac{1}{2}v^2 = -4.5x^2 + 18x - 13.5$

Now, differentiate with respect to $x$ : $a = \frac{d}{dx}(-4.5x^2 + 18x - 13.5)$ $a = -9x + 18$ $a = -9(x - 2)$

Since $a$ is in the form $-n^2(x-c)$ where $n^2 = 9 > 0$ , the motion is Simple Harmonic.

(b) Find the characteristics: From $a = -9(x - 2)$ , we can identify:

$n^2 = 9 \implies n = 3$
Centre of motion $c = 2$ .
Period $T = \frac{2\pi}{n} = \frac{2\pi}{3}$ seconds.

To find the amplitude, we know that at the extremities of the motion, the particle stops momentarily, meaning $v = 0$ . Set $v^2 = 0$ : $-9x^2 + 36x - 27 = 0$ Divide by $-9$ : $x^2 - 4x + 3 = 0$ $(x-1)(x-3) = 0$ $x = 1, x = 3$

The particle oscillates between $x = 1$ and $x = 3$ . The centre is at $x=2$ . The amplitude is the distance from the centre to an extremity: $a = 3 - 2 = 1$ metre.

mini-FAQ page

Q: Does it matter if I use sine or cosine to model SHM? A: Both are mathematically valid, but your choice depends on the initial conditions. If the particle starts at the extremity (maximum displacement), a cosine function ( $x = a\cos(nt)$ ) is easier because $\cos(0) = 1$ . If it starts at the centre of motion, a sine function ( $x = a\sin(nt)$ ) is easier because $\sin(0) = 0$ .

Q: Can period be negative? A: No, time is strictly positive in these contexts. $n$ is generally taken as the positive square root of $n^2$ , so $T = \frac{2\pi}{n}$ remains positive.

Common mistakes to avoid

Forgetting the negative sign in the proof: If you derive $a = 9(x - 2)$ , this is not SHM. The acceleration must be oppositely directed to the displacement, so the constant outside the bracket must be negative.
Confusing frequency and period: In physics, you might use $f = 1/T$ . The syllabus focuses primarily on the period $T = \frac{2\pi}{n}$ . Do not mistakenly substitute $n$ for frequency.
Assuming the centre of motion is always 0: In many questions, the particle oscillates around a point other than the origin. Always factorize the acceleration equation fully to reveal $(x-c)$ .

Practice on Vu's Maths Hub

Simple Harmonic Motion relies on a deep understanding of trigonometry and calculus. Put your skills into practice:

Access dozens of SHM exam questions and step-by-step solutions in our HSC Mechanics Booklet.
Brush up on your sine and cosine wave graphing skills with the HSC Trigonometry Booklet.
Challenge yourself with complex, multi-stage oscillator problems in the HSC Last Resorts Booklet.

Connect with me

Want to master SHM and lock in top marks for HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, detailed video explanations, and expert advice to help you ace your exams!