Understanding Simple Harmonic Motion Revision Notes | A Level AQA Physics

1. What Makes an Oscillation 'Simple Harmonic'?

Picture a mass hanging on a spring. Pull it down a little and release it — it bobs up and down, passing through the same central point repeatedly. That rhythmic, back-and-forth motion is an oscillation. The question is: what mathematical rule governs it?

Not every oscillation is simple harmonic. SHM is a very specific type, and the AQA specification requires you to identify it by a precise condition. The key is the relationship between acceleration and displacement — and once you see it, everything else in this topic falls into place.

The negative sign is not cosmetic — it carries the entire physics. When displacement $x$ is positive (object displaced to the right), the acceleration $a$ is negative (directed to the left, back towards equilibrium). The restoring force always opposes the displacement. Remove the negative sign and you no longer have SHM.

2. The Defining Equation: $a = -\omega^2 x$

The proportionality $a \propto -x$ becomes an equation when you introduce the constant of proportionality. In SHM, that constant is $\omega^2$, where $\omega$ is the angular frequency of the oscillation. This gives the defining equation of SHM:

Notice that $\omega^2$ is always positive, so the sign of $a$ is always opposite to the sign of $x$. This is the mathematical expression of the restoring nature of SHM. The larger the displacement, the greater the magnitude of the acceleration pulling the object back.

Angular frequency $\omega$ links to the period $T$ and frequency $f$ of the oscillation through $\omega = 2\pi f = \frac{2\pi}{T}$. A higher $\omega$ means a faster oscillation and, for the same displacement, a larger acceleration — which makes physical sense.

3. Characteristics of SHM: Displacement, Velocity, and Acceleration

Now that the defining equation is established, it is worth building a clear picture of how the three key quantities — displacement, velocity, and acceleration — behave throughout one complete oscillation. This is where the physics becomes vivid.

Consider a mass on a spring oscillating vertically. Take the equilibrium position as $x = 0$, and let the amplitude (maximum displacement) be $A$. The motion can be described starting from the maximum positive displacement.

• At maximum displacement ($x = +A$): velocity is zero (the object momentarily stops); acceleration is maximum in magnitude and directed towards equilibrium ($a = -\omega^2 A$, i.e. negative).
• At the equilibrium position ($x = 0$): velocity is at its maximum magnitude; acceleration is zero (no net restoring force at the centre).
• At maximum negative displacement ($x = -A$): velocity is zero again; acceleration is maximum in magnitude and now directed in the positive direction ($a = +\omega^2 A$).

This pattern reveals a crucial feature: acceleration and displacement are always in antiphase — when one is at its positive peak, the other is at its negative peak. Velocity, by contrast, is a quarter-cycle out of phase with displacement, reaching its maximum when displacement is zero.

The graph makes the antiphase relationship between $a$ and $x$ unmistakable. Wherever the blue displacement curve peaks, the red acceleration curve troughs by exactly the same amount — scaled by $\omega^2$. This is $a = -\omega^2 x$ made visible.

4. Applying the Equation: Worked Calculations

The equation $a = -\omega^2 x$ is the engine of most SHM calculation questions. You need to be comfortable rearranging it to find $\omega$, $x$, or $a$, and then linking $\omega$ back to period or frequency. Here is a full worked example that mirrors the style of AQA exam questions.

Worked example: Calculating acceleration and angular frequency in SHM

A particle oscillates with SHM. Its displacement from equilibrium varies with time. At one instant, the displacement is $x = 0.030$ m and the acceleration is $a = -0.48$ m s$^{-2}$. (a) Calculate the angular frequency $\omega$ of the oscillation. (b) Calculate the period $T$. (c) State the acceleration when the displacement is $x = -0.030$ m.

1. $a = -\omega^2 x$ — This is the AQA defining equation for SHM. Identify the known quantities: $a = -0.48$ m s$^{-2}$, $x = 0.030$ m. [M1]
2. $\omega^2 = -\frac{a}{x} = -\frac{-0.48}{0.030}$ — Dividing both sides by $-x$ isolates $\omega^2$. The two negatives cancel, giving a positive result as expected. [M1]
3. $\omega^2 = \frac{0.48}{0.030} = 16$ rad$^2$ s$^{-2}$ — Arithmetic check: $0.48 \div 0.030 = 16$. Units of $\omega^2$ are rad$^2$ s$^{-2}$.
4. $\omega = \sqrt{16} = 4.0$ rad s$^{-1}$ — $\sqrt{16} = 4.0$ exactly. [A1] for $\omega = 4.0$ rad s$^{-1}$.
5. $T = \frac{2\pi}{\omega} = \frac{2\pi}{4.0} = \frac{6.283...}{4.0} \approx 1.57$ s — $2\pi \approx 6.283$. Dividing by 4.0 gives $T \approx 1.57$ s (3 s.f.). [A1]
6. $a = -\omega^2 x = -(16)(-0.030) = +0.48$ m s$^{-2}$ — The magnitude is unchanged but the sign reverses — the acceleration now points in the positive direction. This confirms the restoring nature of SHM. [B1]

Answer: (a) $\omega = 4.0$ rad s$^{-1}$; (b) $T \approx 1.57$ s; (c) $a = +0.48$ m s$^{-2}$ (directed towards equilibrium, in the positive direction)

Step 6 is worth pausing on. The displacement changed sign but the magnitude stayed the same, so the acceleration magnitude is identical — only the direction flips. This is a direct consequence of $a = -\omega^2 x$ and is a favourite AQA 'state' question worth one mark.

5. Physical Examples: Mass-Spring and Simple Pendulum

The condition $a \propto -x$ is not just an abstract rule — it emerges naturally from the physics of real systems. Two systems you must be able to discuss are the mass-spring system and the simple pendulum.

For a mass $m$ on a spring with spring constant $k$, Hooke's law gives the restoring force as $F = -kx$. Applying Newton's second law: $ma = -kx$, so $a = -\frac{k}{m}x$. Comparing with $a = -\omega^2 x$ directly gives $\omega^2 = \frac{k}{m}$, and therefore $T = 2\pi\sqrt{\frac{m}{k}}$.

For a simple pendulum of length $L$ (with small-angle approximation, typically $\theta \lesssim 10°$, though the approximation improves for smaller angles), the restoring component of gravity gives $a \approx -\frac{g}{L}x$, where $x$ is the arc displacement. This yields $\omega^2 = \frac{g}{L}$ and $T = 2\pi\sqrt{\frac{L}{g}}$. Both systems satisfy $a \propto -x$ and therefore undergo SHM.

6. Exam Technique and Common Mistakes

SHM questions on AQA papers follow predictable patterns. The marks are there to collect if you know what examiners are looking for — and where students routinely drop them.

When asked to state the condition for SHM, you must write $a \propto -x$ and explain in words that the acceleration is directed towards the equilibrium position. Writing only 'acceleration is proportional to displacement' without the negative sign or the direction qualifier will not score the mark.

When asked to show that a system undergoes SHM, the expected structure is: derive an expression for acceleration, show it has the form $a = -\omega^2 x$, and explicitly identify $\omega^2$ as a positive constant. This three-step argument earns full marks; stopping after the derivation without the identification step does not.

Bringing the lesson together: SHM is defined by the condition $a \propto -x$, made quantitative by the equation $a = -\omega^2 x$. The characteristics — zero acceleration at equilibrium, maximum acceleration at amplitude, velocity and acceleration a quarter-cycle apart — all follow directly from this single equation. Master the equation and the characteristics follow automatically.