use a = –ω2x and recall and use, as a solution to this equation, x = x0 sin ωt

Simple Harmonic Oscillations (SHO) – Cambridge A‑Level Physics 9702

⚡ What is SHM? A motion where the restoring force is proportional to the displacement and opposite in direction. Think of a playground swing that returns to its rest position with a force that gets stronger the farther it is pulled. The key equation that describes this motion is:

$a = -\omega^2 x$
where $a$ is acceleration, $x$ is displacement from the equilibrium, and $\omega$ is the angular frequency. This simple differential equation leads to a sinusoidal motion.

Deriving the Solution

The equation $a = -\omega^2 x$ can be written as a second‑order differential equation: $$\frac{d^2x}{dt^2} = -\omega^2 x$$ The general solution of this equation is: $$x(t) = A \sin(\omega t) + B \cos(\omega t)$$ By choosing the phase such that the motion starts from the maximum displacement ($x(0)=x_0$) and with zero initial velocity, we set $B = 0$ and $A = x_0$. Hence the simple form used in exams: $$x(t) = x_0 \sin(\omega t)$$ 📚 Tip: Remember that $\omega = 2\pi f$ and $T = \frac{2\pi}{\omega}$.

Common Physical Systems

• Mass‑spring system: $m\frac{d^2x}{dt^2} = -kx$ → $\omega = \sqrt{\frac{k}{m}}$
• Simple pendulum (small angles): $l\frac{d^2\theta}{dt^2} = -g\theta$ → $\omega = \sqrt{\frac{g}{l}}$
• LC circuit (electrical analogue): $L\frac{d^2q}{dt^2} = -\frac{q}{C}$ → $\omega = \frac{1}{\sqrt{LC}}$

Key Parameters – Quick Reference

Parameter	Symbol	Relation
Amplitude	$x_0$	Maximum displacement
Frequency	$f$	$f = \frac{\omega}{2\pi}$
Period	$T$	$T = \frac{2\pi}{\omega}$
Angular Frequency	$\omega$	$\omega = 2\pi f$

Exam Tip Box

🎯 Remember: When asked for the maximum speed, use $v_{\text{max}} = \omega x_0$. For maximum acceleration, use $a_{\text{max}} = \omega^2 x_0$. Always check the initial conditions to decide whether the solution involves $\sin$ or $\cos$. Use the small‑angle approximation $\sin\theta \approx \theta$ only for pendulums with $\theta < 10^\circ$.

Quick Practice Problem

A 0.5 kg mass is attached to a spring with $k = 200\,\text{N/m}$.

1. Find $\omega$.
2. Write the equation of motion $x(t)$ if the mass is pulled to $x_0 = 0.02\,\text{m}$ and released from rest.
3. What is the period $T$?

Solution: $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20\,\text{rad/s}$ $x(t) = 0.02\,\sin(20t)$ $T = \frac{2\pi}{\omega} = \frac{2\pi}{20} \approx 0.314\,\text{s}$

Final Thought

🎉 SHM is all around us – from the swing in the park to the oscillation of a tuning fork. By mastering the simple equation $a = -\omega^2 x$ and its sinusoidal solution, you’ll be ready to tackle any question in the Cambridge A‑Level Physics exam. Good luck, and keep swinging! 🚀