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)=x0\$) and with zero initial velocity, we set \$B = 0\$ and \$A = x0\$.

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

ParameterSymbolRelation
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:

  1. When asked for the maximum speed, use \$v{\text{max}} = \omega x0\$.

  2. For maximum acceleration, use \$a{\text{max}} = \omega^2 x0\$.

  3. Always check the initial conditions to decide whether the solution involves \$\sin\$ or \$\cos\$.

  4. 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! 🚀