understand that simple harmonic motion occurs when acceleration is proportional to displacement from a fixed point and in the opposite direction

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Simple Harmonic Oscillations

Simple Harmonic Oscillations

Learning Objective

Understand that simple harmonic motion (SHM) occurs when the acceleration of a particle is directly proportional to its displacement from a fixed point and acts in the opposite direction.

1. Definition of Simple Harmonic Motion

Simple harmonic motion is a type of periodic motion where the restoring force (or acceleration) is proportional to the displacement from an equilibrium position and always directed towards that equilibrium.

Mathematically, this condition is expressed as:

\$\mathbf{a} = -\omega^{2}\,\mathbf{x}\$

where 𝑎 is the acceleration, 𝑥 is the displacement from equilibrium, and ω is the angular frequency (rad s⁻¹).

2. Derivation from Hooke’s Law

For a mass‑spring system the restoring force obeys Hooke’s law:

\$F = -k\,x\$

Using Newton’s second law \$F = m a\$, we obtain:

\$m a = -k x \quad\Rightarrow\quad a = -\frac{k}{m}\,x\$

Comparing with the SHM condition \$a = -\omega^{2}x\$, we identify

\$\omega = \sqrt{\frac{k}{m}}\$

3. Equations of Motion

The displacement, velocity and acceleration as functions of time are:

  • Displacement: \$x(t)=A\cos(\omega t+\phi)\$

  • Velocity: \$v(t) = -A\omega\sin(\omega t+\phi)\$

  • Acceleration: \$a(t) = -A\omega^{2}\cos(\omega t+\phi) = -\omega^{2}x(t)\$

Here \$A\$ is the amplitude and \$\phi\$ the phase constant.

4. Energy in Simple Harmonic Motion

The total mechanical energy \$E\$ of an ideal SHM system is constant and is the sum of kinetic and potential energies:

\$E = \frac{1}{2}kA^{2} = \frac{1}{2}m\omega^{2}A^{2}\$

At any instant:

  • Kinetic energy: \$K = \frac{1}{2}m v^{2} = \frac{1}{2}m\omega^{2}A^{2}\sin^{2}(\omega t+\phi)\$

  • Potential energy: \$U = \frac{1}{2}k x^{2} = \frac{1}{2}kA^{2}\cos^{2}(\omega t+\phi)\$

5. Common Examples of SHM

  1. Mass‑spring system: A mass \$m\$ attached to a spring of constant \$k\$ on a frictionless surface.

  2. Simple pendulum (small angles): For small angular displacements \$\theta\$, the restoring torque is \$-\frac{mgL}{\ell}\theta\$, giving \$\omega = \sqrt{g/L}\$.

  3. Vibrating tuning fork: The prongs execute SHM about their equilibrium positions.

6. Conditions for Simple Harmonic Motion

SHM is observed only when the following conditions are satisfied:

  • The restoring force is linear with displacement (Hooke’s law).

  • There is no energy loss (no damping).

  • The motion is one‑dimensional and the equilibrium position is fixed.

7. Summary Table of Key Quantities

QuantitySymbolExpressionUnits
Angular frequency\$\omega\$\$\sqrt{k/m}\$ (mass‑spring) or \$\sqrt{g/L}\$ (pendulum)rad s⁻¹
Period\$T\$\$2\pi/\omega\$s
Frequency\$f\$\$1/T\$Hz
Maximum speed\$v_{\max}\$\$A\omega\$m s⁻¹
Maximum acceleration\$a_{\max}\$\$A\omega^{2}\$m s⁻²
Total energy\$E\$\$\frac{1}{2}kA^{2}\$J

8. Suggested Diagram

Suggested diagram: A mass \$m\$ attached to a horizontal spring of constant \$k\$, displaced a distance \$x\$ from equilibrium, showing the restoring force \$F = -kx\$ and direction of acceleration \$a = -\omega^{2}x\$.

9. Quick Check Questions

  1. Derive the expression for the period \$T\$ of a mass‑spring system using \$T = 2\pi\sqrt{m/k}\$.

  2. Explain why a simple pendulum only exhibits SHM for small angular displacements.

  3. Calculate the maximum speed of a 0.5 kg mass attached to a spring with \$k = 200\ \text{N m}^{-1}\$ when the amplitude is 0.10 m.

10. Further Reading

For deeper insight, consult the Cambridge International AS & A Level Physics (9702) textbook, chapter on oscillations, and review past examination questions on SHM.