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

Published by Patrick Mutisya · 8 days ago

Cambridge A-Level Physics 9702 – Simple Harmonic Oscillations

Simple Harmonic Oscillations

Simple harmonic motion (SHM) describes the oscillatory motion of a system where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. The defining differential equation for SHM is

\$\frac{d^{2}x}{dt^{2}} = -\omega^{2}x\$

where:

  • \$x\$ – displacement from the equilibrium position (m)

  • \$\omega\$ – angular frequency (rad s⁻¹)

  • \$t\$ – time (s)

Derivation of the General Solution

Integrating the second‑order differential equation yields a sinusoidal solution. One convenient form is

\$x(t) = x_{0}\sin(\omega t + \phi)\$

where:

  • \$x_{0}\$ – amplitude (maximum displacement) (m)

  • \$\phi\$ – phase constant (rad), determined by initial conditions

If the motion starts from the equilibrium position with maximum velocity, the phase constant \$\phi\$ can be set to zero, giving the simpler expression

\$x(t) = x_{0}\sin(\omega t)\$

Key Relationships

QuantityExpressionPhysical Meaning
Angular frequency\$\omega = 2\pi f = \sqrt{\dfrac{k}{m}}\$Rate of oscillation; \$f\$ is frequency (Hz), \$k\$ is spring constant (N m⁻¹), \$m\$ is mass (kg)
Period\$T = \dfrac{2\pi}{\omega}\$Time for one complete oscillation (s)
Velocity\$v(t) = \dfrac{dx}{dt} = \omega x_{0}\cos(\omega t)\$Instantaneous speed; maximum \$v{\max}= \omega x{0}\$
Acceleration\$a(t) = \dfrac{d^{2}x}{dt^{2}} = -\omega^{2}x_{0}\sin(\omega t) = -\omega^{2}x(t)\$Restoring acceleration; proportional to displacement
Energy\$E = \dfrac{1}{2}k x{0}^{2} = \dfrac{1}{2}m\omega^{2}x{0}^{2}\$Total mechanical energy (constant for an ideal SHM system)

Applying the Equation \$a = -\omega^{2}x\$

  1. Identify the system that exhibits a linear restoring force (e.g., mass‑spring, simple pendulum for small angles).

  2. Write the expression for the restoring force \$F = -kx\$ (Hooke’s law) or the equivalent torque for rotational systems.

  3. Use Newton’s second law \$F = ma\$ to obtain \$ma = -kx\$ → \$a = -(k/m)x\$.

  4. Recognise that \$-(k/m)\$ plays the role of \$-\omega^{2}\$, so \$\omega = \sqrt{k/m}\$.

  5. Insert \$\omega\$ into the solution \$x(t) = x_{0}\sin(\omega t + \phi)\$ to describe the motion completely.

Example Problem

Problem: A 0.5 kg mass is attached to a horizontal spring with \$k = 200\ \text{N m}^{-1}\$. The mass is pulled 0.10 m from equilibrium and released from rest. Determine the displacement \$x\$ after \$t = 0.05\ \text{s}\$.

Solution:

  1. Calculate the angular frequency:

    \$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20\ \text{rad s}^{-1}\$

  2. Since the mass is released from rest at maximum displacement, the phase constant \$\phi = 0\$ and the motion follows

    \$x(t) = x_{0}\sin(\omega t)\$

    with \$x_{0}=0.10\ \text{m}\$.

  3. Evaluate at \$t = 0.05\ \text{s}\$:

    \$x(0.05) = 0.10\sin(20 \times 0.05) = 0.10\sin(1.0) \approx 0.10 \times 0.8415 = 0.084\ \text{m}\$

Thus the displacement after 0.05 s is approximately \$0.084\ \text{m}\$.

Common Misconceptions

  • Sign of acceleration: In SHM the acceleration is always opposite to the displacement, hence the negative sign in \$a = -\omega^{2}x\$.

  • Frequency vs. angular frequency: \$f\$ (Hz) and \$\omega\$ (rad s⁻¹) are related by \$\omega = 2\pi f\$, not interchangeable.

  • Phase constant: Ignoring \$\phi\$ can lead to incorrect initial conditions; always determine \$\phi\$ from the given start state.

Suggested Diagram

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