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AS & A-Level Physics 17 — Oscillations and Resonance

AS & A-Level Physics 17 — Oscillations and Resonance

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Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 17 of 26: Oscillations and resonance. Concepts, worked applications and misconception checks.

Physics EN A-Level
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The Defining Equation of SHM and its Sinusoidal Solution

The defining condition for Simple Harmonic Motion, where acceleration is proportional to and opposite to displacement, is mathematically expressed as a=−ω2xa = -\omega^2x. Here, aa is the acceleration, xx is the displacement from equilibrium, and ω\omega (omega) is the angular frequency of the oscillation. The constant ω2\omega^2 ensures that the proportionality constant is always positive, preserving the significance of the negative sign. A common solution to this differential equation, which describes how displacement varies with time, is x=x0sin⁡ωtx = x_0 \sin \omega t. In this equation, x0x_0 represents the amplitude (maximum displacement), and tt is time. This solution assumes that the oscillating object starts at its equilibrium position (x=0x=0) at t=0t=0. If the object starts at maximum displacement (x=x0x=x_0) at t=0t=0, the solution would be x=x0cos⁡ωtx = x_0 \cos \omega t.

Key points

  • Defining equation for SHM: a=−ω2xa = -\omega^2x.
  • ω\omega is the angular frequency (rad s−1^{-1}).
  • A common solution for displacement is x=x0sin⁡ωtx = x_0 \sin \omega t (starting at x=0x=0 at t=0t=0).
  • Alternatively, x=x0cos⁡ωtx = x_0 \cos \omega t (starting at x=x0x=x_0 at t=0t=0).
  • x0x_0 is the amplitude, the maximum displacement.

Worked example

Question

A displacement sensor records a slider at its left turning point, x = −20 mm, at t = 0. It next reaches the right turning point at t = 0.25 s. Assuming SHM, construct x(t) with x in metres and find the acceleration when x = +10 mm.

Solution

1. Opposite turning points are half a period apart, so T=0.50T=0.50 s and ω=2π/T=4π\omega=2\pi/T=4\pi rad s⁻¹. The amplitude is 0.020 m.
2. The initial negative turning point requires x=−0.020cos⁡(4πt)x=-0.020\cos(4\pi t) m.
3. Use a=−ω2xa=-\omega^2x. At x = +0.010 m, a=−(4π)2(0.010)=−1.58a=-(4\pi)^2(0.010)=-1.58 m s⁻², towards equilibrium.

x=−0.020cos⁡(4πt)x=-0.020\cos(4\pi t) m; the requested acceleration is −1.58 m s⁻².

Common pitfalls

  • Forgetting that the argument of the sine/cosine function (ωt\omega t) must be in radians for calculations, not degrees.
  • Incorrectly identifying ω\omega from the given equation, especially if the equation is not in the standard x=x0sin⁡ωtx = x_0 \sin \omega t form (e.g., x=x0sin⁡(kt)x = x_0 \sin(kt) where kk is actually ω\omega).
  • Misinterpreting the negative sign in a=−ω2xa = -\omega^2x as a negative magnitude; it signifies direction.

Prerequisites

  • Understanding of angular frequency, amplitude, and period.
  • Understanding the fundamental definition of SHM.
  • Familiarity with sine and cosine functions and their properties.
Further resources