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What defines simple harmonic motion, and how do displacement, velocity and acceleration vary throughout an oscillation?

Define simple harmonic motion by its defining equation, and describe the variation of displacement, velocity and acceleration with time and with displacement

A focused answer to the H2 Physics learning outcome on simple harmonic motion. The defining condition a = -omega^2 x, the displacement, velocity and acceleration relations, and the period of a mass-spring and pendulum system.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to define simple harmonic motion (SHM) through its defining equation a=ω2xa = -\omega^2 x, and to describe how displacement, velocity and acceleration vary both with time and with displacement. SHM is the model for any system with a linear restoring force, from a mass on a spring to a vibrating molecule.

The answer

The defining condition

A body performs simple harmonic motion if its acceleration is proportional to its displacement from a fixed point and always directed toward that point:

a=ω2xa = -\omega^2 x

Here xx is the displacement from equilibrium and ω\omega is the angular frequency. The negative sign is the heart of SHM: the acceleration (and the restoring force) always points back toward equilibrium, which is what sustains the oscillation.

Angular frequency, period and frequency

The angular frequency relates to the period TT and frequency ff by:

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}

For SHM, the period is independent of the amplitude (this is called isochronism), a defining and useful property.

Variation with time

Taking x=x0x = x_0 at t=0t = 0, the displacement varies as:

x=x0cos(ωt)x = x_0\cos(\omega t)

Differentiating gives the velocity and acceleration:

v=x0ωsin(ωt),a=x0ω2cos(ωt)=ω2xv = -x_0\omega\sin(\omega t), \qquad a = -x_0\omega^2\cos(\omega t) = -\omega^2 x

The velocity leads the displacement by a quarter period, and the acceleration is exactly out of phase with the displacement.

Variation with displacement

Eliminating time gives the velocity directly in terms of displacement:

v=±ωx02x2v = \pm\omega\sqrt{x_0^2 - x^2}

This shows:

  • maximum speed vmax=ωx0v_{\max} = \omega x_0 at the equilibrium position (x=0x = 0),
  • zero speed at the extremes (x=±x0x = \pm x_0),
  • maximum acceleration amax=ω2x0a_{\max} = \omega^2 x_0 at the extremes,
  • zero acceleration at equilibrium.

Standard oscillators

Two systems on the syllabus obey SHM:

  • A mass mm on a spring of constant kk: ω=km\omega = \sqrt{\dfrac{k}{m}}, so T=2πmkT = 2\pi\sqrt{\dfrac{m}{k}}.
  • A simple pendulum of length LL (small angles): ω=gL\omega = \sqrt{\dfrac{g}{L}}, so T=2πLgT = 2\pi\sqrt{\dfrac{L}{g}}.

Examples in context

Example 1. A mass on a vertical spring. Hanging a mass on a spring and displacing it slightly produces SHM about the new equilibrium, with T=2πm/kT = 2\pi\sqrt{m/k}. Plotting T2T^2 against mm gives a straight line of gradient 4π2/k4\pi^2/k, a standard way to measure the spring constant and confirm the SHM model.

Example 2. A tuning fork. The prongs of a struck tuning fork vibrate with SHM at a fixed frequency set by their stiffness and mass, independent of how hard the fork is struck (the amplitude). This isochronism is why tuning forks give a stable reference pitch.

Try this

Q1. State the defining equation of SHM and explain what each symbol means. [2 marks]

  • Cue. a=ω2xa = -\omega^2 x: aa acceleration, xx displacement from equilibrium, ω\omega angular frequency; the minus sign means acceleration is toward equilibrium.

Q2. A pendulum has length 0.99 m0.99\ \text{m}. Find its period (g=9.81 m s2g = 9.81\ \text{m s}^{-2}). [2 marks]

  • Cue. T=2πL/g=2π0.99/9.81=2.0 sT = 2\pi\sqrt{L/g} = 2\pi\sqrt{0.99/9.81} = 2.0\ \text{s}.

Q3. Sketch how the velocity of a body in SHM varies with its displacement, and mark where speed is greatest. [3 marks]

  • Cue. An ellipse-like curve v=±ωx02x2v = \pm\omega\sqrt{x_0^2 - x^2}; speed greatest at x=0x = 0 (equilibrium), zero at x=±x0x = \pm x_0.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA mass on a spring oscillates with simple harmonic motion of amplitude 0.080 m0.080\ \text{m} and period 0.50 s0.50\ \text{s}. (a) Find the angular frequency. (b) Find the maximum speed. (c) Find the maximum acceleration.
Show worked answer →

(a) Angular frequency: ω=2πT=2π0.50=12.6 rad s1\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.50} = 12.6\ \text{rad s}^{-1}.

(b) Maximum speed occurs at the equilibrium position: vmax=ωx0=12.6×0.080=1.01 m s1v_{\max} = \omega x_0 = 12.6 \times 0.080 = 1.01\ \text{m s}^{-1}.

(c) Maximum acceleration occurs at maximum displacement: amax=ω2x0=(12.6)2×0.080=158×0.080=12.7 m s2a_{\max} = \omega^2 x_0 = (12.6)^2 \times 0.080 = 158 \times 0.080 = 12.7\ \text{m s}^{-2}.

Markers reward ω=2π/T\omega = 2\pi/T, vmax=ωx0v_{\max} = \omega x_0 at the centre, and amax=ω2x0a_{\max} = \omega^2 x_0 at the extremes, each with units.

Original4 marks(a) State the defining equation of simple harmonic motion and explain the significance of the negative sign. (b) A particle in SHM has amplitude 0.10 m0.10\ \text{m} and angular frequency 5.0 rad s15.0\ \text{rad s}^{-1}. Find its speed when its displacement is 0.060 m0.060\ \text{m}.
Show worked answer →

(a) Defining equation: a=ω2xa = -\omega^2 x. The acceleration is proportional to the displacement from equilibrium and always directed toward equilibrium. The negative sign shows the acceleration (and hence the restoring force) is opposite in direction to the displacement, which is what makes the motion oscillate.

(b) Use v=±ωx02x2v = \pm\omega\sqrt{x_0^2 - x^2}: v=5.00.1020.0602=5.00.010.0036=5.00.0064=5.0×0.080=0.40 m s1v = 5.0\sqrt{0.10^2 - 0.060^2} = 5.0\sqrt{0.01 - 0.0036} = 5.0\sqrt{0.0064} = 5.0 \times 0.080 = 0.40\ \text{m s}^{-1}.

Markers reward the defining equation, the explanation that the negative sign means the restoring acceleration opposes displacement, and correct use of the velocity-displacement relation.

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