17.1 - Simple harmonic displacement and acceleration

Describing an oscillation

A swing stops for an instant at the end of its travel. Is its acceleration zero there too? Being momentarily at rest does not tell you whether velocity is changing. To answer, distinguish where the oscillator is from the way its motion changes.

An oscillation is repeated motion about an equilibrium position. Define displacement $x$ from that equilibrium, with one direction positive. Displacement is signed; the amplitude $x_0$ is the maximum magnitude of displacement, so it is a positive length. The two extreme positions are $+x_0$ and $-x_0$; their separation is $2x_0$.

The period $T$ is the time for one complete oscillation, returning to the same phase of motion. The frequency $f$ is the number of complete oscillations per unit time. Passing the equilibrium position twice does not necessarily count as two complete oscillations: the direction of motion matters.

Phase identifies the stage of a cycle, including the direction of motion. Two oscillations at the same frequency may be in phase, in antiphase or separated by another phase difference. For a time separation $\Delta t$, the magnitude of the phase difference is $\Delta\phi=2\pi\Delta t/T$ in radians; this relation is not printed on the data sheet. State which motion leads or lags if that is required. Compare like points, such as two positive-going equilibrium crossings.

Apply both timing and phase to a new set of readings.

Restoring acceleration and small angles

A displaced oscillator needs a resultant force towards equilibrium if it is to turn back. Simple harmonic motion makes a more precise requirement: acceleration is proportional to displacement and oppositely directed.

“Restoring force” names the resultant responsible for this acceleration; it is not another force to add to a force diagram. For a horizontal ideal spring, with vertical weight and support balanced, the horizontal spring force is the restoring resultant. Hooke's law gives $F=-kx$, so $a=-(k/m)x$ and $\omega^2=k/m$. Here spring constant $k$ is in $\mathrm{N\,m^{-1}}$ and mass $m$ in kilograms. This spring result is derived from the laws, rather than printed separately on the data sheet.

[DIAGRAM: asset_name: Acceleration points back towards equilibrium; asset_slug: oscillations-restoring-spring; caption: Positive displacement has negative acceleration and negative displacement has positive acceleration; description: Two views of the same horizontal spring-mass system show the mass displaced equally to the right and left of the equilibrium marker. Each acceleration arrow begins at the mass and points towards equilibrium with equal length. Weight and support are stated to balance vertically.]

For a $0.25\,\mathrm{kg}$ mass on a $16\,\mathrm{N\,m^{-1}}$ spring, $\omega=\sqrt{16/0.25}=8.0\,\mathrm{rad\,s^{-1}}$. At $x=+0.030\,\mathrm{m}$, $a=-(8.0)^2(0.030)=-1.92\,\mathrm{m\,s^{-2}}$. At the opposite displacement, acceleration has the opposite sign. At equilibrium, acceleration is zero even though speed can be large.

Why a small pendulum swing is approximately harmonic

A pendulum's tangential restoring acceleration is $-g\sin\theta$, where $\theta$ is its angular displacement from the downward vertical. Its arc displacement is $x=l\theta$ for length $l$ and $\theta$ in radians. The acceleration is not exactly proportional to $x$, because sine is not exactly proportional to its angle.

For a small angle measured in radians,