CP 9.8 - Damped and forced oscillations

Modelling resistance and classifying damping

Does adding more resistance always bring a displaced object back sooner? It suppresses oscillations, but sufficiently strong resistance can also make the eventual return slower. The roots of a differential equation explain both effects.

Measure displacement $x$ from equilibrium. A linear restoring force is $-kx$, while resistance proportional to velocity is $-c\dot x$, with $k>0$, $c\ge0$. The resistance changes direction automatically when velocity changes sign. With an applied force $F(t)$, Newton's second law gives the model.

The model assumes constant coefficients, a linear restoring force and resistance proportional to velocity. Any constant force already balanced at equilibrium disappears when displacement is measured from that equilibrium. Large-amplitude real motion may violate the linear restoring-force assumption.

For free motion, use $mr^2+cr+k=0$. With $c>0$, the sign of $c^2-4mk$ gives three cases.

Here $\omega_{\mathrm d}=\sqrt{k/m-c^2/(4m^2)}$ is the light-damping angular frequency. These classification and frequency results are not supplied in the formula booklet. The case $c=0$ is undamped, not lightly damped.

Distinguish a repeated root from a complex pair in this check.

Light damping and a decaying envelope

Complex auxiliary roots retain the oscillation, while their negative real part supplies exponential decay. Include the exponential factor when differentiating for initial velocity.

For $\ddot x+2\dot x+5x=0$, the roots are $-1\pm2\mathrm{i}$, giving $x=\mathrm{e}^{-t}(A\cos2t+B\sin2t)$. The oscillatory factor has amplitude $\sqrt{A^2+B^2}$, so the envelope curves $x=\pm\sqrt{A^2+B^2}\,\mathrm{e}^{-t}$ bound the displacement. Actual extrema need not occur where the trigonometric factor alone is maximal.

Worked example 2 changes only the initial velocity. The decay rate and oscillatory frequency belong to the equation, so neither changes.

The oscillation period of the trigonometric factor is $\pi$, but the full displacement is not periodic: successive cycles have smaller sizes. Distinguish the time between corresponding phases from a function repeating exactly.

Use another initial velocity with the same damped equation.

Critical and heavy damping

A repeated negative root gives critical damping; two distinct negative roots give heavy damping. Both responses tend to equilibrium without a continuing oscillation. Initial velocity can still affect whether an individual response crosses equilibrium, so do not infer every detail from the classification alone.

[DIAGRAM: asset_name: Light critical and heavy damping from the same initial data; asset_slug: 9fm0-cp98-damping-comparison; caption: With the same mass and restoring coefficient, increasing damping removes oscillation but can lengthen the late return; description: A time-displacement graph compares solutions of x double dot plus c x dot plus 4x equals zero with x(0)=1 and x dot(0)=0. The light curve c=2 crosses the axis, the critical curve c=4 equals (1+2t) exp(-2t), and the heavy curve c=5 equals (4 exp(-t)-exp(-4t))/3. All approach zero, with the heavy curve above the critical curve at later times.]