Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 11

Assume $k>0$, $m>0$ and $\gamma \ge 0$. The spring constant $k$ measures restoring force per displacement from the equilibrium position $x_0$. The term $-\gamma mv$ is linear drag, with damping rate $\gamma$ in inverse-time units. Writing ${\omega }_0=\sqrt{k/m}$, the equation isFor small positive damping it gives underdamped motion: oscillations about $x_0$ with a decaying amplitude. If $\gamma =0$, the motion is undamped simple harmonic motion.

Set ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}>0$. The real solution of the damped harmonic oscillator equation is $x-x_0=e^{-\gamma t/2}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$. The initial displacement fixes $A=a-x_0$, and release from rest gives $B=\gamma A/(2{\omega }_d)$. ThereforeThe exponential decay rate of the amplitude is $\gamma /2$, while the weak-damping energy decay rate is $\gamma$.

The amplitude envelope falls by $e^{-1/2}$ when $e^{-\gamma t/2}=e^{-1/2}$, so $t=1/\gamma$. Dividing by the period $2\pi /{\omega }_d$ givesThis is a cycle-count estimate in the small-damping regime, not necessarily an integer. With zero damping there is no amplitude decay and no finite such time.

The weak-damping mechanical energy is proportional to the squared oscillation amplitude, so it decays as $e^{-\gamma t}$. Over one period $P=2\pi /{\omega }_d$, the fractional loss is $1-e^{-\gamma P}\simeq \gamma P$. HenceFor the release-from-rest solution, comparison at successive corresponding turning points gives the more precise ratio $E/\Delta E=(1-e^{-\gamma P}{)}^{-1}$. The quality factor is $Q\simeq 2\pi E/\Delta E={\omega }_0/\gamma$; the requested energy-to-loss ratio itself has the extra factor $1/(2\pi )$.

The moving equilibrium produces a forced harmonic oscillator:A particular solution is $F{e}^{-\alpha t}$, where $F={\omega }_0^{2}d/({\alpha }^2-\gamma \alpha +{\omega }_0^2)$. The denominator equals $(\alpha -\gamma /2{)}^2+{\omega }_d^2>0$, so it cannot vanish for real $\alpha$ under the given underdamping condition. Imposing both initial conditions givesFor $\alpha =\gamma$, $F=d$, so the entire oscillatory term vanishes:The body is tracking an exponentially moving equilibrium without oscillation. There is no spring extension in this special motion: linear drag alone supplies $m\ddot{x}={\gamma }^{2}md{e}^{-\gamma t}$. An underdamped homogeneous equation does not require every forced solution to contain an oscillatory transient.