Damped and forced oscillations

Damped oscillations occur when, besides the restoring force $F=-kx$, a resistance proportional to velocity acts: $F_{\text{drag}}=-bv$. Newton's second law then gives $m\ddot{x}+b\dot{x}+kx=0$, whose weak-damping solution is $x(t)=A_0e^{-\gamma t}\cos(\omega t+\varphi)$ with damping rate $\gamma=b/(2m)$. Energy decays twice as fast as amplitude: $E(t)\approx\frac12 kA_0^2e^{-2\gamma t}$, and the fraction lost per period is $\Delta E/E\approx 2\pi/Q$, where the quality factor is $Q=m\omega_0/b=\omega_0/(2\gamma)$. Forced oscillations obey $m\ddot{x}+b\dot{x}+kx=F_0\cos\omega t$; after the transient dies out, the system oscillates at the drive frequency with steady amplitude $A(\omega)=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2}}.$ The maximum of this curve is resonance: for light damping $\omega_{\text{res}}\approx\omega_0$, and the peak height $\sim F_0/(2m\gamma\omega_0)$ grows as $b$ decreases. Practical analysis therefore reduces to two questions: how fast the system loses energy, and how large its amplitude becomes under external drive.

An oscillator with mass $m=0.50$ kg on a spring $k=20$ N/m has drag coefficient $b=0.40$ kg/s. Find $\omega_0$, $\gamma$, $Q$, and the time for the amplitude to fall to half its initial value.

The same oscillator ($m=0.50$ kg, $k=20$ N/m, $b=0.40$ kg/s) is driven by $F(t)=F_0\cos\omega t$ with $F_0=1.0$ N. Find the steady-state amplitude at resonance, the stored energy, and the energy dissipated per cycle; compare the last quantity with the estimate $2\pi/Q$.

A car body of mass $m=400$ kg (per wheel) rides on a spring $k=2.0\cdot 10^4$ N/m. Road ripples act as a driving force with amplitude $F_0=200$ N. Choose a damping coefficient $b$ such that the steady-state amplitude at resonance does not exceed $5.0$ cm, and report the resulting $Q$ and bandwidth $\Delta\omega$.

Resonance simulation (safe self-check). Open an interactive driven-oscillator simulation (for example "Masses and Springs" or any driven damped oscillator model).

1. Set $m$, $k$, and two values of $b$: a small one and one four times larger; compute $\gamma$, $Q$, and $\Delta\omega\approx2\gamma$ in advance.
2. Slowly sweep the drive frequency from $0.5\omega_0$ to $1.5\omega_0$ in steps of $0.1\omega_0$, waiting for steady state each time, and record the amplitude.
3. Plot $A(\omega/\omega_0)$ for both values of $b$ and compare the measured peak height with $F_0/(2m\gamma\omega_0)$ and the width with $2\gamma$.
4. Switch off the drive and read the half-amplitude time from the graph; compare it with $\ln 2/\gamma$.

Optional (safe measurements only): hang a pendulum and, next to it, an identical pendulum with a small cardboard vane attached; release both and record the amplitude every 5 oscillations. The vane increases air drag, so its amplitude decays faster: estimate $\gamma$ from $A(t)=A_0e^{-\gamma t}$. Do not drive real structures or induce resonance in household appliances.

Show how damping both lowers and broadens the resonance peak, and connect energy loss with the decay envelope.

Graph: horizontal axis is the dimensionless frequency $\omega/\omega_0$, vertical axis is the normalized amplitude $A/(F_0/k)$; three curves for $Q=5$, $20$, and $100$ obtained from $A(\omega)=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2}},$ with the peak positions $\omega_{\text{res}}\approx\omega_0$ marked and the widths $\Delta\omega\approx2\gamma$ indicated. Inset: a decay sketch — the graph of $x(t)$ with envelopes $\pm A_0e^{-\gamma t}$ and the annotation of the energy fraction lost per period $\Delta E/E\approx2\pi/Q$.

A $1500\ \text{kg}$ car crosses a speed bump at $30\ \text{km/h}$ and joins a concrete motorway whose expansion joints are $20\ \text{m}$ apart. The combined stiffness of the four springs is $k=1.2\times10^{5}\ \text{N/m}$, so the body bouncing on them has a natural frequency $f_0=\frac{1}{2\pi}\sqrt{k/m}=1.4\ \text{Hz}$. At $90\ \text{km/h}=25\ \text{m/s}$ the wheels strike a joint every $20/25=0.8\ \text{s}$, that is at $1.25\ \text{Hz}$ — close to $f_0$ — and the body starts rocking visibly. Hydraulic shock absorbers stop this: with damping ratio $\zeta=0.3$ the amplitude falls by a factor $e^{2\pi\zeta/\sqrt{1-\zeta^2}}\approx7$ during one free period, so after a single bump the body settles in about $1\ \text{s}$.

The free bounce of the body obeys $m\ddot{x}+c\dot{x}+kx=0$ with $c=2\zeta\sqrt{km}$: the amplitude decays as $A(t)=A_0e^{-\gamma t}$ with $\gamma=\zeta\omega_0$ (here $\gamma\approx2.6\ \text{s}^{-1}$). The road profile plays the role of the periodic driving force $F_0\cos\omega t$: the response peaks near $\omega\approx\omega_0$, that is at $f\approx1.4\ \text{Hz}$, which is why the body rocks at $90\ \text{km/h}$.

A $30\ \text{kg}$ child rides a swing whose ropes are $2.5\ \text{m}$ long: $f_0=\frac{1}{2\pi}\sqrt{g/l}=0.31\ \text{Hz}$, $T=3.2\ \text{s}$. An adult pushes once per period, at the lowest point, in the direction of motion: a force of $20\ \text{N}$ acting over $0.3\ \text{m}$ does $6\ \text{J}$ of work, about $12\ \text{J}$ per cycle (average power $3.8\ \text{W}$). At a $30^\circ$ amplitude the stored energy is $E=mgl(1-\cos 30^\circ)\approx99\ \text{J}$, and roughly a tenth of it is lost each cycle to air drag and pivot friction, so $Q=2\pi E/\Delta E\approx50$; the pushes exactly replace this loss. Push out of step, say once every $2\ \text{s}$ ($0.5\ \text{Hz}$, $1.6$ times $f_0$), and the amplitude stays small.

This is forced oscillation: the work done by the driving force per cycle is positive only when the force is in phase with the velocity, which happens near $\omega_0$ — at the top of the resonance curve $A(\omega)$. Without pushes the swing is a damped oscillator with $A(t)=A_0e^{-\gamma t}$, and the steady state sets in when the work of the pushes per cycle equals the dissipation per cycle.

A washing machine drum with wet laundry, total mass $25\ \text{kg}$, spins at $1200\ \text{rpm}=20\ \text{Hz}$. The drum is carried on springs inside the frame, and this suspension has a natural frequency of only about $2\ \text{Hz}$. During the $30\ \text{s}$ spin-up the rotation rate passes through $2\ \text{Hz}$: that is when the frame shakes hardest, and only friction dampers keep the swing at a few centimetres instead of letting it grow by a factor $Q\approx10$. A lump of wet cloth $m=0.2\ \text{kg}$ at radius $r=0.25\ \text{m}$ makes a rotating force $F=mr\omega^2=0.2\cdot0.25\cdot(2\pi\cdot20)^2\approx790\ \text{N}$ at full speed — yet at $\omega/\omega_0\approx10$ the frame barely moves at all.

Both halves of the theory are at work here: the amplitude–frequency response of a forced oscillator with its maximum at $\omega_0=\sqrt{k/m}$ (the passage through resonance during spin-up) and its fall-off at $\omega\gg\omega_0$ (the calm spin at $20\ \text{Hz}$). The damping ratio $\zeta$ sets the height and width of the resonance peak, and the energy balance of the forced oscillator says that per cycle the imbalance feeds in work $\propto F\cdot A$, which the dampers carry away as heat.