damping forcing resonance and self-oscillation

A free harmonic oscillator obeys $m\ddot{x}+kx=0$, but real systems lose energy. Damping is introduced by a resistive force, often $F_{\mathrm{drag}}=-b\dot{x}$, so the equation becomes $m\ddot{x}+b\dot{x}+kx=0$; the dimensionless parameter $\zeta=b/(2\sqrt{km})$ separates underdamped, critically damped, and overdamped regimes. Under periodic external driving $F(t)=F_0\cos\omega t$, the solution contains a decaying transient part and a steady-state part at the driving frequency. The steady-state amplitude is $A(\omega)=F_0/\sqrt{(k-m\omega^2)^2+(b\omega)^2}$, and resonance occurs near $\omega\approx\omega_0=\sqrt{k/m}$; with damping, the maximum shifts to $\omega_r=\sqrt{\omega_0^2-2\beta^2}$, where $\beta=b/(2m)$. Self-oscillation is not external forcing but maintenance of oscillation by the system itself through feedback and negative damping, leading to a limit cycle; examples include a clock pendulum, a violin string, and an electronic oscillator.

An oscillator has $m=1\ \mathrm{kg}$, $k=100\ \mathrm{N/m}$, and $b=8\ \mathrm{kg/s}$. Determine $\omega_0$, $\beta$, $\zeta$, and $\omega_d$; classify the regime.

For the system $m=1\ \mathrm{kg}$, $k=100\ \mathrm{N/m}$, add driving $F(t)=F_0\cos\omega t$ with $F_0=10\ \mathrm{N}$ and $b=2\ \mathrm{kg/s}$. Find the steady-state amplitude at $\omega=8\ \mathrm{rad/s}$ and at $\omega=\omega_0=10\ \mathrm{rad/s}$. Compare them.

A van der Pol self-oscillator is described by $\ddot{x}-\mu(1-x^2)\dot{x}+\omega_0^2 x=0,\qquad \mu=0.5,\ \omega_0=2\ \mathrm{rad/s}.$ Determine the sign of the effective damping at $x=0.5$ and $x=1.5$. Explain why the system reaches a limit cycle without an external periodic force, and how this differs from resonance.

Carry out safe self-study with an interactive simulation (for example, PhET “Resonance” or spreadsheet plotting). Fix $m$ and $k$, vary $b$ and the driving frequency $\omega$, and record the steady-state amplitude and phase lag; plot a family of $A(\omega)$ curves and mark resonance. Then in a separate negative-damping simulation (van der Pol model), observe the limit cycle in the $x$–$\dot{x}$ phase plane. Do not use loud sound, high voltage, or fragile mechanical structures to achieve resonance.

The visualization aims to show how damping limits and shifts the resonance peak, how the phase changes, and how the limit cycle of self-oscillation differs from forced resonance.

A graph of normalized amplitude $A/(F_0/k)$ versus $\omega/\omega_0$ for several values of $\zeta$; next to it, a phase diagram of $\dot{x}$ versus $x$ for a negative-damping oscillator, showing the limit cycle.

Station wagon: sprung mass on one wheel $m = 350\ \text{kg}$, spring stiffness $k = 30\ \text{kN/m}$, shock-absorber damping coefficient $c = 1500\ \text{N·s/m}$. The suspension's own frequency is $\omega_0 = \sqrt{k/m} = \sqrt{30000/350} \approx 9.3\ \text{rad/s}$, about $1.5\ \text{Hz}$. After a speed bump the body nods for roughly $1\ \text{s}$ and settles.

This is the solution of the damped-oscillator equation $m\ddot{x} + c\dot{x} + kx = 0.$ With $\zeta = c/(2\sqrt{km}) \approx 0.23 < 1$ the system is underdamped: $x(t) = A e^{-\zeta\omega_0 t}\cos(\omega_d t + \varphi)$, where $\omega_d = \omega_0\sqrt{1-\zeta^2} \approx 9.1\ \text{rad/s}$ and the decay time is $1/(\zeta\omega_0) \approx 0.47\ \text{s}$ — that is why one bump gives one or two visible nods instead of a long bounce.

Rope length $L = 2.5\ \text{m}$, so $\omega_0 = \sqrt{g/L} = \sqrt{9.81/2.5} \approx 1.98\ \text{rad/s}$ and the period is $T = 2\pi/\omega_0 \approx 3.2\ \text{s}$. An adult pushes once per period with a force of about $20\ \text{N}$ lasting $0.3\ \text{s}$. The arc grows to about $35^\circ$ and then stops growing.

This is the driven oscillator $m\ddot{x} + c\dot{x} + kx = F_0\cos\omega t,$ whose steady-state amplitude $A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + (2\zeta\omega_0\omega)^2}}$ peaks near $\omega = \omega_0$. Pushing once every $3.2\ \text{s}$ puts $\omega$ right on the peak (resonance), and the limiting arc is set by the balance between the work of one push per cycle and the losses to air drag and friction.

A pendulum of length $L = 1.0\ \text{m}$ has $T = 2\pi\sqrt{L/g} \approx 2.0\ \text{s}$; the $m = 0.5\ \text{kg}$ bob swings with an amplitude of $2^\circ \approx 0.035\ \text{rad}$. A $4\ \text{kg}$ weight descends $1.0\ \text{m}$ per day, releasing $4 \times 9.81 \times 1.0 \approx 39\ \text{J}$ per day, an average of about $0.45\ \text{mW}$. The amplitude stays constant for weeks.

This is self-oscillation: the escapement delivers a short impulse at each half-period, tied to the pendulum's own motion rather than to an external force of some other frequency. The system sets the period $T$ itself. The constant amplitude settles where the energy fed in per cycle by the falling weight equals the losses to friction and air drag — the limit cycle of a non-linear oscillator with feedback.