Question bank — Q factor — quality of oscillator

This page belongs to Q factor — quality of oscillator and leans on Damped Harmonic Motion, Simple Harmonic Motion, Resonance & Forced Oscillations, RLC Circuits and Energy in Oscillations.

Where $Q={\omega }_0/\gamma$ comes from, and exactly when it holds

Before trusting any trap below, see why the clean formula $Q={\omega }_0/\gamma$ is only a light-damping statement.

Start from the energy definition and the ring-down envelope $A(t)=A_0{e}^{-\gamma t/2}$ (from Damped Harmonic Motion). Since energy $\propto$ amplitude${}^2$ (from Energy in Oscillations), $E(t)=E_0{e}^{-\gamma t}.$ Over one period $T$ the fraction of energy lost is $\frac{\Delta E}{E}=1-e^{-\gamma T}.$ Why an approximation now? For light damping the exponent $\gamma T$ is tiny, so the exponential's Taylor expansion $e^{-\gamma T}\approx 1-\gamma T$ gives $\frac{\Delta E}{E}\approx \gamma T=\gamma \cdot \frac{2\pi }{{\omega }_0}.$ This is where light damping enters, and it is worth pinning down how light. Dropping the next term $(\gamma T{)}^2/2$ is only legitimate when $\gamma T=\gamma \cdot \frac{2\pi }{{\omega }_0}\ll 1⟹\gamma \ll \frac{{\omega }_0}{2\pi }⟹Q=\frac{{\omega }_0}{\gamma }\gg 2\pi \approx \mathrm{6.3.}$ So the approximation genuinely wants $Q\gg 2\pi$ (certainly $Q\gg 1$) — not merely $Q\gg \frac{1}{2}$. The value $Q=\frac{1}{2}$ is a different landmark: it marks where oscillation itself disappears (critical damping), not where the light-damping expansion becomes accurate. Plug in: $Q=2\pi \cdot \frac{E}{\Delta E}\approx 2\pi \cdot \frac{1}{\gamma T}=2\pi \cdot \frac{{\omega }_0}{2\pi \gamma }=\frac{{\omega }_0}{\gamma }.$

The figure above shows the ring-down envelope: the amplitude falls to $1/e$ after $\tau =2/\gamma$, which is ${\omega }_0\tau /2\pi =Q/\pi$ oscillations.

Why the resonance width is exactly $\gamma$ (visual derivation)

The bandwidth face of $Q$ deserves its own picture, not a bald assertion. Drive the oscillator with a force at frequency $\omega$; from Resonance & Forced Oscillations the steady amplitude squared (which sets the power the oscillator soaks up) is the Lorentzian $P(\omega )\propto \frac{1}{({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2}.$ Why amplitude squared? Power delivered goes as velocity${}^2$, hence as amplitude${}^2$ — the same "square" that made energy decay twice as fast as amplitude (Energy in Oscillations).

Now find where this curve drops to half its peak. The peak sits essentially at $\omega \approx {\omega }_0$, where the first bracket vanishes and $P_{\max }\propto 1/({\gamma }^2{\omega }_0^2)$. Write $\omega ={\omega }_0+\delta$ for a small offset $\delta$ from the peak. Then ${\omega }_0^2-{\omega }^2=({\omega }_0-\omega )({\omega }_0+\omega )\approx (-\delta )(2{\omega }_0)=-2{\omega }_0\delta .$ Substitute into the Lorentzian and set it equal to half the peak: $\frac{1}{(2{\omega }_0\delta {)}^2+{\gamma }^2{\omega }_0^2}=\frac{1}{2}\cdot \frac{1}{{\gamma }^2{\omega }_0^2}.$ Cross-multiplying, the half-power condition is $(2{\omega }_0\delta {)}^2={\gamma }^2{\omega }_0^2$, i.e. $\delta =\pm \frac{\gamma }{2}.$ So the curve falls to half its height at ${\omega }_0-\gamma /2$ on the left and ${\omega }_0+\gamma /2$ on the right. The full width between those two points is $\Delta \omega =\left({\omega }_0+\frac{\gamma }{2}\right)-\left({\omega }_0-\frac{\gamma }{2}\right)=\gamma .$ That is the whole reason $\Delta \omega =\gamma$, and hence $Q={\omega }_0/\Delta \omega$. The figure below marks the peak, the half-power line, and the two edges at ${\omega }_0\pm \gamma /2$:

Why "cycles" stop making sense at heavy damping — a walk through the root plane

To see the underdamped/critical/overdamped boundary geometrically, put $x=e^{\lambda t}$ into the damped ODE (Damped Harmonic Motion). This gives ${\lambda }^2+\gamma \lambda +{\omega }_0^2=0$, whose two roots are $\lambda =-\frac{\gamma }{2}\pm \sqrt{{\left(\frac{\gamma }{2}\right)}^2-{\omega }_0^2}.$ Read each part with a picture in mind. The real part $-\gamma /2$ is the decay rate (how fast the envelope shrinks). The imaginary part, when it exists, is the ringing frequency ${\omega }_d$. The figure below plots both roots as $\gamma$ grows from $0$:

Follow the annotated arrows on the figure:

• $\gamma =0$ (top): roots sit on the imaginary axis at $\pm i{\omega }_0$ — pure oscillation, no decay, infinite $Q$.
• $0<\gamma <2{\omega }_0$ (blue path): both roots move left (faster decay) while staying complex; the imaginary part shrinks along the blue curve, so the system still rings but at ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}<{\omega }_0$.
• $\gamma =2{\omega }_0$ ($Q=\frac{1}{2}$): the square root hits zero, the two roots collide on the real axis. No imaginary part remains ⇒ no oscillation ⇒ "counting cycles" becomes meaningless.
• $\gamma >2{\omega }_0$ (red points): the roots split apart along the real axis — two pure decays, the overdamped creep-back. There is no ringing to count at all.

The value ${\omega }_d<{\omega }_0$ visible in the shrinking imaginary part is exactly why the observed ring-down frequency is always a touch below the natural one.

True or false — justify

Each answer must give the reason, not just "true/false".

Spot the error

Each line states a flawed claim; the answer names the flaw and repairs it.

Why questions

Edge cases

Connections

• Q factor — quality of oscillator — parent topic these traps target.
• Damped Harmonic Motion — where $\gamma$ and the underdamped/critical/overdamped cases come from.
• Simple Harmonic Motion — the $b=0$, $Q\to \infty$ limit.
• Resonance & Forced Oscillations — bandwidth and build-up-time trade-offs.
• RLC Circuits — the $Q=\frac{1}{R}\sqrt{L/C}$ electrical analogue.
• Energy in Oscillations — why energy decays twice as fast as amplitude.