Q factor — quality of oscillator: Worked examples

This page is a drill hall. The parent note Q factor — quality of oscillator built the three faces of $Q$ ; here we throw every kind of question at those formulas — small $Q$ , huge $Q$ , the degenerate boundary where oscillation dies, a real-world word problem, and an exam-twist. First we map the whole territory, then we walk each cell.

The scenario matrix

Every $Q$ problem is really one of these cells. The three "input routes" are the three faces of the same number.

Cell	What you're given	Which formula	Regime
A	Energy lost per cycle $Δ E / E$	$Q = 2 π \frac{E}{Δ E}$	light damping
B	$m, k, b$ (mechanical parts)	$Q = \frac{mk}{b}$	light damping
C	Resonance width $f_{0}, Δ f$	$Q = \frac{f _{0}}{Δ f}$	light damping
D	Ring-down count (swings to $1/ e$ )	$Q = π N_{amp}$	light damping
E	Electrical parts $R, L, C$	$Q = \frac{1}{R} L / C$	light damping
F	Scaling / "what if I change one part"	proportionalities	light damping
G	Degenerate: $Q \leq \frac{1}{2}$	no oscillation — check regime	critical / over
H	Word problem (real device)	pick the right face	light damping
I	Exam twist (two faces combined)	chain two formulas	light damping

We now hit A → I, one example each (plus one extra so both mechanical and electrical get their own worked case). Every cell of the matrix gets covered.

Before the examples, one line of vocabulary we will lean on constantly:

Definition The symbols, in plain words

Numbers you compute or read off:

$Q$ — a plain number (no units). Big = ringy, small = dead-thud.
$ω_{0}$ — the "wants-to-swing" angular rate, in radians per second, $ω_{0} = k / m$ .
$f_{0}$ — the ordinary frequency in cycles per second (hertz), tied to $ω_{0}$ by $f_{0} = ω_{0} /2 π$ (there are $2 π$ radians in one cycle, so dividing converts radians-per-second into cycles-per-second).
$γ$ — the "friction eats speed" rate, in per-seconds, $γ = b / m$ .
$Δ E / E$ — the fraction of energy gone after one full back-and-forth.
$Δ f$ — the width of the resonance hump, measured where the power is half its peak.

Physical parts (the hardware):

$m$ — mass (kg); $k$ — spring stiffness (N/m); $b$ — mechanical damping coefficient (kg/s). These are the three parts of a mass–spring–damper.
$R$ — resistance (Ω); $L$ — inductance (H); $C$ — capacitance (F). These are the electrical twins: $L$ acts like $m$ , $1/ C$ like $k$ , $R$ like $b$ (see RLC Circuits).

Counts (how many swings):

$N_{amp}$ — number of full oscillations for the amplitude to fall to $1/ e$ ; equals $Q / π$ .
$N_{energy}$ — number of full oscillations for the energy to fall to $1/ e$ ; equals $Q /2 π$ (half of $N_{amp}$ , because energy $\propto$ amplitude $^{2}$ decays twice as fast).

All of these were derived in the parent; here we only use them.

Here is the whole terrain at a glance — the same $Q$ reached from four different doors.

Figure — Q factor — quality of oscillator

Intuition How to read the terrain figure (Cell → door map)

The amber box in the centre is the single number $Q$ . Four arrows point into it, one from each corner — each corner is a different thing you were handed that lets you compute the very same $Q$ :

Top-left door → Cell A: you were handed the energy lost per cycle $Δ E / E$ , so you enter via $Q = 2 π E /Δ E$ .
Top-right door → Cells B and E: you were handed the physical parts — mechanical $m, k, b$ (Cell B) or their electrical twins $R, L, C$ (Cell E) — so you enter via $Q = mk / b$ (or $\frac{1}{R} L / C$ ).
Bottom-left door → Cell C: you were handed the resonance width $f_{0}, Δ f$ , so you enter via $Q = f_{0} /Δ f$ .
Bottom-right door → Cell D: you were handed a ring-down count $N$ , so you enter via $Q = π N$ .

The point of the picture: four utterly different measurements, one destination. As you meet each example below, first ask "which door am I standing at?" — that single question picks your formula. (Cells F, G, H, I are not new doors; they re-use these four doors in trickier ways: scaling one part, refusing the formula, a real device, and chaining two doors together.)

Cell A — given energy lost per cycle

Worked example A · A gong that loses 1% per swing

(This is the top-left door of the terrain figure above.) A struck gong oscillates and loses $1%$ of its stored energy each full period. Find $Q$ .

Forecast: Losing only a hundredth each cycle sounds very stingy — guess whether $Q$ lands nearer $6$ , $60$ , or $600$ before reading on.

Step 1 — write the fraction. $\frac{Δ E}{E} = 0.01$ . Why this step? This is the raw input to the energy face of $Q$ ; nothing to compute yet.

Step 2 — apply the energy definition. $Q = 2 π \cdot \frac{E}{Δ E} = 2 π \cdot \frac{1}{0.01} = 200 π \approx 628.$ Why this step? The definition $Q = 2 π E /Δ E$ inverts the ratio: a small loss fraction lands in the denominator, so it makes $Q$ large — exactly the "stingy = high $Q$ " idea.

Verify: Sanity — losing $1%$ per cycle is ten times stingier than the parent's $2%$ pendulum ( $Q \approx 314$ ), and indeed $628 = 2 \times 314$ . Halving the loss fraction doubles $Q$ . Units: pure number ✓.

Cell B — given the mechanical parts $m, k, b$

Worked example B · A car suspension element

(This is the top-right door of the terrain figure — "parts in, $Q$ out".) $m = 2.0 kg$ , $k = 800 N/m$ , $b = 8.0 kg/s$ . Find $Q$ and say whether it oscillates.

Forecast: Car suspensions are deliberately near-dead so you don't bounce. Guess: is $Q$ big or small?

Step 1 — natural frequency. $ω_{0} = k / m = 800/2 = 400 = 20 rad/s .$ Why this step? $ω_{0}$ is the "wants-to-swing" rate; every mechanical $Q$ needs it.

Step 2 — damping rate. $γ = b / m = 8.0/2.0 = 4.0 s^{- 1} .$ Why this step? $γ$ is the friction-eats-speed rate, the denominator of $Q = ω_{0} / γ$ .

Step 3 — combine. $Q = \frac{ω _{0}}{γ} = \frac{20}{4.0} = 5.0.$ Why this step? The boxed formula turns the two rates into the quality number.

Verify: Cross-check with the one-shot form $Q = mk / b = 2 \cdot 800 /8 = 1600 /8 = 40/8 = 5.0$ ✓. Is it underdamped? $Q = 5 > \frac{1}{2}$ , so yes it oscillates — but only mildly ringy, which is what a suspension should be.

Cell C — given resonance width

Worked example C · A tuning-fork resonance curve

(This is the bottom-left door of the terrain figure — "curve width in, $Q$ out".) A tuning fork resonates at $f_{0} = 440 Hz$ ; its response drops to half-power at $439.6 Hz$ and $440.4 Hz$ . Find $Q$ .

Forecast: The peak is razor-thin (only $\pm 0.4$ Hz). Sharp peak ⇒ high or low $Q$ ?

Step 1 — read off the full width. $Δ f = 440.4 - 439.6 = 0.8 Hz .$ Why this step? Bandwidth is the total width between the two half-power points, not the half-width — a classic slip.

Step 2 — apply the bandwidth face. $Q = \frac{f _{0}}{Δ f} = \frac{440}{0.8} = 550.$ Why this step? The bandwidth definition converts "how narrow" directly into "how ringy". A narrow $Δ f$ divides into a large $Q$ .

Verify: A tuning fork should ring for a long time — $Q = 550$ means about $Q / π \approx 175$ audible swings before the amplitude falls to $1/ e$ ✓, consistent with a fork humming for a good fraction of a second.

Cell D — given the ring-down count (a figure cell)

Worked example D · Counting swings on a decay trace

(This is the bottom-right door of the terrain figure — "swing count in, $Q$ out".) An oscilloscope shows a freely decaying oscillation whose amplitude falls to $1/ e$ of its start after 20 full oscillations. Estimate $Q$ .

Forecast: More swings before dying = higher or lower $Q$ ?

Look at the envelope in the figure below — the dashed amber curve is $A_{0} e^{- γ t /2}$ , and the horizontal amber line marks the $1/ e$ level.

Step 1 — link count to $Q$ . The parent showed amplitude reaches $1/ e$ after about $N_{amp} = Q / π$ oscillations (this $N_{amp}$ is the count defined in the vocabulary block above). Why this step? That relation is exactly the "ring-down" face; we invert it to solve for $Q$ .

Step 2 — invert. $Q = π N_{amp} = π \times 20 \approx 62.8.$ Why this step? Multiplying the counted swings by $π$ undoes the $Q / π$ relation.

Verify: Energy (which goes as amplitude $^{2}$ ) should reach $1/ e$ twice as fast: after $N_{energy} = Q /2 π = 62.8/6.28 = 10$ oscillations. Ten is half of twenty ✓ — consistent because energy decays at twice the amplitude rate.

Cell E — the electrical analogue $R, L, C$

Worked example E · An RLC resonant filter

(Same top-right door as Cell B — "parts in, $Q$ out" — but now the parts are electrical.) A series RLC circuit has $L = 50 μ H$ , $C = 200 pF$ , $R = 5 Ω$ . Find $Q$ and the resonant frequency $f_{0}$ .

Forecast: Radio front-ends want sharp tuning. Expect $Q$ in the tens, hundreds, or thousands?

Step 1 — quality from parts. $Q = \frac{1}{R} \frac{L}{C} = \frac{1}{5} \frac{50 \times 1 0 ^{- 6}}{200 \times 1 0 ^{- 12}} .$ Why this step? This is the electrical twin of $Q = mk / b$ : $L$ plays $m$ , $1/ C$ plays $k$ , $R$ plays $b$ . See RLC Circuits.

Step 2 — crunch the ratio. $\frac{L}{C} = \frac{50 \times 1 0 ^{- 6}}{200 \times 1 0 ^{- 12}} = 2.5 \times 1 0^{5}$ , so $\dots = 500$ . Then $Q = 500/5 = 100.$ Why this step? Keep the powers of ten explicit so the square root is clean.

Step 3 — resonant frequency (bonus). $ω_{0} = 1/ L C = 1/ 50 \times 1 0^{- 6} \cdot 200 \times 1 0^{- 12} = 1/ 1 0^{- 14} = 1 0^{7} rad/s$ , so $f_{0} = ω_{0} /2 π \approx 1.59 MHz .$ Why this step? $ω_{0} = 1/ L C$ is the electrical $k / m$ — needed if you later want the bandwidth $Δ f = f_{0} / Q$ .

Verify: Bandwidth would be $Δ f = f_{0} / Q = 1.59 MHz /100 \approx 15.9 kHz$ — a plausibly narrow AM-band filter ✓. Units of $L / C$ : $Ω \cdot s \div s /Ω = Ω$ , cancels the $R$ in ohms, leaving a pure number ✓.

Cell F — scaling / "what if I change one part"

Worked example F · Triple the stiffness

A lightly damped mass–spring has $Q_{1} = 30$ . Keeping $m$ and $b$ fixed, you replace the spring with one three times stiffer ( $k \to 3 k$ ). Find the new $Q_{2}$ .

Forecast: Stiffer spring swings faster. Does that make it ringier (higher $Q$ ) or not?

Step 1 — isolate the $k$ -dependence. From $Q = mk / b$ , only $k$ changes; $m, b$ are constant. Why this step? Turning a formula into a proportionality $Q \propto k$ lets us skip re-computing $m$ and $b$ .

Step 2 — apply the ratio. $\frac{Q _{2}}{Q _{1}} = \frac{3 k}{k} = 3, Q_{2} = 303 \approx 51.96.$ Why this step? Dividing the two $Q$ s cancels everything unchanged, leaving only the stiffness ratio under the root.

Verify: $3 \approx 1.732$ , and $30 \times 1.732 = 51.96$ ✓. Reasonableness: stiffer ⇒ faster swing ⇒ more swings squeezed in before friction (unchanged) wins ⇒ higher $Q$ , matching the forecast that should have been "higher".

Cell G — the degenerate case (no oscillation)

Worked example G · When the formula must be refused

$m = 1.0 kg$ , $k = 4.0 N/m$ , $b = 5.0 kg/s$ . A student computes $Q$ and reports "the mass rings for $\approx Q / π$ swings." What is wrong?

Forecast: Big $b$ next to small $k$ — do you expect any bouncing at all?

Step 1 — compute the number anyway. $Q = mk / b = 1 \cdot 4 /5 = 2/5 = 0.4.$ Why this step? We must see the number cross the danger threshold before we can flag it.

Step 2 — test the regime. Underdamped (oscillating) needs $γ < 2 ω_{0}$ , i.e. $Q > \frac{1}{2}$ . Here $Q = 0.4 < 0.5$ . Why this step? The whole ring-down interpretation assumes underdamping. The decay roots are $r = - \frac{γ}{2} \pm \frac{γ ^{2}}{4} - ω_{0}^{2}$ ; the quantity under that root, $\frac{γ ^{2}}{4} - ω_{0}^{2}$ , is the discriminant. When it is positive the two roots are real (no cosine ⇒ no swings); when negative they are complex (genuine oscillation).

Step 3 — evaluate the discriminant. $ω_{0} = 2$ , $γ = 5$ ; discriminant $\frac{γ ^{2}}{4} - ω_{0}^{2} = \frac{25}{4} - 4 = 6.25 - 4 = 2.25 > 0$ . Positive ⇒ real roots ⇒ overdamped. Why this step? This is the direct check from Damped Harmonic Motion: a positive discriminant means no imaginary part, hence no cosine, hence no swings.

Verify: So $Q = 0.4$ is not wrong as a number, but "rings for $Q / π \approx 0.13$ swings" is meaningless — there are zero swings. The correct statement: overdamped, the mass returns to rest sluggishly without crossing zero ✓. This is the boundary cell every exam loves.

Cell H — a real-world word problem

Worked example H · How long does a wine glass sing?

Tap a crystal wine glass; it rings at $f_{0} = 1000 Hz$ and you can hear it for about $t_{1/ e} = 2.0 s$ before it fades to $1/ e$ of its start (amplitude). Estimate $Q$ .

Forecast: Two whole seconds of ringing at a thousand hertz is a lot of cycles. Rough $Q$ ?

Step 1 — count the oscillations in that time. $N_{amp} = f_{0} t_{1/ e} = 1000 \times 2.0 = 2000$ full oscillations. Why this step? Frequency times time = number of cycles; this converts the audible duration into a swing count, which is the ring-down face's input.

Step 2 — convert count to $Q$ . $Q = π N_{amp} = π \times 2000 \approx 6283.$ Why this step? Same $Q = π N_{amp}$ ring-down relation as Cell D — we just fed it a count derived from a clock instead of read off a screen.

Verify: Here is an independent route to the same number, built from the two definitions in the vocabulary block.

The amplitude $1/ e$ -time is the time for $e^{- γ t /2}$ to reach $e^{- 1}$ , i.e. $γ τ /2 = 1$ , so $τ = 2/ γ$ . (The factor of 2 comes from the amplitude envelope carrying $γ /2$ , not $γ$ , in its exponent — energy would use $γ$ and give half this time.)
Combining $Q = ω_{0} / γ$ with $τ = 2/ γ$ : solve the second for $γ = 2/ τ$ , then substitute into the first — $Q = \frac{ω _{0}}{γ} = \frac{ω _{0}}{2/ τ} = \frac{ω _{0} τ}{2}$ . Why this chaining works: both formulas contain the same $γ$ , so eliminating $γ$ links $Q$ directly to the measured ring-time $τ$ .
Plug in $ω_{0} = 2 π f_{0} = 2 π (1000)$ and $τ = 2.0 s$ : $Q = \frac{ω _{0} τ}{2} = \frac{( 2 π \cdot 1000 ) ( 2.0 )}{2} = 2 π \cdot 1000 \approx 6283$ ✓ — same as Step 2. A crystal glass with $Q \sim 6000$ is realistic — good crystal has very low internal friction.

Cell I — the exam twist (chain two faces)

Worked example I · From bandwidth to how long it rings

A microwave cavity resonates at $f_{0} = 10 GHz$ with measured bandwidth $Δ f = 1.0 MHz$ . After you cut the drive, how many oscillations pass before the stored energy falls to $1/ e$ ?

Forecast: A GHz cavity with only MHz width is extremely sharp. Expect thousands, or tens of thousands, of cycles?

Step 1 — get $Q$ from the bandwidth face. $Q = \frac{f _{0}}{Δ f} = \frac{10 \times 1 0 ^{9}}{1.0 \times 1 0 ^{6}} = 1 0^{4} = 10000.$ Why this step? The measurement is a resonance width, so face C (the bottom-left door) is the natural way in.

Step 2 — convert $Q$ to an energy ring-down count. Energy reaches $1/ e$ after $N_{energy} = Q /2 π$ oscillations (the count defined in the vocabulary block). $N_{energy} = \frac{10000}{2 π} \approx 1592.$ Why this step? This is the twist — the question asks about energy, whose decay is twice as fast as amplitude, so we use $Q /2 π$ , not $Q / π$ . Mixing these up is the trap. We walked in through the bandwidth door and out through the ring-down door — two faces chained.

Verify: Cross-check via amplitude count $N_{amp} = Q / π = 3183$ ; energy should reach $1/ e$ in half as many cycles, and $3183/2 = 1592$ ✓ — the factor-of-two between amplitude and energy decay holds, confirming we chained the two faces correctly. See Resonance & Forced Oscillations and Energy in Oscillations.

Pulling the nine cells together

Intuition The whole matrix in one breath

Every one of the nine cells reached (or refused) the same number $Q$ . Reading them as a story: A hands you a loss fraction, B and E hand you the physical parts (mechanical or electrical — same formula, different hardware), C hands you a resonance width, D hands you a swing count. Those are the four doors of the terrain figure. Then the harder cells re-use those doors: F changes one part and rides a proportionality, G computes the number but refuses the story because the system is overdamped ( $Q < \frac{1}{2}$ , positive discriminant, zero swings), H dresses the ring-down door as a real singing glass, and I walks in one door (bandwidth) and out another (ring-down), chaining two faces — while never forgetting energy dies twice as fast as amplitude. Master the one question "what was I handed?" and every $Q$ problem collapses to picking the right door.

Recall Did every cell get a home?

Match each example to its matrix row. A energy-per-cycle ::: gong, $Q = 200 π \approx 628$ B mechanical parts ::: suspension, $Q = 5$ C resonance width ::: tuning fork, $Q = 550$ D ring-down count (figure) ::: scope trace, $Q \approx 62.8$ E electrical parts ::: RLC filter, $Q = 100$ F scaling one part ::: $3 \times$ stiffer, $Q = 303 \approx 52$ G degenerate / no oscillation ::: overdamped, $Q = 0.4 < \frac{1}{2}$ H real-world word problem ::: wine glass, $Q \approx 6283$ I exam twist (chain two faces) ::: cavity, $N_{energy} \approx 1592$

Connections

Q factor — quality of oscillator — the parent that derived every formula used here.
Damped Harmonic Motion — the $Q > \frac{1}{2}$ underdamped test used in Cell G.
Simple Harmonic Motion — $ω_{0} = k / m$ baseline.
Resonance & Forced Oscillations — bandwidth face (Cells C, I).
RLC Circuits — electrical analogue (Cell E).
Energy in Oscillations — why energy decays twice as fast as amplitude (Cell I).