Forced oscillations — driving frequency: Worked examples

Before we start, here are the only two tools we need, copied so nothing is assumed:

Two symbols still need a plain-words home before we use them:

One more quantity shows up in Example 2, so let us pin it down now, in plain words:

The scenario matrix

Every question this topic can ask falls into one of these cells. Each worked example below is tagged with the cell it fills.

Cell	What varies	What's tricky	Example
A. Slow drive ( $ω_{d} ≪ ω_{0}$ )	limiting value	must reduce to Hooke's law $A \to F_{0} / k$	Ex 1
B. Exact resonance denominator ( $ω_{d} = ω_{0}$ )	mismatch term = 0	only damping survives; $ϕ = 9 0^{\circ}$	Ex 2
C. True amplitude peak ( $ω_{d} = ω_{r es}$ )	peak is below $ω_{0}$	must use $ω_{0}^{2} - 2 γ^{2}$	Ex 3
D. Phase below resonance ( $ω_{d} < ω_{0}$ )	$ω_{0}^{2} - ω_{d}^{2} > 0$	$tan ϕ > 0$ , $ϕ$ acute	Ex 4
E. Phase above resonance ( $ω_{d} > ω_{0}$ )	$ω_{0}^{2} - ω_{d}^{2} < 0$	$tan ϕ < 0$ — quadrant fix needed	Ex 5
F. Fast drive ( $ω_{d} ≫ ω_{0}$ )	limiting value	inertia wins, $A \to F_{0} / (m ω_{d}^{2})$ , $ϕ \to 18 0^{\circ}$	Ex 6
G. Zero damping ( $γ = 0$ )	degenerate input	$A \to \infty$ at $ω_{0}$ ; $ϕ$ jumps $0 \to 18 0^{\circ}$	Ex 7
H. Real-world word problem	modelling	translate words → symbols	Ex 8
I. Exam twist (find $ω_{d}$ from a given $A$ )	inverse problem	solve for the unknown inside the root	Ex 9

We use one base system for the number-crunching so results are comparable: $m = 0.5 kg$ , $k = 200 N/m$ , $b = 2 kg/s$ , $F_{0} = 10 N$ . Then $ω_{0} = 200/0.5 = 20 rad/s$ and $2 γ = b / m = 4 \Rightarrow γ = 2 s^{- 1}$ .

Figure — Forced oscillations — driving frequency

Figure s01 — Amplitude resonance curve (alt-text). The horizontal axis is driving frequency $ω_{d}$ from $0$ to $45$ rad/s; the vertical axis is steady-state amplitude $A$ from $0$ to about $0.28$ m. A blue curve starts near the static value $0.05$ m (dashed gray line, labelled $A \to F_{0} / k$ ), rises to a sharp red peak of $\approx 0.251$ m at $ω_{r es} \approx 19.8$ rad/s (a touch left of the dotted orange line marking $ω_{0} = 20$ ), then falls toward zero for fast drives. Coloured dots mark where each cell lives: green "A slow" low on the left rise, orange "B $ω_{0}$ " just right of the peak, red "C peak" at the very top, gray "F fast" far down the right tail.

The figure above is the amplitude-vs-frequency curve for this base system, with the cells A–G marked where they live on the curve. Keep glancing back at it: each example is just "read off one point of this curve" or "read off one point of the phase graph".

Cell A — Slow drive (static limit)

Worked example Ex 1 · Cell A ·

$ω_{d} = 2$ rad/s (very slow) Drive the base system slowly at $ω_{d} = 2$ rad/s. Find $A$ , and confirm it is heading toward the Hooke's-law value $F_{0} / k$ .

Forecast: guess before computing — will $A$ be near $F_{0} / k = 10/200 = 0.05$ m, or much larger?

Mismatch term $ω_{0}^{2} - ω_{d}^{2} = 400 - 4 = 396$ . Why this step? This is the "how far from resonance" number; large here because $2 ≪ 20$ .
Damping term $2 γ ω_{d} = 4 \times 2 = 8$ . Why this step? Second piece of the denominator; small because $ω_{d}$ is small.
Denominator $39 6^{2} + 8^{2} = 156816 + 64 = 156880 \approx 396.08$ . Why this step? Plugging both pieces into $(\dots)^{2} + (\dots)^{2}$ .
Amplitude $A = \frac{F _{0} / m}{396.08} = \frac{20}{396.08} \approx 0.0505$ m. Why this step? $F_{0} / m = 10/0.5 = 20$ ; divide by the denominator.

Verify: the pure static value is $F_{0} / k = 0.05$ m. Our answer $0.0505$ m sits a hair above it — exactly what "slow drive $\approx$ static" predicts. Units: $\frac{N}{kg} \div \frac{1}{s ^{2}} = \frac{N \cdot s ^{2}}{kg} = m$ . ✓

Cell B — Exact resonance denominator ( $ω_{d} = ω_{0}$ )

Worked example Ex 2 · Cell B ·

$ω_{d} = 20$ rad/s Drive the base system at exactly $ω_{d} = ω_{0} = 20$ rad/s. Find $A$ and the phase lag $ϕ$ .

Forecast: the mismatch term is now zero. Do you expect this to be the biggest $A$ possible, or just close? And what should $ϕ$ be exactly when the frequencies match?

Mismatch term $ω_{0}^{2} - ω_{d}^{2} = 400 - 400 = 0$ . Why this step? When $ω_{d} = ω_{0}$ the frequencies match, so this contribution vanishes.
Denominator $0^{2} + (2 γ ω_{d})^{2} = 2 γ ω_{d} = 4 \times 20 = 80$ . Why this step? With the mismatch gone, only the damping cost limits the response.
Amplitude $A = \frac{20}{80} = 0.25$ m. Why this step? $F_{0} / m = 20$ , divide by $80$ .
Phase lag $tan ϕ = \frac{2 γ ω _{d}}{ω _{0}^{2} - ω _{d}^{2}} = \frac{80}{0} \to + \infty$ , so $ϕ = 9 0^{\circ} = \frac{π}{2}$ rad. Why this step? When the denominator of $tan ϕ$ hits zero from above, $tan ϕ$ blows up, and the angle whose tangent is $+ \infty$ is exactly $9 0^{\circ}$ . This is the physically special case: the mass runs a quarter-cycle behind the force, which puts the force in phase with the velocity → maximum power input. Recall the units rule: we compute with SI numbers, then quote the angle in degrees.

Verify: $0.25$ m is much larger than the static $0.05$ m — a factor of $5$ boost from resonance. That factor is exactly the quality factor $Q = ω_{0} / (2 γ) = 20/4 = 5$ (amplitude at $ω_{0}$ is $Q$ times the static value $F_{0} / k$ ). And $ϕ = 9 0^{\circ}$ sits exactly between Ex 4's $7. 6^{\circ}$ (below) and Ex 5's $166. 5^{\circ}$ (above) — the phase graph crosses $9 0^{\circ}$ right at $ω_{0}$ . ✓ (This point is not quite the amplitude peak — see Ex 3.)

Cell C — The true amplitude peak ( $ω_{d} = ω_{r es}$ )

Worked example Ex 3 · Cell C · find

$ω_{r es}$ and the peak $A$ For the base system, find the frequency where $A$ is genuinely largest, and the amplitude there. Show it beats Ex 2.

Forecast: will the peak sit above, at, or below $ω_{0} = 20$ ?

Peak location $ω_{r es} = ω_{0}^{2} - 2 γ^{2} = 400 - 2 (4) = 392 \approx 19.799$ rad/s. Why this step? The parent note minimised the denominator by differentiating; the result is this shifted-down peak. Damping drags the peak below $ω_{0}$ .
Mismatch there $ω_{0}^{2} - ω_{r es}^{2} = 400 - 392 = 8 = 2 γ^{2}$ . Why this step? Sanity: by construction the mismatch at the peak equals $2 γ^{2}$ .
Denominator $8^{2} + (2 γ ω_{r es})^{2} = 64 + (4 \times 19.799)^{2} = 64 + 6272 = 6336 \approx 79.60$ .
Amplitude $A_{p e ak} = \frac{20}{79.60} \approx 0.2513$ m.

Verify: $0.2513 > 0.25$ from Ex 2 by a whisker — the true peak really is a touch higher than the value at $ω_{0}$ , and sits at $19.80 < 20$ rad/s. Both facts match "peak below $ω_{0}$ , only equal when $γ \to 0$ ". ✓

Cell D — Phase lag below resonance ( $ω_{d} < ω_{0}$ )

Worked example Ex 4 · Cell D ·

$ω_{d} = 10$ rad/s, find $ϕ$ Base system driven at $ω_{d} = 10$ rad/s (below $ω_{0}$ ). Find the phase lag $ϕ$ .

Forecast: below resonance the mass roughly follows the force. Is $ϕ$ near $0$ , near $9 0^{\circ}$ , or near $18 0^{\circ}$ ?

Mismatch $ω_{0}^{2} - ω_{d}^{2} = 400 - 100 = 300$ (positive). Why this step? Positive mismatch means the denominator of $tan ϕ$ is positive — the tell-tale of "below resonance".
Numerator $2 γ ω_{d} = 4 \times 10 = 40$ .
$tan ϕ = 40/300 = 0.1 \overline{3}$ , so $ϕ = arctan (0.1333) \approx 7.5 9^{\circ}$ . Why this step? $arctan$ answers "which angle has this tan?" Because both numerator and denominator are positive, $ϕ$ lands in the first quadrant ( $0$ – $9 0^{\circ}$ ) — no fix needed. (Compute the ratio from SI numbers; convert the answer to degrees at the end.)

Verify: $ϕ \approx 7. 6^{\circ}$ is tiny → the mass moves almost in phase with the force. Consistent with "slow-ish drive follows the force". ✓

Cell E — Phase lag above resonance (the quadrant trap)

Worked example Ex 5 · Cell E ·

$ω_{d} = 30$ rad/s, find $ϕ$ Base system driven at $ω_{d} = 30$ rad/s (above $ω_{0}$ ). Find the phase lag $ϕ$ — and watch the sign.

Forecast: above resonance the mass fights the force. Should $ϕ$ be less than $9 0^{\circ}$ or more than $9 0^{\circ}$ ?

Mismatch $ω_{0}^{2} - ω_{d}^{2} = 400 - 900 = - 500$ (negative). Why this step? Above resonance the mismatch flips sign — this is the crucial difference from Ex 4.
Numerator $2 γ ω_{d} = 4 \times 30 = 120$ (always positive — $γ, ω_{d} > 0$ ).
$tan ϕ = 120/ (- 500) = - 0.24$ . A naive $arctan (- 0.24) = - 13. 5^{\circ}$ . Why this step? $arctan$ only ever returns answers between $- 9 0^{\circ}$ and $+ 9 0^{\circ}$ , because $tan$ repeats every $18 0^{\circ}$ . A negative answer here is a red flag, not the physics: a phase lag must lie between $0$ and $18 0^{\circ}$ .
Quadrant fix: the physical lag lives in $(9 0^{\circ}, 18 0^{\circ})$ whenever the mismatch is negative, so add $18 0^{\circ}$ : $ϕ = - 13. 5^{\circ} + 18 0^{\circ} = 166. 5^{\circ}$ . Why this step? We know $ϕ$ must increase past $9 0^{\circ}$ as we cross resonance (it slides $0 \to 9 0^{\circ} \to 18 0^{\circ}$ ). The numerator being positive tells us $sin ϕ > 0$ ; the denominator being negative tells us $cos ϕ < 0$ ; both together pin $ϕ$ into the second quadrant ( $9 0^{\circ}$ – $18 0^{\circ}$ ).

Verify: $166. 5^{\circ}$ is close to $18 0^{\circ}$ — expected, because $ω_{d} = 30$ is well above $ω_{0} = 20$ , heading toward the "moves opposite the force" limit. ✓ Compare Ex 4's $7. 6^{\circ}$ (below) with this $166. 5^{\circ}$ (above): crossing resonance swings the lag from near- $0$ to near- $18 0^{\circ}$ , passing $9 0^{\circ}$ exactly at $ω_{0}$ . See the figure below.

Figure s02 — Phase lag vs frequency (alt-text). The horizontal axis is driving frequency $ω_{d}$ from $0$ to $45$ rad/s; the vertical axis is phase lag $ϕ$ in degrees from $0$ to $180$ . A blue S-shaped curve starts near $0^{\circ}$ for slow drives, rises steeply through the dashed gray $9 0^{\circ}$ line exactly at the dotted orange $ω_{0} = 20$ , then levels off toward $18 0^{\circ}$ for fast drives. The lower-left band (below $9 0^{\circ}$ ) is tinted green and labelled "below res: $ϕ$ in (0,90)"; the upper band is tinted red and labelled "above res: $ϕ$ in (90,180)". A green dot marks Ex 4 ( $ω_{d} = 10$ , $ϕ = 7. 6^{\circ}$ ); a red dot marks Ex 5 ( $ω_{d} = 30$ , $ϕ = 166. 5^{\circ}$ , annotated "add 180!").

Cell F — Fast drive (inertia limit)

Worked example Ex 6 · Cell F ·

$ω_{d} = 200$ rad/s (very fast) Base system hammered at $ω_{d} = 200$ rad/s. Find $A$ and confirm it approaches $F_{0} / (m ω_{d}^{2})$ .

Forecast: ten times above $ω_{0}$ — do you expect a large or a tiny amplitude?

Mismatch $ω_{0}^{2} - ω_{d}^{2} = 400 - 40000 = - 39600$ . Why this step? Enormous negative mismatch — the drive is nowhere near the natural frequency.
Damping term $2 γ ω_{d} = 4 \times 200 = 800$ .
Denominator $(- 39600)^{2} + 80 0^{2} = 1, 568, 160, 000 + 640, 000 \approx 39608.1$ .
Amplitude $A = \frac{20}{39608.1} \approx 5.049 \times 1 0^{- 4}$ m.

Verify: the inertia-limit prediction is $A \to \frac{F _{0}}{m ω _{d}^{2}} = \frac{10}{0.5 \times 40000} = 5.0 \times 1 0^{- 4}$ m. Our exact answer $5.049 \times 1 0^{- 4}$ m matches to three figures. ✓ And it is tiny — the mass simply cannot keep up. Its phase lag $ϕ$ (numerator $+$ , denominator $-$ ) is near $18 0^{\circ}$ : motion nearly opposite the force.

Cell G — Zero damping (degenerate input)

Worked example Ex 7 · Cell G · set

$γ = 0$ Remove all damping ( $b = 0 \Rightarrow γ = 0$ ), keep $m, k, F_{0}$ . What happens to $A$ near $ω_{d} = ω_{0}$ ? What is $ϕ$ just below and just above $ω_{0}$ ?

Forecast: with nothing to bleed off energy, what is the amplitude exactly at resonance?

Amplitude formula collapses: with $γ = 0$ , $A = \frac{F _{0} / m}{( ω _{0}^{2} - ω _{d}^{2} ) ^{2} + 0} = \frac{F _{0} / m}{∣ ω _{0}^{2} - ω _{d}^{2} ∣} .$ Why this step? The damping term $(2 γ ω_{d})^{2}$ is now literally $0$ , so the root is just the absolute value of the mismatch.
At $ω_{d} = ω_{0}$ : denominator $= ∣400 - 400∣ = 0$ , so $A \to \infty$ . Why this step? No damping means every push adds energy forever — the textbook "infinite resonance". Physically the spring would break first; mathematically it diverges.
Phase just below ( $ω_{d} \to ω_{0}^{-}$ ): mismatch $\to 0^{+}$ , and with $γ = 0$ the numerator of $tan ϕ$ is $0$ , so $tan ϕ = 0^{+} / 0^{+} \to 0 \Rightarrow ϕ = 0$ .
Phase just above ( $ω_{d} \to ω_{0}^{+}$ ): mismatch $\to 0^{-}$ , so $tan ϕ = 0/ 0^{-}$ from the negative side $\Rightarrow ϕ = 18 0^{\circ}$ . Why this step? With no damping the phase does not glide through $9 0^{\circ}$ — it jumps discontinuously from $0$ to $18 0^{\circ}$ as you cross $ω_{0}$ . Damping is what makes that transition smooth (Ex 4→5).

Verify: check with a slightly off-resonance number, $ω_{d} = 19$ rad/s, $γ = 0$ : $A = \frac{20}{∣400 - 361∣} = \frac{20}{39} \approx 0.513$ m — finite and large, and it blows up as $ω_{d} \to 20$ . ✓ The $0 \to 18 0^{\circ}$ phase jump is the light-damping limit of the smooth curve in figure s02.

Cell H — Real-world word problem

Worked example Ex 8 · Cell H · a washing machine

A washing machine drum of mass $m = 8$ kg is mounted on springs of total stiffness $k = 3200$ N/m with damping $b = 64$ kg/s. During spin, an off-centre wet towel produces a periodic shaking force of amplitude $F_{0} = 40$ N. At a spin rate giving $ω_{d} = 20$ rad/s, how far does the drum shake? Is this spin rate near resonance?

Forecast: guess whether the shaking is millimetres or centimetres before computing.

Translate words to symbols: natural frequency $ω_{0} = k / m = 3200/8 = 400 = 20$ rad/s. Why this step? "Springs + mass" is exactly a driven oscillator; $ω_{0} = k / m$ is its preferred rhythm.
Damping rate $2 γ = b / m = 64/8 = 8 \Rightarrow γ = 4$ s⁻¹. Why this step? Same standardising as the parent note.
Notice $ω_{d} = 20 = ω_{0}$ : the spin is driving right at resonance. Mismatch $ω_{0}^{2} - ω_{d}^{2} = 400 - 400 = 0$ . Why this step? This is the dangerous case — washing machines "walk" across the floor when a spin rate hits resonance. It also means the amplitude formula collapses to just the damping term.
Amplitude $A = \frac{F _{0} / m}{2 γ ω _{d}} = \frac{40/8}{8 \times 20} = \frac{5}{160} = 0.03125$ m $\approx 3.1$ cm. Why this step? With the mismatch gone, the denominator collapses to just $2 γ ω_{d}$ ; plug in $F_{0} / m = 40/8 = 5$ over $8 \times 20 = 160$ .
Phase check: at resonance $ϕ = 9 0^{\circ}$ (Cell B), so the shaking runs a quarter-cycle behind the towel's force — the force is in phase with the drum's velocity, pumping in maximum energy. That is why the shake is so large. Why this step? Confirms the resonance diagnosis and explains the violence physically.

Verify: $3.1$ cm of shaking at resonance is exactly why machines vibrate violently mid-spin, then calm down as they speed past $ω_{0}$ . Units: $\frac{N / kg}{s ^{- 1} \cdot s ^{- 1}} = m$ . ✓ Real machines add extra damping/counterweights to shrink this. See energy for why resonance pumps in so much.

Cell I — Exam twist (inverse problem)

Worked example Ex 9 · Cell I · find the driving frequency from a given amplitude

Base system ( $m = 0.5$ , $ω_{0} = 20$ , $γ = 2$ , $F_{0} = 10$ ). At what drive frequency below resonance does the amplitude equal $A = 0.10$ m?

Forecast: we already know $A = 0.05$ m at $ω_{d} \to 0$ and $A = 0.25$ m at $ω_{0}$ . So the target $0.10$ m must sit somewhere between $0$ and $20$ rad/s. Guess roughly where.

Invert the amplitude formula. From $A = \frac{F _{0} / m}{D}$ with $D = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}$ : $D = \frac{F _{0} / m}{A} = \frac{20}{0.10} = 200 \Rightarrow D = 40000.$ Why this step? We're told $A$ and want $ω_{d}$ , so we peel back the formula: square both sides to free the frequency hiding inside the root.
Write $D$ out. Let $u = ω_{d}^{2}$ . Then $(400 - u)^{2} + (2 γ ω_{d})^{2} = (400 - u)^{2} + 16 u = 40000.$ Why this step? $(2 γ ω_{d})^{2} = (4 ω_{d})^{2} = 16 ω_{d}^{2} = 16 u$ . Using $u = ω_{d}^{2}$ turns this into a quadratic in $u$ .
Expand: $160000 - 800 u + u^{2} + 16 u = 40000 \Rightarrow u^{2} - 784 u + 120000 = 0$ .
Solve the quadratic: $u = \frac{784 \pm 78 4 ^{2} - 4 ( 120000 )}{2} = \frac{784 \pm 614656 - 480000}{2} = \frac{784 \pm 134656}{2} = \frac{784 \pm 366.96}{2} .$ So $u = 575.5$ or $u = 208.5$ .
Pick the physical root. $u = ω_{d}^{2}$ , so $ω_{d} = 575.5 \approx 23.99$ (above resonance) or $ω_{d} = 208.5 \approx 14.44$ (below). Why this step? Two frequencies give the same amplitude — one on each shoulder of the peak. We were asked for the one below resonance: $ω_{d} \approx 14.44$ rad/s.

Verify: plug $ω_{d} = 14.44$ back: mismatch $= 400 - 208.5 = 191.5$ ; damping $= 4 \times 14.44 = 57.76$ ; denominator $= 191. 5^{2} + 57.7 6^{2} = 36672 + 3336 = 40008 \approx 200.0$ ; $A = 20/200 = 0.10$ m. ✓ The other root $23.99$ also gives $0.10$ m — the resonance curve is symmetric-ish in $ω_{d}^{2}$ .

Recall Quick self-test across the matrix

Slow drive limit of $A$ ? ::: $F_{0} / k$ (Hooke's law), Cell A. Amplitude at $ω_{d} = ω_{0}$ ? ::: $F_{0} / (2 mγ ω_{0}) = Q \cdot F_{0} / k$ , Cell B. Phase lag at $ω_{d} = ω_{0}$ ? ::: $ϕ = 9 0^{\circ} = \frac{π}{2}$ rad — force in phase with velocity, max power, Cell B. True peak location? ::: $ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ , slightly below $ω_{0}$ , Cell C. Sign of $tan ϕ$ above resonance, and the fix? ::: Negative; add $18 0^{\circ}$ so $ϕ \in (9 0^{\circ}, 18 0^{\circ})$ , Cell E. What is $A$ as $ω_{d} \to \infty$ ? ::: $F_{0} / (m ω_{d}^{2}) \to 0$ , Cell F. What happens to $A$ at $ω_{0}$ when $γ = 0$ ? ::: Diverges to infinity; phase jumps $0 \to 18 0^{\circ}$ , Cell G. Inverse problem: how many drive frequencies give one below-peak amplitude? ::: Two — one on each shoulder; solve a quadratic in $ω_{d}^{2}$ , Cell I.

See also: Complex Exponential Method for the slick way to get $A$ and $ϕ$ together, Damped Oscillations for the transient that these steady states sit on top of, and Simple Harmonic Motion for the $b = 0, F_{0} = 0$ ancestor of everything here.