Forced oscillations — driving frequency: Exercises

Before we start, one shared toolbox — everything on this page uses only these, and nothing else:

Intuition When does the resonance peak even exist?

The formula $ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ only gives a real (physical) frequency when what is under the square root is positive: $ω_{0}^{2} - 2 γ^{2} > 0$ , i.e. $== γ < ω_{0} / 2 ==$ .

Light damping ( $γ < ω_{0} / 2$ ): the amplitude curve genuinely humps up, and its top sits at $ω_{r es}$ , slightly below $ω_{0}$ .
Heavy damping ( $γ \geq ω_{0} / 2$ ): the square root would be of a negative number — no real $ω_{r es}$ . Physically this means the amplitude curve has no hump; it is largest at $ω_{d} = 0$ (the static value $F_{0} / k$ ) and just slides downhill as $ω_{d}$ grows. There is no resonance to speak of — friction is so strong it smothers the peak before it can form. Every worked system on this page uses light damping, so the peak always exists; but you must check this condition before quoting $ω_{r es}$ .

Level 1 — Recognition

L1.1 — Which frequency wins?

A driven-damped mass has natural frequency $ω_{0} = 8$ rad/s and is driven by a force oscillating at $ω_{d} = 5$ rad/s. After a long time (transient gone), at what frequency does the mass oscillate?

Recall Solution

WHAT: Identify the steady-state frequency. WHY: A periodic force of frequency $ω_{d}$ can only sustain a response at the same frequency — the system is "enslaved" to the driver once the transient $e^{- γ t}$ dies. Answer: $ω_{d} = 5$ rad/s. The natural $ω_{0} = 8$ rad/s only governs how large the response is, not the frequency. WHAT IT LOOKS LIKE: picture the mass tracing a cosine wave whose spacing between crests is set by the pusher's rhythm, not the spring's — a swing forced to move at whatever pace the person pushes.

L1.2 — Read off $ω_{0}$ and $γ$

A mass $m = 2$ kg on a spring $k = 200$ N/m has damping constant $b = 8$ kg/s. Find $ω_{0}$ and $γ$ .

Recall Solution

WHAT: Convert the raw constants into the standardised pair. WHY: Every formula above is written in $ω_{0}, γ$ , not $k, b$ — so we translate first.

$ω_{0} = k / m = 200/2 = 100 = 10$ rad/s.
$2 γ = b / m = 8/2 = 4$ , so $γ = 2$ s⁻¹. WHAT IT LOOKS LIKE: $ω_{0}$ is the crest-to-crest rhythm of the undriven spring; $γ$ is the rate at which a free wobble's height shrinks — a fatter $γ$ means the free wiggle fades to nothing faster.

L1.3 — Static limit

For the system in L1.2 with $F_{0} = 20$ N, what amplitude does an extremely slow drive ( $ω_{d} \to 0$ ) produce?

Recall Solution

WHAT: Take $ω_{d} \to 0$ in the amplitude formula. WHY: At $ω_{d} = 0$ the drive is essentially a constant push; the mass just sits displaced by Hooke's law $F_{0} = k x$ . $A \to \frac{F _{0} / m}{( ω _{0}^{2} ) ^{2}} = \frac{F _{0} / m}{ω _{0}^{2}} = \frac{F _{0}}{m ω _{0}^{2}} = \frac{F _{0}}{k} = \frac{20}{200} = 0.1 m .$ WHAT IT LOOKS LIKE: no wobble at all — the mass just holds a fixed offset, like a hand slowly leaning on a spring and pinning it a little way from rest.

Level 2 — Application

L2.1 — Amplitude off resonance

System: $m = 0.5$ kg, $k = 200$ N/m, $b = 2$ kg/s, $F_{0} = 10$ N, driven at $ω_{d} = 10$ rad/s. Find $A$ .

Recall Solution

WHAT: Plug numbers into $A (ω_{d})$ . WHY: This is direct substitution, but note the two pieces of the denominator: the mismatch $(ω_{0}^{2} - ω_{d}^{2})$ and the damping cost $2 γ ω_{d}$ .

$ω_{0} = 200/0.5 = 20$ rad/s, $2 γ = b / m = 4$ .
Mismatch: $ω_{0}^{2} - ω_{d}^{2} = 400 - 100 = 300$ .
Damping term: $2 γ ω_{d} = 4 \times 10 = 40$ .
Denominator: $30 0^{2} + 4 0^{2} = 90000 + 1600 = 91600 \approx 302.66$ .
Numerator: $F_{0} / m = 10/0.5 = 20$ . $A = \frac{20}{302.66} \approx 0.0661 m .$ WHAT IT LOOKS LIKE: a small wobble — because $ω_{d} = 10$ is only half of $ω_{0} = 20$ , the big mismatch (300) dominates the denominator, so the driven swing stays modest, only a bit above the static offset.

L2.2 — Phase lag off resonance

Same system, same $ω_{d} = 10$ rad/s. Find the phase lag $ϕ$ .

Recall Solution

WHAT: Use $tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2})$ . WHY: $ϕ$ is the angle whose tangent is (damping term)/(mismatch term). "arctan" answers: which angle has this tan?

$tan ϕ = 40/300 = 0.1333$ .
Mismatch is positive (we are below resonance, $ω_{d} < ω_{0}$ ), so $ϕ$ lies in the first quadrant — a small positive lag. $ϕ = arctan (0.1333) \approx 7.5 9^{\circ} \approx 0.1325 rad .$ WHAT IT LOOKS LIKE: the mass's peak arrives just barely after the force's peak — almost in step.

L2.3 — Amplitude exactly at resonance-of-drive $ω_{d} = ω_{0}$

Same system, driven at $ω_{d} = ω_{0} = 20$ rad/s. Find $A$ .

Recall Solution

WHAT: Set the mismatch to zero. WHY: When $ω_{d} = ω_{0}$ , the term $(ω_{0}^{2} - ω_{d}^{2}) = 0$ vanishes; only damping survives in the denominator, so $A$ is nearly at its largest.

Denominator $= 2 γ ω_{d} = 4 \times 20 = 80$ .
$A = \frac{20}{80} = 0.25$ m. WHAT IT LOOKS LIKE: a big wobble — nearly four times the off-resonance swing of L2.1 (0.066 m). With the mismatch gone, only friction holds the amplitude back, so the swing balloons up to its near-maximum height.

Level 3 — Analysis

L3.1 — Where is the amplitude peak?

System: $ω_{0} = 20$ rad/s, $γ = 2$ s⁻¹. Find the drive frequency $ω_{r es}$ that maximises amplitude, and compare to $ω_{0}$ .

Recall Solution

WHAT: First check the peak exists, then use $ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ . WHY: Amplitude is largest when the denominator is smallest. Minimising $(ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}$ (a calculus step done in the parent note) gives this. The $- 2 γ^{2}$ shows damping pulls the peak below $ω_{0}$ — but only if the square root stays real.

Existence check: is $γ < ω_{0} / 2$ ? Here $ω_{0} / 2 = 20/1.414 \approx 14.14$ , and $γ = 2 ≪ 14.14$ . Yes — light damping, a real peak exists.
$ω_{r es} = 400 - 2 (4) = 400 - 8 = 392 \approx 19.799$ rad/s.
It sits just below $ω_{0} = 20$ — because damping resists motion, the "best" drive is a touch slower than the bare natural frequency. WHAT IT LOOKS LIKE: Picture (and see figure s01) a smooth curve of amplitude $A$ plotted upward against drive frequency $ω_{d}$ rightward. Starting from the static value at the left, it rises to a single rounded hump and then falls away toward zero. The top of that hump is $ω_{r es} \approx 19.8$ ; a vertical reference line drawn at $ω_{0} = 20$ sits just to the right of the hump's top by a hair — that tiny gap is the $- 2 γ^{2}$ shift made visible.

Figure — Forced oscillations — driving frequency

Figure s01 — Amplitude $A$ (vertical, metres) versus drive frequency $ω_{d}$ (horizontal, rad/s) for $ω_{0} = 20$ , $γ = 2$ . The curve humps to a single maximum (marked with a dot) at $ω_{r es} \approx 19.8$ , just left of the dashed vertical reference line at $ω_{0} = 20$ . Left end approaches the static value $F_{0} / k$ ; right tail decays toward zero.

L3.2 — Peak amplitude value

For the L3.1 system with $F_{0} = 10$ N, $m = 0.5$ kg, find the amplitude at $ω_{r es}$ .

Recall Solution

WHAT: Substitute $ω_{d} = ω_{r es}$ back into $A$ . WHY: Now both terms are nonzero, so we compute the full denominator.

$ω_{d}^{2} = 392$ , so mismatch $= 400 - 392 = 8$ ; squared $= 64$ .
Damping term $2 γ ω_{d} = 4 \times 19.799 = 79.196$ ; squared $= 6272$ .
Denominator $= 64 + 6272 = 6336 \approx 79.599$ .
$A = \frac{20}{79.599} \approx 0.2513$ m. WHAT IT LOOKS LIKE: this is the height of the hump's top on figure s01 — barely taller than the $A = 0.25$ m value found at $ω_{d} = ω_{0}$ in L2.3. For light damping the peak is only a whisker higher, confirming $ω_{r es} \approx ω_{0}$ .

L3.3 — Phase quadrant above resonance

System $ω_{0} = 20$ , $γ = 2$ , driven at $ω_{d} = 30$ rad/s. Find $ϕ$ and say which quadrant it lives in.

Recall Solution

WHAT: Compute $tan ϕ$ , then place the angle correctly. WHY: Above resonance the mismatch $ω_{0}^{2} - ω_{d}^{2}$ goes negative, so a raw arctan gives a negative angle — but physically the mass lags the force by more than 90°. We must add $180°$ .

Mismatch: $400 - 900 = - 500$ (negative).
Damping: $2 γ ω_{d} = 4 \times 30 = 120$ (always positive).
$tan ϕ = 120/ (- 500) = - 0.24$ . Bare $arctan (- 0.24) = - 13. 5^{\circ}$ .
The correct physical lag is $ϕ = 18 0^{\circ} - 13. 5^{\circ} = 166. 5^{\circ}$ (second quadrant), because the lag always grows from $0$ toward $18 0^{\circ}$ as $ω_{d}$ increases through resonance. $ϕ \approx 166. 5^{\circ} \approx 2.906 rad .$ WHAT IT LOOKS LIKE: In figure s02 the phase-lag curve climbs from $0^{\circ}$ at the left, crosses $9 0^{\circ}$ right at resonance, and keeps rising toward $18 0^{\circ}$ on the right. At $ω_{d} = 30$ we read a value near $16 6^{\circ}$ — the mass now moves almost opposite the force, its inertia unable to keep up, so it flips.

Figure s02 — Phase lag $ϕ$ (vertical, degrees, running $0$ to $180$ ) versus drive frequency $ω_{d}$ (horizontal, rad/s). The curve rises S-shaped from near $0^{\circ}$ (in phase, below resonance), passes through exactly $9 0^{\circ}$ at $ω_{d} = ω_{0} = 20$ (dashed vertical reference), and climbs toward $18 0^{\circ}$ (anti-phase) at high $ω_{d}$ .

Level 4 — Synthesis

L4.1 — Design for a target amplitude

You have $m = 1$ kg, $k = 100$ N/m, $F_{0} = 5$ N. You want the amplitude at resonance ( $ω_{d} = ω_{0}$ ) to equal exactly $0.5$ m. What damping constant $b$ achieves this?

Recall Solution

WHAT: Work backwards from $A$ at $ω_{d} = ω_{0}$ to solve for $b$ . WHY: At resonance the mismatch vanishes, so $A = \frac{F _{0} / m}{2 γ ω _{0}}$ . One unknown ( $γ$ , hence $b$ ), one equation.

$ω_{0} = 100/1 = 10$ rad/s.
$A = \frac{F _{0} / m}{2 γ ω _{0}} \Rightarrow 0.5 = \frac{5/1}{2 γ \cdot 10} = \frac{5}{20 γ}$ .
$20 γ \times 0.5 = 5 \Rightarrow 10 γ = 5 \Rightarrow γ = 0.5$ s⁻¹.
$b = 2 γ m = 2 (0.5) (1) = 1$ kg/s. Check: with $b = 1$ , $2 γ ω_{0} = 2 (0.5) (10) = 10$ , so $A = 5/10 = 0.5$ m. ✓ (Also $γ = 0.5 < ω_{0} / 2 \approx 7.07$ , so a real peak exists — the design is physical.) WHAT IT LOOKS LIKE: we are dialling the friction knob until the resonant hump lands at exactly half a metre — less friction would make the hump taller, more friction would flatten it; $b = 1$ kg/s is the setting that hits the target height.

L4.2 — Two contributions to the denominator, balanced

For $ω_{0} = 10$ rad/s, $γ = 3$ s⁻¹, find the drive frequency at which the mismatch term $(ω_{0}^{2} - ω_{d}^{2})$ has the same magnitude as the damping term $2 γ ω_{d}$ . (This is where the amplitude curve's shoulder sits.)

Recall Solution

WHAT: Set $∣ ω_{0}^{2} - ω_{d}^{2} ∣ = 2 γ ω_{d}$ and solve for $ω_{d}$ (take the below-resonance root, mismatch positive). WHY: Where the two "legs" of the Pythagorean denominator are equal, the phase lag is exactly $4 5^{\circ}$ (since $tan ϕ = 1$ ) — a physically meaningful marker.

$ω_{0}^{2} - ω_{d}^{2} = 2 γ ω_{d} \Rightarrow 100 - ω_{d}^{2} = 6 ω_{d}$ .
$ω_{d}^{2} + 6 ω_{d} - 100 = 0$ .
$ω_{d} = \frac{- 6 + 36 + 400}{2} = \frac{- 6 + 436}{2} = \frac{- 6 + 20.881}{2} \approx 7.44$ rad/s. Check phase: $tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}) = 1 \Rightarrow ϕ = 4 5^{\circ}$ . ✓ (We keep the positive root since $ω_{d} > 0$ .) WHAT IT LOOKS LIKE: on a phase curve like s02, this $ω_{d} \approx 7.44$ is the point halfway up the climb — right where the mass's peak sits exactly one-eighth of a cycle behind the force ( $4 5^{\circ}$ ), the balance point between spring-controlled and damping-controlled behaviour.

Level 5 — Mastery

L5.1 — Power delivered at resonance

For $m = 0.5$ kg, $k = 200$ N/m, $b = 2$ kg/s, $F_{0} = 10$ N, driven at $ω_{d} = ω_{0} = 20$ rad/s, find the average power the driver delivers.

Recall Solution

WHAT: Compute time-averaged power from force times velocity, using only the single phase $ϕ$ already defined. WHY: Instantaneous power is $P = F \overset{x}{˙}$ (force times velocity). We need to know how the velocity lines up with the force, because if they point the same way the push does positive work.

Setting up the two waves (using only $ϕ$ ):

Force: $F (t) = F_{0} cos (ω_{d} t)$ — our reference, peaks at $t = 0$ .
Displacement lags the force by $ϕ$ : $x (t) = A cos (ω_{d} t - ϕ)$ .
Velocity is the time-derivative: $\overset{x}{˙} (t) = - A ω_{d} sin (ω_{d} t - ϕ) = A ω_{d} cos (ω_{d} t - ϕ + 9 0^{\circ})$ . Differentiating a cosine advances its peak by $9 0^{\circ}$ , so velocity leads displacement by exactly $9 0^{\circ}$ — this is a fact about $cos$ , needing no new symbol.
Therefore the velocity peak sits at phase angle $- ϕ + 9 0^{\circ}$ relative to the force. The gap between force ( $0^{\circ}$ ) and velocity ( $- ϕ + 9 0^{\circ}$ ) is $(9 0^{\circ} - ϕ)$ , expressed entirely through the one phase $ϕ$ .

Averaging over a cycle: the time-average of $cos (ω_{d} t) \cdot cos (ω_{d} t + θ)$ over one full period is $\frac{1}{2} cos θ$ . Here $θ = (9 0^{\circ} - ϕ)$ is the force-to-velocity gap, so $\overset{ˉ}{P} = \frac{1}{2} F_{0} (A ω_{d}) cos (9 0^{\circ} - ϕ) = \frac{1}{2} F_{0} A ω_{d} sin ϕ,$ using $cos (9 0^{\circ} - ϕ) = sin ϕ$ . Everything is in terms of $ϕ$ — no second phase symbol was needed.

At resonance $ϕ = 9 0^{\circ}$ , so $sin ϕ = 1$ (force and velocity perfectly aligned):

From L2.3, $A = 0.25$ m, so peak velocity $A ω_{d} = 0.25 \times 20 = 5$ m/s.
$\overset{ˉ}{P} = \frac{1}{2} F_{0} A ω_{d} sin 9 0^{\circ} = \frac{1}{2} (10) (5) (1) = 25 W .$ WHAT IT LOOKS LIKE: every push lands exactly while the mass is already moving the same way — like shoving a swing right at the bottom of its arc every single pass. No effort is wasted fighting the motion, so the energy pumped in per second is at its ceiling: maximum power transfer.

L5.2 — Full comparison across all three regimes

For $ω_{0} = 20$ , $γ = 2$ , $F_{0} = 10$ , $m = 0.5$ : tabulate $A$ and $ϕ$ at $ω_{d} = 2$ (slow), $ω_{d} = 20$ (matched), $ω_{d} = 200$ (fast). Confirm the "Slow Spring, Match Max, Fast Faint" mnemonic.

Recall Solution

WHAT: Evaluate $A$ and $ϕ$ at three well-separated frequencies. WHY: To see the full curve — spring-controlled, resonant, inertia-controlled — and verify limiting behaviour.

Slow ( $ω_{d} = 2$ ):

Mismatch $= 400 - 4 = 396$ ; damping $= 4 \times 2 = 8$ .
$A = 20/ 39 6^{2} + 8^{2} = 20/ 156816 + 64 = 20/396.08 \approx 0.0505$ m. ( $\approx F_{0} / k = 0.05$ m ✓, spring-dominated.)
$tan ϕ = 8/396 = 0.0202 \Rightarrow ϕ \approx 1.1 6^{\circ} \approx 0$ → in phase (mass follows force).

Matched ( $ω_{d} = 20$ ):

$A = 0.25$ m (from L2.3) — the peak. $ϕ = 90°$ → force in phase with velocity, max power.

Fast ( $ω_{d} = 200$ ):

Mismatch $= 400 - 40000 = - 39600$ ; damping $= 4 \times 200 = 800$ .
$A = 20/ 3960 0^{2} + 80 0^{2} = 20/ 1568160000 + 640000 = 20/39608.1 \approx 0.000505$ m.
Compare $F_{0} / (m ω_{d}^{2}) = 10/ (0.5 \times 40000) = 0.0005$ m ✓ — inertia-dominated, tiny.
$tan ϕ = 800/ (- 39600) = - 0.0202 \Rightarrow ϕ = 180° - 1.16° \approx 178.8°$ → nearly anti-phase (flipped).

Verdict: Slow → spring, in-phase, $A \approx F_{0} / k$ . Matched → maximum $A$ , $ϕ = 90°$ . Fast → faint & flipped, $A \to 0$ , $ϕ \to 180°$ . The mnemonic holds on every count. ✓ WHAT IT LOOKS LIKE: these three points trace out the whole s01/s02 story — the tiny left-end offset, the tall central hump at $ϕ = 90°$ , and the vanishing right tail where the curve flattens to the axis and the phase pins at $18 0^{\circ}$ .

Active recall

Recall One-line answers

Steady-state frequency of a driven oscillator ::: the driving frequency $ω_{d}$ , never $ω_{0}$ . Denominator of $A$ combines its two terms how ::: Pythagorean — $mismatch^{2} + damping^{2}$ , not a plain sum. Sign fix for $ϕ$ above resonance ::: add $18 0^{\circ}$ to the calculator's arctan (mismatch is negative there). Condition for a real resonance peak ::: $γ < ω_{0} / 2$ ; heavier damping gives no hump (largest $A$ at $ω_{d} = 0$ ). Why $A$ is finite at resonance ::: damping term $2 γ ω_{d} \neq = 0$ keeps the denominator away from zero. Average power formula in terms of $ϕ$ ::: $\overset{ˉ}{P} = \frac{1}{2} F_{0} A ω_{d} sin ϕ$ ; at resonance $sin ϕ = 1$ so $\overset{ˉ}{P} = 25$ W for L5.1.

Related: Resonance and Quality Factor · Damped Oscillations · Energy in Oscillations · Simple Harmonic Motion · Complex Exponential Method