Forced oscillations — driving frequency: Question bank

Before the traps, we pin down every formula the questions lean on, so nothing is used before it is earned.

Notation reminder (so nothing here is used before it is earned):

$ω_{d}$ = driving frequency, set by the external push.
$ω_{0} = k / m$ = natural frequency, the system's own preferred rhythm.
$γ = b / (2 m)$ = damping rate; $b$ is the drag constant.
$A (ω_{d})$ = displacement amplitude; $A_{v} (ω_{d}) = ω_{d} A$ = velocity amplitude; $ϕ$ = phase lag of displacement behind the force.
$P = F \overset{x}{˙}$ = instantaneous power the driver feeds the mass: force times velocity, i.e. how fast the push does work on it. It is large when force and velocity point the same way.

Figure 1 — the amplitude curve the "true or false" items describe

Figure — Forced oscillations — driving frequency

Figure 2 — the phase curve behind TF4, SE3, SE5 and every "phase" question

Figure 3 — the right triangle that produces $tan ϕ$

Where $ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ actually comes from

Several questions quote this. Here is the one-paragraph derivation so no one takes it on faith. $A = \frac{F _{0} / m}{D}$ is a positive constant $F_{0} / m$ divided by $D$ . Because that numerator does not contain $ω_{d}$ at all, changing $ω_{d}$ cannot move it — it is just an overall scale factor. So the $ω_{d}$ that makes $A$ largest is precisely the $ω_{d}$ that makes the denominator smallest; the numerator drops out of the extremum condition entirely. We therefore minimise $D (ω_{d}) = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2} .$ Treat $u = ω_{d}^{2}$ so $D = (ω_{0}^{2} - u)^{2} + 4 γ^{2} u$ . Set $\frac{d D}{d u} = 0$ : $\frac{d D}{d u} = - 2 (ω_{0}^{2} - u) + 4 γ^{2} = 0 \Rightarrow u = ω_{0}^{2} - 2 γ^{2} .$ Hence $ω_{r es}^{2} = ω_{0}^{2} - 2 γ^{2}$ , i.e. $ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ — the amplitude peak sits below $ω_{0}$ .

True or false — justify

TF1. In steady state a driven oscillator vibrates at its natural frequency

ω_{0}

False. After the transient dies, a periodic force of frequency

ω_{d}

can only sustain a response at the same frequency, so the motion is at

ω_{d}

;

ω_{0}

only controls how big that response is.

TF2. The amplitude peak sits exactly at

ω_{d} = ω_{0}

False. Minimising the denominator

D

gives the peak at

ω_{r es} = ω_{0}^{2} - 2 γ^{2}

, pushed below

ω_{0}

by damping; only in the undamped limit

γ \to 0

do they coincide.

TF3. Doubling the driving force

F_{0}

shifts the resonance to a higher frequency.

False.

F_{0}

appears only in the numerator of

A

, so it scales amplitude linearly and never enters

ω_{r es} = ω_{0}^{2} - 2 γ^{2}

; the peak's location is fixed by

ω_{0}

and

γ

alone.

TF4. At resonance the driving force is in phase with the displacement.

False. At resonance

ϕ = π /2

(90°): the force leads the displacement by 90°, which places it exactly in phase with the velocity — so the power

P = F \overset{x}{˙}

(force times velocity) is maximal because both point the same way at every instant.

TF5. Increasing damping always lowers the amplitude at every driving frequency.

Mostly true, but sharpest near resonance. More

γ

raises the

2 γ ω_{d}

term in the denominator, lowering

A

; the effect is tiny far from resonance (where the mismatch term dominates) and huge at the peak.

TF6. Below resonance the mass moves in phase with the driving force.

Approximately true only as $ω_{d} \to 0$ . In the slow limit

tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}) \to 0

ϕ \to 0

(spring-dominated); as

ω_{d}

climbs toward

ω_{0}

ϕ

grows continuously toward

π /2

(90°), so "in phase" is a limiting statement, not a below-resonance rule.

TF7. A forced oscillator can have larger amplitude than a free one with the same energy input.

True at resonance. Because energy is injected every cycle in step with the velocity (power

P = F \overset{x}{˙}

stays positive), amplitude builds until the per-cycle input equals the per-cycle damping loss — a balance a single free release never reaches.

TF8. With zero damping the steady-state amplitude at

ω_{d} = ω_{0}

is finite.

False. Setting

γ = 0

makes the denominator

(ω_{0}^{2} - ω_{d}^{2})^{2}

, which vanishes at

ω_{d} = ω_{0}

, so

A \to \infty

— the idealised runaway that real damping tames.

Spot the error

SE1. "The transient dies out, therefore the natural frequency stops existing in the system."

The natural frequency still governs the transient's oscillation rate while it lasts and still sets where the amplitude peaks; only the transient's contribution to the motion decays as

e^{- γ t}

, not

ω_{0}

itself.

SE2. "Since

A = F_{0} / k

when

ω_{d} \to 0

, the amplitude only depends on the spring."

That is only the static limit. In

A = F_{0} / (k - m ω_{d}^{2})^{2} + (b ω_{d})^{2}

both inertia (

m ω_{d}^{2}

) and damping (

b ω_{d}

) enter for general

ω_{d}

; the spring-only formula is the special case where the drive is too slow for those terms to matter.

SE3. "Resonance means the force and displacement build up together, so they must be in phase."

They build up together in magnitude, but not in phase — at resonance

tan ϕ \to \infty

so the force is

π /2

(90°) ahead of displacement. Confusing "growing together" with "phase-aligned" is the trap.

SE4. "

ω_{r es} < ω_{0}

, so the velocity amplitude also peaks below

ω_{0}

The displacement amplitude

A

peaks at

ω_{r es} = ω_{0}^{2} - 2 γ^{2}

, but the velocity amplitude

A_{v} = ω_{d} A

carries an extra factor

ω_{d}

that pushes its maximum up to exactly

ω_{d} = ω_{0}

; different quantities peak at different frequencies.

SE5. "Because the driver enslaves the system, the phase lag

ϕ

is always zero."

Enslavement fixes the frequency, not the phase. Since

ϕ = arctan (2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}))

, damping makes

ϕ

run from

0

(slow) through

π /2

(resonance) to

π

(fast).

SE6. "At

ω_{d} ≫ ω_{0}

the amplitude approaches

F_{0} / k

just like the slow limit."

No — in the fast limit the

m ω_{d}^{2}

term dominates the denominator and

A \to F_{0} / (m ω_{d}^{2}) \to 0

. Only the slow limit gives

F_{0} / k

; the two extremes are opposite (large vs vanishing amplitude).

SE7. "Sharper resonance peaks mean the system absorbs less total energy."

A sharper peak means a higher quality factor and a taller response at resonance — it absorbs energy very efficiently in a narrow band, not less. See Resonance and Quality Factor.

Why questions

WHY1. Why must the steady-state response have the same frequency as the drive?

The equation of motion is linear; a sinusoidal input at

ω_{d}

can only produce an output at

ω_{d}

(linear systems don't create new frequencies), so any lasting response is locked to

ω_{d}

WHY2. Why does damping shift the amplitude peak below

ω_{0}

rather than above?

Setting

d D / d u = 0

with

u = ω_{d}^{2}

gives

u = ω_{0}^{2} - 2 γ^{2}

; the damping term

4 γ^{2} u

grows with frequency and penalises higher

ω_{d}

, so the minimum of the denominator lands below

ω_{0}

WHY3. Why is power transfer maximal exactly at

ω_{0}

and not at

ω_{r es}

Power

P = F \overset{x}{˙}

is maximal when force and velocity are in phase, which is

ϕ = π /2

(90°); that phase condition holds when

ω_{0}^{2} - ω_{d}^{2} = 0

, i.e. exactly at

ω_{d} = ω_{0}

, regardless of the displacement peak's location. (The velocity-maximisation in the boxed intuition above confirms the same

ω_{0}

WHY4. Why does a phase lag

ϕ

exist at all?

Damping means the system can't respond instantaneously; the

tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2})

formula shows the lag is proportional to the damping term, so with

γ > 0

the displacement always trails the force by a growing angle.

WHY5. Why does the amplitude fall off on both sides of resonance?

Look at

D = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}

. As

ω_{d}

moves either below or above

ω_{0}

, the difference

ω_{0}^{2} - ω_{d}^{2}

leaves zero; because it is squared, its sign is irrelevant and

D

grows in both directions, so

A = (F_{0} / m) / D

shrinks on both sides.

WHY6. Why can we ignore the transient when computing steady-state amplitude?

The transient decays as

e^{- γ t}

and is gone after a few time constants; the amplitude formula describes the surviving driven part, which is what any long-time measurement sees.

WHY7. Why is

ω_{0}

still worth knowing if the system oscillates at

ω_{d}

Because

ω_{0}

sets the scale of the response: through

D

it fixes where amplitude peaks, how sharp the peak is, and via

tan ϕ

the sign of the phase lag — the whole shape of the

A (ω_{d})

curve is anchored to it. See Simple Harmonic Motion.

Edge cases

EC1. What is the amplitude as

ω_{d} \to 0

, and physically why?

A = F_{0} / (k - m ω_{d}^{2})^{2} + (b ω_{d})^{2}

the denominator

\to k

, so

A \to F_{0} / k

: an infinitely slow drive is a constant force and the mass just sits displaced by Hooke's law.

EC2. What is the phase lag

ϕ

ω_{d} \to \infty

, and what does it look like?

ϕ \to π

(180°): the displacement is exactly opposite the force. Watch out for the arctan trap — the raw ratio

tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2})

approaches

0

from the negative side (denominator is a large negative number), and a naive

arctan

would read that as

ϕ \to 0

; the correct branch (denominator negative ⇒ add

π

, see EC7) gives

ϕ \to π

, matching the physics of opposite motion.

EC3. What happens to the amplitude curve when

2 γ^{2} \geq ω_{0}^{2}

ω_{r es}^{2} = ω_{0}^{2} - 2 γ^{2} \leq 0

becomes non-positive, so there is no real peak frequency. Tracing

D = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}

: its derivative

d D / d u = - 2 (ω_{0}^{2} - u) + 4 γ^{2}

is already positive at

u = 0

when

2 γ^{2} \geq ω_{0}^{2}

, so

D

only increases and

A

falls monotonically from its static value

F_{0} / k

. This is the heavily-overdamped regime, linked to Damped Oscillations.

EC4. In the undamped limit

γ \to 0

, where is the peak and how tall is it?

ω_{r es} = ω_{0}^{2} - 0 = ω_{0}

and the denominator hits zero there, so

A \to \infty

— the mathematical signature of true resonance with nothing to limit it.

EC5. If

ω_{d} = ω_{0}

but damping is heavy, is the system "at resonance"?

You're at the frequency of maximum power transfer (velocity/

A_{v}

peak, always at

ω_{0}

), but not at the amplitude peak (which is at

ω_{r es} < ω_{0}

); "resonance" needs you to name the specific quantity you mean.

EC6. At exactly

ω_{d} = ω_{0}

, what is

tan ϕ

, and how do we read

ϕ

there?

The mismatch

ω_{0}^{2} - ω_{d}^{2} = 0

, so

tan ϕ = 2 γ ω_{d} /0 \to + \infty

, forcing

ϕ = π /2

(90°) exactly — the crossover where the response switches from spring-led to inertia-led.

EC7. The naive

ϕ = arctan (2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}))

returns values in

(- π /2, π /2)

. How do we get the true lag that runs

0 \to π

Below

ω_{0}

the denominator

ω_{0}^{2} - ω_{d}^{2} > 0

arctan

gives the correct acute lag; above

ω_{0}

it turns negative and

arctan

jumps to a negative angle, so we add $π$ to keep

ϕ

increasing smoothly through

π /2

(90°) toward

π

(180°). This is the standard arctan branch-fix — the phase physically never goes backward. See Complex Exponential Method.

EC8. What is the steady-state amplitude if the driving force is switched off (

F_{0} = 0

A = 0

: with

F_{0} = 0

the numerator vanishes, so there is no steady state — only the free transient that decays away, consistent with Energy in Oscillations losing energy to damping.

Recall One-line self-test

Cover every answer above, sweep top to bottom, and flag any where you said "true/false" without the because — and check that your reasoning cites the actual formula ( $D$ , $tan ϕ$ , or $A_{v} = ω_{d} A$ ), not just words. The reasoning is the exam answer; the verdict alone earns nothing.