Forced oscillations — driving frequency: Foundations

Before you can read a single line of the parent note, you need to earn every symbol it throws at you. This page builds each one from nothing — plain words first, then the picture, then why the topic can't live without it. Read top to bottom; each block leans on the one above.

0. The stage: a mass on a spring

Everything starts with one picture — a block that can slide left and right, tied to a wall by a spring.

Figure — Forced oscillations — driving frequency

1. Rates of change — the dots $\overset{x}{˙}$ and $\overset{x}{¨}$

The parent writes $\overset{x}{˙}$ and $\overset{x}{¨}$ without explaining the dots. Here is where they come from.

2. The three forces and their symbols

The parent adds three forces. Let's define every letter in each. All forces are measured in newtons (N).

3. Angular frequency $ω$ — the heart of the topic

The symbol $ω$ (Greek "omega") appears everywhere: $ω_{d}$ , $ω_{0}$ , $ω_{r es}$ . Nail it here.

Recall Why is

$ω_{0} = k / m$ ? For a plain spring-mass, Newton gives $m \overset{x}{¨} = - k x$ , i.e. $\overset{x}{¨} = - (k / m) x$ . Anything obeying $\overset{x}{¨} = - ω^{2} x$ wiggles at rate $ω$ , so matching $ω^{2} = k / m$ gives $ω_{0} = k / m$ . Stiffer or lighter → faster natural rhythm.

4. The damping rate $γ$

5. Amplitude $A$ and phase lag $ϕ$

The two answers the topic computes.

Common mistake "Just take

$ϕ = arctan (2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}))$ and you're done." Why it feels right: the parent quotes a single $tan ϕ$ formula, so one $arctan$ button seems enough. Fix: plain $arctan$ only returns angles between $- 90°$ and $+ 90°$ , but the physical lag $ϕ$ runs all the way from $0$ up to $180°$ as you speed the driver up. The catch is the sign of the adjacent side $(ω_{0}^{2} - ω_{d}^{2})$ :

Below resonance ( $ω_{d} < ω_{0}$ ): mismatch is positive, damping cost positive → both triangle sides positive → $ϕ$ is a small acute angle, $0 < ϕ < 90°$ . Nearly in step with the push.
At resonance ( $ω_{d} = ω_{0}$ ): mismatch is zero, so the triangle is purely vertical → $ϕ = 90°$ exactly. (The formula's denominator hits $0$ ; don't panic — the angle is simply a right angle.)
Above resonance ( $ω_{d} > ω_{0}$ ): mismatch goes negative (adjacent side points backwards) → $ϕ$ is obtuse, $90° < ϕ < 180°$ . The response is now on the opposite side of the push. The safe recipe is the two-argument arctangent: take $ϕ = atan2 (2 γ ω_{d}, ω_{0}^{2} - ω_{d}^{2})$ , which uses both signs to land $ϕ$ in the intended range $0 < ϕ < π$ . A single $arctan$ would wrongly report a negative lag above resonance.

6. The zero-damping edge case

What happens to everything above if friction vanishes? This is the case the mnemonics quietly skip.

Intuition

$γ \to 0$ : undamped resonance blows up Look at the amplitude denominator $(ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}$ . The second term $(2 γ ω_{d})^{2}$ is the only thing keeping the denominator away from zero when you drive exactly at $ω_{d} = ω_{0}$ .

With damping ( $γ > 0$ ): at $ω_{d} = ω_{0}$ the first term is $0$ but the damping term survives, so $A$ is large but finite — a rounded peak.
With no damping ( $γ = 0$ ): at $ω_{d} = ω_{0}$ both terms are zero, the denominator is $0$ , and $A \to \infty$ . The push adds energy every single cycle with nothing to bleed it away, so the amplitude grows without bound (a real spring would break first).
Phase in this limit: $ϕ$ jumps abruptly from $0$ (below) straight to $180°$ (above), with the $90°$ value only touched exactly at $ω_{0}$ — the smooth $atan2$ transition of Section 5 collapses to a sudden flip. This is why the topic needs damping $b$ : it's what makes resonance a tall-but-finite peak instead of a blow-up. See Resonance and Quality Factor.

The prerequisite map

Read this map as the assembly order of the topic: the leftmost boxes are raw ingredients you must already own; arrows show which symbol feeds into which. Notice that Newton's second law is the funnel — mass $m$ , spring $k$ , damping $b$ and the drive all pour into it, and out the other side come the two headline answers $A$ and $ϕ$ . If any upstream box is fuzzy, the box it points to will be too, so use this to locate exactly which foundation to revisit.

For the shortcut everyone uses to do the substitution step painlessly, see Complex Exponential Method. To see where this all leads once oscillations travel through space, see Waves and Standing Waves. The full topic lives at the parent note.

Equipment checklist

Test yourself — say the answer out loud before revealing.

What does

m

mean and what are its units?

Mass of the block (its sluggishness); units kilograms (kg).

What does

x

mean, in one phrase, and its units?

The block's distance from equilibrium; units metres (m).

What does one dot (

\overset{x}{˙}

) mean? Two dots (

\overset{x}{¨}

One dot = velocity in m/s; two dots = acceleration in m/s².

Why does the restoring force

- k x

carry a minus sign, and what are

k

's units?

It always points back toward home, opposite the displacement;

k

is in N/m.

Why is drag written

- b \overset{x}{˙}

, and what are

b

's units?

Faster motion means more drag and the minus opposes velocity;

b

is in N·s/m = kg/s.

What are the three pieces of

F_{0} cos (ω_{d} t)

F_{0}

= peak push strength (N);

cos

= the smooth back-and-forth shape;

ω_{d}

= how fast you push (rad/s).

What is

ω

physically, its units, and why radians/second?

How fast the rhythm cycles, in rad/s; radians because

cos

eats radians, keeping

cos (ω t)

clean.

Difference between

ω_{d}

and

ω_{0}

ω_{d}

is the driver's rhythm (you choose it);

ω_{0} = k / m

is the system's own preferred rhythm.

How is

γ

related to

b

, and what are its units?

γ = b / (2 m)

; units 1/s, the same as

ω

so they compare directly.

What do

A

and

ϕ

each answer?

A

= how big the steady swing is (m);

ϕ

= how far (as an angle) the motion lags behind the push.

When taking

ϕ

from

tan ϕ

, why can't you use plain

arctan

Plain

arctan

only spans

- 90°

+ 90°

; use

atan2

so the sign of

(ω_{0}^{2} - ω_{d}^{2})

places

ϕ

correctly in

0

180°

In the zero-damping limit

γ \to 0

, what happens at

ω_{d} = ω_{0}

The denominator hits zero and the amplitude

A

diverges to infinity — undamped resonance blows up.