Visual walkthrough — Forced oscillations — driving frequency

We will use a few ideas from the vault as we go: the spring picture from Simple Harmonic Motion, the drag idea from Damped Oscillations, the phasor shortcut in Complex Exponential Method, and — at the peak in Step 7 — the sharpness language of Resonance and Quality Factor.

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Step 1 — Draw every force pushing the mass

WHAT. Put a block of mass $m$ on a spring. Three arrows act on it. We add them up — that is all Newton's second law says: (mass) × (acceleration) = (sum of arrows).

WHY. Before any symbol means anything, we must know what is physically shoving the block. There are exactly three pushes and no more, so the whole rest of the page is bookkeeping on these three arrows.

PICTURE. Look at figure — the block sits at position $x$ measured from its rest spot (the vertical dashed line).

The three arrows, read off the picture:

• Spring pull $-kx$. The spring always points the block back to the dashed line, so when $x$ is positive (block to the right) this arrow points left → the minus sign. $k$ is the spring's stiffness (from Simple Harmonic Motion).
• Drag $-b\dot{x}$. Here $\dot{x}$ (read "$x$-dot") means the block's velocity — how fast $x$ is changing. Drag opposes motion, so it fights the velocity $\dot{x}$ → another minus. $b$ is the drag strength (from Damped Oscillations).
• The hand $F_0\cos ({\omega }_{d}t)$. An outside agent shoves back and forth. $F_0$ is the shove's strength, ${\omega }_d$ is how often the hand cycles, and $\cos ({\omega }_{d}t)$ makes it swing smoothly from $+F_0$ to $-F_0$ and back.

Adding them:

$\underset{\text{mass}\times \text{accel.}}{\underbrace{m\ddot{x}}}=\underset{\text{spring back}}{\underbrace{-kx}}\underset{\text{drag fights }\dot{x}}{\underbrace{-b\dot{x}}}+\underset{\text{the hand}}{\underbrace{F_0\cos ({\omega }_{d}t)}}$

Here $\ddot{x}$ ("$x$-double-dot") is the acceleration — how fast the velocity itself is changing.

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Step 2 — Tidy the equation into pure physics constants

WHAT. Divide the whole line by $m$. Then rename the leftover clumps of constants.

WHY. The raw numbers $k,b,m$ hide the two things that actually control behaviour: how fast the spring wants to bounce and how quickly friction eats energy. We repackage them so those two ideas stand out.

PICTURE. In the figure, the same three arrows are relabelled by the tidy constants — nothing physical moved, only the names.

Define two new symbols by looking at what survives division by $m$:

• ${\omega }_0^2=\frac{k}{m}$, so ${\omega }_0=\sqrt{k/m}$ — the natural frequency. Because $k$ and $m$ are both positive, ${\omega }_0$ is a real positive number (we always take the positive root). ${\omega }_0$ (omega-zero) is the rhythm the block would bounce at with no friction and no hand. Stiffer spring or lighter mass → faster natural rhythm.
• $2\gamma =\frac{b}{m}$ — the damping rate. $\gamma$ (gamma) tells how fast the free wobble dies as $e^{-\gamma t}$. The "$2$" is a convenience that makes later algebra land cleanly.

The tidy equation:

$\ddot{x}+\underset{\text{friction rate}}{\underbrace{2\gamma }}\dot{x}+\underset{{\text{natural rhythm}}^2}{\underbrace{{\omega }_0^2}}x=\underset{\text{drive per kg}}{\underbrace{\frac{F_0}{m}}}\cos ({\omega }_{d}t)$

Every term is now an acceleration (units of ${\text{m/s}}^2$). The left side is the oscillator's own personality; the right side is the outside world pushing it.

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Step 3 — The transient fades; only the driven part remains

WHAT. The full motion is a sum of two pieces: a transient (the block's own damped wobble) and a steady state (the response to the hand). We watch the transient die and throw it away.

WHY. A periodic hand cannot support a response at any frequency except its own. Anything at the natural frequency ${\omega }_0$ has no one feeding it, so friction drains it via $e^{-\gamma t}$. Wait long enough and only the fed motion is left.

PICTURE. The figure stacks three curves: the dying transient (fades to a flat line), the never-dying steady state, and their sum (starts messy, settles into a clean repeat).

For the common underdamped case (light friction, $\gamma <{\omega }_0$) the transient is a decaying wobble:

$x(t)=\underset{\text{transient}\to 0}{\underbrace{C{e}^{-\gamma t}\cos ({\omega }_{1}t+\psi )}}+\underset{\text{steady state, survives}}{\underbrace{x_{ss}(t)}}$

Reading the transient's symbols:

• $C$ — the transient's starting size, and $\psi$ (psi) its starting phase. These two are not set by the hand; they are fixed by how the block was released (its initial position and velocity). They only affect the messy first moments, never the final motion.
• ${\omega }_1=\sqrt{{\omega }_0^2-{\gamma }^2}$ — the damped wobble frequency. Friction slows the free bounce slightly below ${\omega }_0$; that shifted rhythm is ${\omega }_1$ (this comes straight from Damped Oscillations). It is real only when $\gamma <{\omega }_0$, which is exactly the underdamped assumption.

The envelope $e^{-\gamma t}$ (the orange fading curtain in the figure) squashes the transient to nothing. From here on we only chase $x_{ss}$.

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Step 4 — Guess the response shape, with a lag

WHAT. Propose that the steady state looks like the drive but shifted: $x_{ss}(t)=A\cos ({\omega }_{d}t-\varphi )$

WHY. Since Step 3 forces frequency ${\omega }_d$, the only freedoms left are how big the swing is ($A$) and how far behind the hand it trails ($\varphi$). A pure cosine at ${\omega }_d$ with two dials is the most general such motion.

PICTURE. The figure overlays the hand's push and the block's response. The block's peak arrives later — that time gap, converted to an angle, is $\varphi$.

• $A$ — amplitude, the height of the response (what we want to solve for).
• $\varphi$ — phase lag (phi). The hand peaks; a moment later the block peaks. The minus in $({\omega }_{d}t-\varphi )$ means the block runs behind. Damping causes this delay: friction stops the block from answering instantly. We keep $\varphi$ in the range $0\le \varphi \le \pi$ (a response can lag from "in step" to "opposite", never lead).

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Step 5 — Substitute, match term-by-term, read off a right triangle

WHAT. Put the guess into the tidy equation. We need ${\dot{x}}_{ss}$ and ${\ddot{x}}_{ss}$: ${\dot{x}}_{ss}=-A{\omega }_d\sin ({\omega }_{d}t-\varphi ),{\ddot{x}}_{ss}=-A{\omega }_d^2\cos ({\omega }_{d}t-\varphi )$ Each dot pulls down a factor of ${\omega }_d$ and rotates cosine → sine → cosine (with signs).

WHY. For the guess to be a solution, the left side must equal the right side at every single instant $t$. A cosine and a sine of the same argument are independent shapes — they never cancel each other — so the equation can only hold for all $t$ if the total cosine coefficient on the left equals the cosine coefficient on the right, AND the total sine coefficient on the left equals the sine coefficient on the right. That "match term-by-term" rule turns one equation-that-holds-always into two ordinary equations, one per shape, which is exactly enough to pin down our two unknowns $A$ and $\varphi$.

Writing everything in the argument $u={\omega }_{d}t-\varphi$, the left side collects to $[({\omega }_0^2-{\omega }_d^2)A]\cos u-[(2\gamma {\omega }_d)A]\sin u,$ and the right side $\frac{F_0}{m}\cos ({\omega }_{d}t)$ expands (with ${\omega }_{d}t=u+\varphi$) into a cosine-part and a sine-part too. Matching the two shapes gives: $\underset{\text{cosine coefficients}}{\underbrace{({\omega }_0^2-{\omega }_d^2)A=\frac{F_0}{m}\cos \varphi }},\underset{\text{sine coefficients}}{\underbrace{(2\gamma {\omega }_d)A=\frac{F_0}{m}\sin \varphi }}.$

PICTURE — why those two equations are the legs of a right triangle. This is the Complex Exponential Method phasor idea, which we state plainly here so it is not taken on faith. A phasor is just an arrow whose length is a term's amplitude and whose direction records its phase; a $\cos$ term points along the horizontal axis, and a $\sin$ term points along the vertical axis (a quarter-turn away), because a sine is a cosine shifted by 90°. So the two matched equations above are literally the horizontal and vertical components of one arrow-sum, and the drive $F_0/m$ is that arrow. Look at the figure:

• The horizontal leg is the cosine equation: $({\omega }_0^2-{\omega }_d^2)A$ — the in-step part, the spring-vs-inertia frequency mismatch.
• The vertical leg is the sine equation: $(2\gamma {\omega }_d)A$ — the quarter-turn damping cost, at right angles because drag follows velocity, which leads displacement by 90°.
• The hypotenuse is the drive strength $F_0/m$.

Square-and-add the two matched equations (using ${\cos }^2\varphi +{\sin }^2\varphi =1$) — that is Pythagoras on the triangle:

${\left(\frac{F_0}{m}\right)}^2=[({\omega }_0^2-{\omega }_d^2)A{]}^2+[(2\gamma {\omega }_d)A{]}^2$

Factor out $A^2$ and take the square root:

$\boxed{A({\omega }_d)=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }_d^2{)}^2+(2\gamma {\omega }_d{)}^2}}}$

Divide the sine equation by the cosine equation ($\tan \varphi =\sin \varphi /\cos \varphi$) — that is the corner angle of the triangle, "opposite over adjacent":

$\tan \varphi =\frac{\text{vertical (damping)}}{\text{horizontal (mismatch)}}=\frac{2\gamma {\omega }_d}{{\omega }_0^2-{\omega }_d^2}$

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Step 6 — Walk all three regimes of the dial ${\omega }_d$

WHAT. Turn the driving-frequency dial from very slow, through matched, to very fast, and watch $A$ and $\varphi$.

WHY. The formula is only trustworthy if it gives sensible pictures at every setting — including the extremes. We check each and confirm nothing breaks.

PICTURE. The figure is the amplitude curve with three flags planted on it; below each flag, a little clock shows the phase lag $\varphi$.

• Slow, ${\omega }_d\ll {\omega }_0$. Mismatch term $\to {\omega }_0^2$, so $A\to \frac{F_0/m}{{\omega }_0^2}=\frac{F_0}{k}$. The block just sits where the steady force stretches it — pure Hooke's law. Lag $\varphi \to 0$: block moves with the hand.
• Matched, ${\omega }_d\approx {\omega }_0$. Horizontal leg of the triangle vanishes, so the denominator is only the damping leg $2\gamma {\omega }_d$ — tiny — and $A$ peaks. The triangle is now purely vertical, so $\varphi \to \pi /2$: the block is a quarter-cycle behind, which means it moves in step with its own velocity → force feeds power every instant (see Energy in Oscillations).
• Fast, ${\omega }_d\gg {\omega }_0$. Mismatch term $\to -{\omega }_d^2$ (huge), so $A\to \frac{F_0/m}{{\omega }_d^2}\to 0$. Inertia wins; the block cannot keep up. The horizontal leg flips sign, so $\varphi \to \pi$: block moves opposite the hand.

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Step 7 — Where exactly is the peak? (the degenerate-damping check)

WHAT. Find the ${\omega }_d$ that maximises $A$. Biggest $A$ means smallest denominator, so minimise the stuff under the root: $D({\omega }_d)=({\omega }_0^2-{\omega }_d^2{)}^2+(2\gamma {\omega }_d{)}^2$ Set its slope to zero, $\frac{dD}{d{\omega }_d}=0$, and solve.

WHY. We use a derivative because it is the tool that answers "where does a curve stop rising and start falling?" — the flat point at the very top. That is precisely the peak we want.

PICTURE. The figure zooms on the peak and marks its true location ${\omega }_{res}$ sitting slightly left of ${\omega }_0$; a dashed light-damping curve peaks almost on top of ${\omega }_0$.

Working the derivative: $\frac{dD}{d{\omega }_d}=2({\omega }_0^2-{\omega }_d^2)(-2{\omega }_d)+2(2\gamma {)}^2{\omega }_d=4{\omega }_d[{\omega }_d^2-{\omega }_0^2+2{\gamma }^2]=0$

This product is zero in exactly two ways, and we must examine both — dividing by $4{\omega }_d$ is only allowed once we have separately accounted for the root it would throw away:

• Root 1: ${\omega }_d=0$. This is a genuine extremum of $D$, not a throwaway. At ${\omega }_d=0$ the drive is static and $D={\omega }_0^4$, giving $A=F_0/k$ (the slow-drive value from Step 6). Checking whether it is a max or a min: for very light damping $D$ dips below ${\omega }_0^4$ near ${\omega }_d={\omega }_0$, so ${\omega }_d=0$ is a local maximum of $D$ — i.e. a local minimum of $A$ on the way up to the resonant bump. So it is a real stationary point, but not the amplitude peak we want.
• Root 2: the bracket $=0$. Now, with ${\omega }_d=0$ already handled, we may divide by $4{\omega }_d$ and set the bracket to zero: ${\omega }_d^2-{\omega }_0^2+2{\gamma }^2=0$, giving

$\boxed{{\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}}$

This second root is the true amplitude peak (it is where $D$ bottoms out). Reading it:

• Light damping ($\gamma \ll {\omega }_0$): ${\omega }_{res}\approx {\omega }_0$. The peak sits essentially on the natural frequency — the textbook slogan.
• Heavy damping ($2{\gamma }^2\ge {\omega }_0^2$): the square root breaks (goes imaginary). Then Root 2 does not exist and only Root 1 survives, so $A$ has no peak at all — it just falls off monotonically from its slow-drive value $F_0/k$. The system is too sluggish to resonate.

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The one-picture summary

This single figure carries the whole page: three force-arrows on the block (Step 1) become two clean constants ${\omega }_0,\gamma$ (Step 2); the transient fades and leaves a response locked at ${\omega }_d$ (Steps 3–4); matching cosine- and sine-parts term-by-term makes a right triangle whose hypotenuse gives $A$ and whose angle gives $\varphi$ (Step 5); turning the dial sweeps the amplitude curve with its peak at ${\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}$ and sharpness set by $Q={\omega }_0/2\gamma$ (Steps 6–7).

Recall Feynman retelling — the whole walk in plain words

A block on a spring feels three shoves: the spring pulling it home, friction dragging on its speed, and a hand pushing back and forth. Write "mass times acceleration equals those three added up." Clean it: the spring's eagerness becomes ${\omega }_0$ (its favourite rhythm) and the friction becomes $\gamma$ (how fast it tires). At first the block does two things at once — its own dying wobble and the hand's driven motion — but the wobble fades (its size and starting phase, $C$ and $\psi$, are just leftovers of how you let go), and forever after the block just marches to the hand's beat at ${\omega }_d$, only trailing a little because friction slows its reflexes. To find how big the march is, we plug the marching guess back in and insist it work at every instant; since wiggles-that-follow-a-cosine and wiggles-that-follow-a-sine are different shapes that never cancel, we must match each shape on its own — that gives two tidy equations, which are exactly the two legs of a right triangle. One leg is how mismatched the hand's speed is from the block's favourite rhythm, the other leg is the friction cost, and the slanted side is the hand's strength. Pythagoras on that triangle gives the amplitude; the triangle's corner angle gives the lag. Finally, turning the hand's speed dial: too slow and the block just follows like a stretched spring; matched and the block swings hugest (resonance); too fast and inertia wins so it barely twitches, moving opposite the hand. The very biggest swing sits a hair below the favourite rhythm — how tall and narrow that swing is depends on $Q={\omega }_0/2\gamma$ — and if friction is enormous, there is no big-swing bump at all.

Recall Quick self-check

Where does the amplitude formula's denominator come from? ::: Pythagoras on the term-balance triangle: horizontal leg = mismatch $({\omega }_0^2-{\omega }_d^2)$, vertical leg = damping $2\gamma {\omega }_d$. Why can we match cosine- and sine-parts separately? ::: Because cosine and sine of the same argument are independent shapes; an equation holding for all $t$ forces each shape's coefficient to balance on its own. Why does the transient disappear? ::: It is built only from decaying exponentials ($e^{-\gamma t}$ etc.), so it dies out — in every damping case. Exact peak location, and both roots of dD/dwd=0? ::: Roots are ${\omega }_d=0$ (a minimum of $A$) and ${\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}$ (the true peak). When is there no resonant peak? ::: When $2{\gamma }^2\ge {\omega }_0^2$ (i.e. $Q\le 1/\sqrt{2}$) — the square root turns imaginary. What does $Q={\omega }_0/2\gamma$ measure? ::: The sharpness/height of the resonance peak: large $Q$ = tall narrow spike, small $Q$ = low broad hump.