Hooke's law — spring force F = −kx: Visual walkthrough

Step 1 — Draw the spring and name its "home"

WHAT. A real spring, lying flat, no one touching it. It has a length it likes — its natural length. Mark that resting position as the number line's zero.

WHY. Every measurement below is "how far from home." Before we can say "how far," we need to agree where home is. Home $= 0$ .

PICTURE. In the figure, the grey block sits at the tick labelled $x = 0$ . Pull it right and the pointer moves to positive numbers; push it left and it moves to negative numbers.

Figure — Hooke's law — spring force F = −kx

Step 2 — Energy lives in a valley

WHAT. Instead of forces, start with energy. Call the energy stored in the spring $U$ . Plot how much energy the spring holds for each position $x$ . That plot is a smooth U-shaped valley with its lowest point at $x = 0$ .

WHY. A spring at home stores nothing; stretch OR compress it and it stores more. "More energy on both sides, least in the middle" is exactly the shape of a valley. Starting from energy (not force) lets us prove that any spring must be linear — we'll cash that in by Step 5.

PICTURE. The green curve dips to a minimum at $x = 0$ . A marble placed on this curve rolls downhill — toward the bottom, toward home. Remember that marble; it is the restoring force.

Step 3 — Force is the steepness of the valley

WHAT. Before any formula, name the thing we are after. Then turn "the marble rolls downhill" into a formula: the force is minus the steepness of the energy curve.

$F = - \frac{d U}{d x}$

WHY this tool — the derivative $\frac{d U}{d x}$ . We need a number that says how tilted the valley is right where the block sits. A single ratio "rise over run" only describes a straight line — but our valley curves. The derivative $\frac{d U}{d x}$ is the tool built exactly for this: it is the slope of the tangent line touching the curve at one point — the steepness right here. No other tool measures local steepness of a curve; that is why it enters.

Term by term:

$F$ — the spring's force on the block, just defined above.
$\frac{d U}{d x}$ — the slope of the energy curve at position $x$ (how fast energy climbs as you move right).
the minus sign — makes force point downhill: where the curve rises to the right, the marble is pushed left.

PICTURE. On the right wall of the valley the tangent tilts up (positive slope), so $- slope$ points left — back toward home. On the left wall the tangent tilts down (negative slope), so $- slope$ points right — again toward home. At the very bottom the tangent is flat: slope $= 0$ , force $= 0$ .

Step 4 — Zoom into the bottom: every valley looks like a parabola

WHAT. Take any smooth valley and zoom in on its lowest point. Up close, it stops looking special — it looks like the simplest possible valley: a parabola, the curve $U = \frac{1}{2} k x^{2}$ .

WHY this tool — the Taylor expansion. We want to know the shape near the bottom, not far away. The Taylor expansion is the tool that rebuilds a smooth curve near one point as a sum of simple powers of $x$ : a constant piece, a straight-line piece ( $x$ ), a bowl piece ( $x^{2}$ ), a gentler-bending piece ( $x^{3}$ ), and so on: $U (x) = height at bottom U (0) + tilt at bottom U^{'} (0) x + curviness \frac{1}{2} U^{''} (0) x^{2} + higher powers \frac{1}{6} U^{'''} (0) x^{3} + \dots$

Term by term, and why most terms vanish or fade:

$U (0)$ — the energy at the bottom. A constant. Shifting all energy up or down changes nothing about forces, so we drop it (set the bottom to $0$ ).
$U^{'} (0) x$ — the tilt at the bottom. But the bottom of a valley is flat: its slope is $0$ . So $U^{'} (0) = 0$ and this whole term disappears. (This is what "equilibrium" means: no force, so no slope.)
$\frac{1}{2} U^{''} (0) x^{2}$ — the curviness. This is the first term that survives, and it is proportional to $x^{2}$ : a parabola.
$\frac{1}{6} U^{'''} (0) x^{3} + \dots$ — the higher powers. Here is the key intuition: for small $x$ , a bigger power is a smaller number. If $x = 0.1$ , then $x^{2} = 0.01$ but $x^{3} = 0.001$ — ten times tinier, and $x^{4}$ tinier still. So the moment you are close enough to home, the $x^{2}$ bowl towers over every higher term. ("Smooth" is precisely the promise that these higher pieces exist and shrink like this — no sudden kinks.) That is why keeping only up to $x^{2}$ is not laziness; near the bottom it is the whole story.

PICTURE. The figure overlays a bumpy real energy curve (blue) and the parabola (yellow) that hugs it near $x = 0$ . Inside the shaded band close to home they are indistinguishable; only far out, where $x^{3}$ and beyond wake up, do they peel apart. That is why every small wobble is a spring.

Step 5 — Slide down the parabola to get $F = - k x$

WHAT. We now have the energy $U = \frac{1}{2} k x^{2}$ and the rule $F = - \frac{d U}{d x}$ . Combine them.

WHY. Step 3 gave us the machine (force = −steepness). Step 4 gave us the valley (a parabola). Feeding one into the other must spit out the force law. This is the payoff.

Differentiate $\frac{1}{2} k x^{2}$ — the slope of a parabola at position $x$ is $k x$ : $F = - \frac{d U}{d x} = - \frac{d}{d x} (\frac{1}{2} k x^{2}) = - k x$

Term by term:

$\frac{d}{d x} (\frac{1}{2} k x^{2}) = k x$ — the parabola's steepness grows straight-line with $x$ : twice as far out, twice as steep.
the minus — inherited from Step 3: force points back down the wall.
result $F = - k x$ — a straight line through the origin with slope $- k$ .

PICTURE. The figure plots $F$ against $x$ : a straight line sloping downward. Its slope is $- k$ . Cross the vertical axis and the force flips sign exactly as $x$ flips sign — the fingerprint of "restoring."

Step 6 — All four sign-cases on one line (nothing is skipped)

WHAT. Check every sign of $x$ against the line $F = - k x$ — a large stretch, a small stretch, the degenerate rest point, and a large compression.

WHY. A law you only tested for "stretched to the right" is a law you don't trust. Walk each sign scenario so the reader never meets an unshown case. The far-out cases ( $∣ x ∣$ large) also show the "farther = harder" magnitude, but their real job here is to confirm the sign still behaves.

Case	$x$	$- k x$ gives	Direction of $F$	Meaning
Big stretch	$x ≫ 0$	negative (large)	points left	strong pull back home
Small stretch	$x > 0$	negative	points left	pulls block back home
At rest	$x = 0$	$0$	no force	equilibrium, bottom of valley
Big compression	$x ≪ 0$	positive (large)	points right	strong push back home

PICTURE. Four little blocks on one number line, each with its force arrow. Notice: every arrow points at the zero mark, and the far-out arrows are the longest. That, drawn, is what "restoring" means.

Step 7 — The energy is the triangle, not the rectangle

WHAT. How much energy do you store by stretching from home ( $x^{'} = 0$ ) out to some positive stretch $x > 0$ ? It is the area under the stretching-force line — a triangle, giving $U = \frac{1}{2} k x^{2}$ .

WHY this tool — the integral (area under a curve). The stretching force grows from $0$ to $k x$ as you go out; it is never constant, so plain (force $\times$ distance) is wrong — that would be the rectangle of the maximum force. To add up a force that changes at every position, you accumulate $F_{ext} d x^{'}$ across the trip: that accumulation is the integral, and geometrically it is the area under the graph. $U = \int_{0}^{x} F_{ext} > 0 k x^{'} d x^{'} = \frac{1}{2} k x^{2}$

Term by term:

$k x^{'}$ — the outward (positive) force your hand applies at each intermediate position $x^{'}$ , with $0 \leq x^{'} \leq x$ .
$\int_{0}^{x} \dots d x^{'}$ — glue up all the thin slivers of (force $\times$ tiny step) from home out to $x$ (limits $0$ to $x$ , both non-negative).
$\frac{1}{2} k x^{2}$ — the area of the triangle: base $x$ , height $k x$ , so $\frac{1}{2} (base) (height)$ .

PICTURE. The blue triangle (correct, $\frac{1}{2} k x^{2}$ ) sits inside the dashed red rectangle (the wrong $k x \cdot x = k x^{2}$ ). The rectangle is exactly twice the triangle — that's the famous factor of $\frac{1}{2}$ .

The one-picture summary

Everything above, compressed: the valley on the left ( $U = \frac{1}{2} k x^{2}$ ) and the line on the right ( $F = - k x$ ) are the same object seen two ways — the line is the slope of the valley, negated. Read left-to-right: shape of energy $\to$ its steepness $\to$ the restoring force.

Recall Feynman retelling — the whole walk in plain words

A spring lives at the bottom of a valley made of energy. Sitting at the bottom, it's happy — nothing pushing. If you drag it up either wall, it wants to roll back down, and the steepness of the wall right under it is how hard it's shoved. Zoom into the bottom of any valley and it always looks like the same smooth bowl — a parabola, $\frac{1}{2} k x^{2}$ — because up close the tiny higher-power wiggles ( $x^{3}$ , $x^{4}$ , …) shrink faster than the $x^{2}$ bowl and simply don't matter. The steepness of that bowl at position $x$ works out to $k x$ , and "roll downhill" flips the sign, so the spring's push-back is $- k x$ : twice as far, twice as hard, always aimed at home — whether you stretched it or squished it. And to store the energy up you add your hand's growing outward push over the whole trip — that's the triangle under the line, $\frac{1}{2} k x^{2}$ , half of what you'd naively guess, because the force started from nothing.

Where does the minus sign in

F = - k x

come from, pictorially?

From

F = - d U / d x

— the force points downhill in the energy valley, opposite to where energy increases.

Why does every stable system look like a spring near rest?

Taylor-expanding its energy minimum, the constant and linear terms drop out and higher powers shrink faster near home, leaving

\frac{1}{2} U^{''} (0) x^{2}

— a parabola — so

F = - U^{''} (0) x = - k x

On an

F

–

x

graph, what is

k

The magnitude of the (downward) slope of the straight line

F = - k x

Why do we integrate

+ k x^{'}

(not

- k x^{'}

) to get the stored energy?

Because we track the work your hand does, which supplies the outward force

F_{ext} = + k x^{'}

opposing the spring; over a stretch

x^{'}

runs from

0

x > 0

, so the integrand is positive.

Connections

Parent topic — the full statement and examples.
Conservative forces and potential energy — the rule $F = - d U / d x$ used in Step 3.
Work done by a variable force — the integral behind the triangle in Step 7.
Elastic potential energy — the $\frac{1}{2} k x^{2}$ reservoir.
Simple Harmonic Motion — feed $F = - k x$ into motion and you get oscillation, $ω = k / m$ .
Conservation of mechanical energy — the valley trades height (PE) for speed (KE).
Interatomic forces — the "generic valley" argument applied to atomic bonds.