Spring Force and Hooke's Law

Pull on a spring and it pulls back. Pull twice as hard and it stretches twice as far. That homely observation — Robert Hooke compressed it into the Latin motto ut tensio, sic vis, "as the extension, so the force" — turns out to be one of the most consequential approximations in all of physics. Not because springs are special, but because nothing is: zoom in on any system resting at a stable equilibrium — a bridge cable, a chemical bond, a tuning fork, the tip of an atomic force microscope — and for small disturbances it behaves exactly like a spring. Same law, same mathematics, different hardware. You arrive here knowing Newton's second law and how forces produce acceleration. Hooke's law hands you your first force that depends on position: gravity near the ground is the same everywhere, friction doesn't care where you are, but a spring's force is dictated entirely by how far it has been stretched or compressed. That one new ingredient — a force that grows with displacement and always points back home — is what turns motion into oscillation, and it opens the door to the most important solvable problem in physics, the harmonic oscillator. Master the content of one minus sign and one proportionality constant here, and you will spend the rest of your physics education cashing it in: simple harmonic motion, molecular vibrations, sound, circuits, even quantum fields all begin with $F = -kx$ .

Key takeaways

An ideal spring exerts a restoring force proportional to its displacement from natural length: $F = -kx$ , with spring constant $k$ in N/m.
The minus sign is the physics: the force always points back toward equilibrium, opposing the displacement — that is what makes oscillation possible.
The energy stored is $U = \tfrac{1}{2}kx^2$ , a parabola; the force is the negative slope of this energy curve.
Feed $F = -kx$ into Newton's second law and the motion is sinusoidal with angular frequency $\omega = \sqrt{k/m}$ — stiffness and mass alone set the clock, not amplitude.
Hooke's law is an approximation, valid below the elastic limit — but a universal one: every stable equilibrium is Hooke-like for small displacements, with $k = U''(x_0)$ .
Springs in series are softer ( $1/k_{\text{eff}} = 1/k_1 + 1/k_2$ ); springs in parallel are stiffer ( $k_{\text{eff}} = k_1 + k_2$ ).
Key idea: linear and restoring — a force proportional to displacement, aimed back at equilibrium, is the universal local behaviour of every stable system.

The core insight

A spring force is linear and restoring, $\vec F = -k\vec x$ ; because this is the leading term in the expansion of any potential about a stable minimum, the spring is the universal local model of equilibrium — its stiffness $k$ is just the curvature $U''$ of the potential energy at the bottom of the well.

Where this lives today

Hooke's law is quietly running inside your pocket. The accelerometer in every phone is a MEMS device: a microscopic proof mass suspended on silicon springs, whose deflection $x = ma/k$ is read out capacitively thousands of times per second — Hooke's law as a measuring instrument, exactly as in a Victorian spring scale but four orders of magnitude smaller. Atomic force microscopes push the same idea to its limit, using calibrated cantilevers ( $k$ of order 0.01–100 N/m) to feel surfaces atom by atom and to unfold single protein molecules, measuring the piconewton forces of biology directly against a known stiffness. The universality theorem (equation 4) is a working tool as much as a slogan. Every infrared and Raman spectrometer reads molecular bond stiffnesses directly: a diatomic bond is a spring with $k = U''(r_0)$ , vibrating at $\omega = \sqrt{k/\mu}$ with $\mu$ the reduced mass, so a measured vibration frequency is a measurement of the curvature of the bonding potential. In every molecular-dynamics simulation of a protein or a battery electrolyte, those bonds are literal harmonic springs, equation (4) executed a trillion times per run. And at the grandest scale, the mirror suspensions of the LIGO gravitational-wave detectors are exquisitely engineered pendulum-and-fibre spring systems, designed so that the thermal jiggling of their Hookean modes stays below the whisper of a passing spacetime ripple.

The physical picture

Put a block on a frictionless table and tie it to the wall with a spring. Leave it alone at the spring's natural length and nothing happens — zero stretch, zero force. Drag it to the right and the stretched spring tugs it back left. Shove it to the left and the compressed spring pushes it back right. In both cases the force points toward the equilibrium position, and its strength grows in exact proportion to how far you've gone. That's the entire law: proportional in magnitude, always aimed back home.

Figure

One spring, three states — the force always points back to natural length

Plot the same story as a graph and the whole law becomes one straight line through the origin with slope $-k$ — and the graph is honest about where the law ends. Stray too far from natural length and the response bends away from the line; push past the elastic limit and the material deforms permanently, never returning home.

Figure

Force vs displacement — the linear window and where it ends

A second picture is worth carrying with you for the rest of physics. The spring's potential energy $U = \tfrac12 kx^2$ is a smooth bowl — a parabola — and the block is a marble rolling in it. The force is minus the slope of the bowl: steep walls (large $k$ ) mean a violent shove back toward the bottom; a shallow bowl (small $k$ ) means a gentle nudge. Now zoom in on any smooth valley — the bottom of a wine glass, the minimum of a molecular bond's energy curve. Close enough to the bottom, every valley looks parabolic, because the first surviving term in the Taylor expansion of a minimum is the quadratic one. Nature is full of springs in disguise, and $k$ is simply how sharply the valley curves.

The law and its energy

Now compress both pictures into symbols. The question the law answers is: given how far a spring has been deformed, what force does it fire back? Hooke's experimental finding is that the response is perfectly linear — deform twice as far, feel exactly twice the force — and always aimed so as to undo the deformation. For an ideal spring displaced a distance $x$ from its natural (unstretched) length, the force the spring exerts on the attached body is

F = -kx \qquad (1)

where: $F$ — force exerted by the spring on the body; $k$ — spring constant (stiffness), in N/m, always positive; $x$ — signed displacement from natural length (positive for stretch, negative for compression). The minus sign makes the force restoring: opposite to the displacement, aimed at equilibrium. In three dimensions the same statement reads $\vec F = -k\vec x$ : the force lies along the displacement and opposes it. Note carefully what equation (1) is not: it is not the force you apply. To hold the spring stretched at $x$ you must supply $F_{\text{applied}} = +kx$ — the spring's pull and your pull are equal and opposite.

Derivation AThe stored energy of a stretched spring

We want the energy banked in a spring held at displacement $x$ — the quantity that will power every energy-conservation argument involving springs. The strategy is to compute the work an external agent must do to stretch the spring slowly from its natural length, then confirm that this work is recoverable by checking that the force is the negative slope of the resulting energy function. The thing to watch is that the force is not constant during the stretch, so "force times distance" fails; the integral is mandatory.

Step 1. Compute the work needed to stretch the spring. Stretching slowly from $0$ to $x$ , your applied force at intermediate stretch $x'$ is $+kx'$ , so

W = \int_0^x kx'\,\mathrm{d}x' = \tfrac{1}{2}kx^2

Why the integral: the force grows with stretch, so the work is the area of a triangle under the force–displacement line — half of the naive force × distance.

Step 2. Identify the work as stored potential energy. The spring force is conservative — the work depends only on the endpoints, not the path or the schedule of the stretch — so the energy is recoverable and we define

U(x) = \tfrac{1}{2}kx^2 \qquad (2)

where: $U$ — elastic potential energy, chosen zero at natural length. Stretch and compression store equally: $U$ depends on $x^2$ , blind to the sign. Why "recoverable": release the spring and exactly this energy flows back into the block as kinetic energy; nothing is lost in the ideal case.

Step 3. Check the round trip. Differentiating, $F = -\mathrm{d}U/\mathrm{d}x = -kx$ — the force is the negative slope of the energy parabola: downhill in energy is the direction the spring pushes.

Step 4. Confirm the total energy is conserved during free motion. For a mass $m$ oscillating on the spring, $E = \tfrac12 m\dot x^2 + \tfrac12 kx^2$ , and

\frac{\mathrm{d}E}{\mathrm{d}t} = m\dot{x}\ddot{x} + kx\dot{x} = \dot{x}\,(m\ddot{x} + kx) = 0

Why the bracket vanishes: $m\ddot x = -kx$ is exactly Newton's second law for the block, so the two terms cancel identically — energy shuttles between kinetic and elastic but the sum never changes. At the turning points ( $\dot x = 0$ , $x = \pm A$ ) it is all elastic, $E = \tfrac12 kA^2$ ; at the centre it is all kinetic.

Go deeper · Derivation B · 1 minWhy every stable equilibrium is a spring

Derivation BWhy every stable equilibrium is a spring

This is the derivation that explains why Hooke's law escapes the springs chapter and colonizes the rest of physics. We show that any smooth potential energy function, examined close enough to a stable minimum, is indistinguishable from a spring potential — so the linear restoring force is not a special property of coiled metal but the universal local behaviour of stable systems. The route is a Taylor expansion about the minimum; the point to watch is which terms survive and why.

Step 1. Set the scene. Let a particle move in an arbitrary smooth potential $U(x)$ with a stable equilibrium at $x_0$ : the slope vanishes there, $U'(x_0) = 0$ , and the curvature is positive, $U''(x_0) > 0$ . Why these conditions: "equilibrium" means zero force, i.e. zero slope; "stable" means the energy rises on both sides, i.e. positive curvature.

Step 2. Taylor-expand about the minimum:

U(x) = U(x_0) + U'(x_0)(x-x_0) + \tfrac{1}{2}U''(x_0)(x-x_0)^2 + \cdots

Why this is legitimate: any smooth function near a point is captured by its Taylor series; "near" is exactly the small-displacement regime we care about.

Step 3. Kill the trivial terms. The constant $U(x_0)$ only shifts the energy zero; the linear term vanishes because $U'(x_0)=0$ . For small displacements the quadratic term dominates everything higher:

U(x) \approx \tfrac{1}{2}\,U''(x_0)\,(x - x_0)^2 \qquad (3)

Step 4. Differentiate to get the force:

F = -\frac{\mathrm{d}U}{\mathrm{d}x} \approx -\,U''(x_0)\,(x - x_0), \qquad k_{\text{eff}} = U''(x_0) \qquad (4)

where: $k_{\text{eff}}$ — the effective spring constant, equal to the curvature of the potential at its minimum. This is exactly Hooke's law, with the displacement measured from the equilibrium point; a system with mass $m$ therefore oscillates at $\omega = \sqrt{U''(x_0)/m}$ for small amplitudes. Why it matters: hand me any stable system — a bond, a pendulum, a floating ship — and equation (4) converts it into a spring problem with a computable stiffness. *Note: stability requires $U''(x_0) > 0$ — if the curvature were negative the equilibrium would be unstable and the "spring" would push away from rather than back toward the minimum.*

This is why Hooke's law refuses to stay in the springs chapter. Chemical bonds, crystal lattices, pendulums near the bottom of their swing, LC circuits — all reduce to equation (4) for small displacements. And the dynamical payoff deserves its own derivation, because it is the single most-reused calculation in physics.

Go deeper · Derivation C · 1 minFrom the spring force to simple harmonic motion

Derivation CFrom the spring force to simple harmonic motion

We now let the spring act: feed $F = -kx$ into Newton's second law and solve for the motion of a mass $m$ on a frictionless surface. The goal is the celebrated result that the motion is sinusoidal with a frequency set only by stiffness and mass — not by how hard you launched it. The strategy is to write the equation of motion, recognize it as a linear ODE with constant coefficients, and try an exponential; the thing to watch is how imaginary roots announce oscillation rather than growth or decay.

Step 1. Apply Newton's second law. The only horizontal force is the spring force, so

m\ddot{x} = -kx

Step 2. Rearrange into standard form and name the combination that will become the frequency:

\ddot{x} + \frac{k}{m}\,x = 0, \qquad \omega^2 \equiv \frac{k}{m}

Why define $\omega$ now: the equation says "acceleration is proportional to minus displacement," and the proportionality constant must have units of 1/s² — it is begging to be a frequency squared.

Step 3. Try $x = e^{\lambda t}$ . The characteristic equation is $\lambda^2 + \omega^2 = 0$ , so $\lambda = \pm i\omega$ . Why imaginary roots matter: real roots would give exponential growth or decay — running away or dying out. Imaginary roots mean the solution circulates: oscillation. This is the minus sign of Hooke's law doing its work; with $+kx$ the roots would be real and the motion would explode.

Step 4. Write the real general solution:

x(t) = C_1\cos\omega t + C_2\sin\omega t = A\cos(\omega t+\phi)

where: $A$ — amplitude, $\phi$ — phase, both fixed by the initial position and velocity.

Step 5. Read off the frequency and period:

\omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}} \qquad (5)

where: $\omega$ — angular frequency (rad/s); $T$ — period of one full cycle. Stiffer springs and lighter masses oscillate faster — and, crucially, $\omega$ is independent of amplitude. Swing it gently or violently: same clock. This isochronism is what made spring-balance watches possible. Quick check: differentiating $x(t) = A\cos(\omega t + \phi)$ twice gives $\ddot x = -\omega^2 x = -(k/m)x$ , which reproduces $m\ddot x = -kx$ exactly — confirming the solution is self-consistent.

Combining springs

Two springs can share the load in two ways, and the rules mirror capacitors in circuits. The results are worth a short derivation because students routinely guess them backwards.

Go deeper · Derivation D · 1 minEffective stiffness of springs in series and in parallel

Derivation DEffective stiffness of springs in series and in parallel

We want a single equivalent spring constant for two springs joined end to end (series) and side by side (parallel), so that any multi-spring contraption can be collapsed to one effective $k$ before doing dynamics. The whole derivation hangs on one physical question per case: what do the two springs share? In series they carry the same force; in parallel they suffer the same stretch. Watch how each sharing rule dictates which quantities add.

Step 1. Series — same force. A massless junction between the springs would accelerate infinitely under any net force, so the tension is common: $F = k_1 x_1 = k_2 x_2$ . Why "same force" and not "same stretch": each spring stretches by however much that shared force dictates — the soft one stretches more.

Step 2. Series — stretches add. The total extension is $x = x_1 + x_2 = F/k_1 + F/k_2$ , so

\frac{1}{k_{\text{series}}} = \frac{1}{k_1} + \frac{1}{k_2} \qquad (6)

where: $k_{\text{series}}$ — the effective constant, always smaller than either spring alone: compliances (inverse stiffnesses) add. Two identical springs of constant $k$ in series give $k/2$ — the same force now buys twice the total stretch, which is why a longer bungee cord feels softer.

Step 3. Parallel — same stretch, forces add. Both springs are anchored to the same body, so $x_1 = x_2 = x$ and the body feels $F = k_1 x + k_2 x$ :

k_{\text{parallel}} = k_1 + k_2 \qquad (7)

where: $k_{\text{parallel}}$ — always larger than either spring alone: each spring contributes its full pull at the shared displacement. Two identical springs side by side give $2k$ . Sanity checks: series is always softer than the softest member alone, and parallel always stiffer than the stiffest — the identical-spring results $k/2$ and $2k$ confirm this at a glance. The rules are also formally identical to those for capacitors: series compliances add just as series capacitances inverse-add.

With the full toolkit on the table, here is the one-glance guide to which equation answers which question:

you are asked for…	reach for	equation
force or acceleration at a given stretch	$F = -kx$ , then $a = F/m$	(1)
speed, distance, or "how far before it stops"	energy bookkeeping with $U = \tfrac12 kx^2$	(2)
the stiffness of a non-spring system	$k_{\text{eff}} = U''(x_0)$	(4)
frequency or period of oscillation	$\omega = \sqrt{k/m}$ , $T = 2\pi\sqrt{m/k}$	(5)
several springs at once	collapse to one $k_{\text{eff}}$ first	(6), (7)

Limiting cases

Case	Result	Meaning
Small displacement	$F \approx -kx$ , $U \approx \tfrac12 kx^2$	the Hookean regime — pure linear response, the universal small-oscillation limit
Larger stretch	$F = -kx - \beta x^3 + \cdots$	anharmonic corrections appear; the response stiffens or softens and the period starts to depend on amplitude — isochronism breaks
Past the elastic limit	irreversible	the material yields plastically: it never returns to natural length, energy is dissipated, Hooke's law fails entirely
$k \to \infty$	$x \to 0$ for any finite force	the spring becomes a rigid rod — an ideal constraint; the normal force and rod tension are exactly this limit

Common trap“x is the stretch from wherever it happens to be hanging”

No — $x$ in $F = -kx$ is measured from the spring's natural length, and confusing reference points is the classic Hooke's-law error. Hang a mass $m$ on a vertical spring and gravity stretches it to a new equilibrium at $x_0 = mg/k$ . Measure displacement $y$ from that shifted point and the net force is $-k(x_0 + y) + mg = -ky$ : the constant gravity term cancels exactly, leaving a pure Hooke's law about the new equilibrium with the same $k$ — and the same frequency $\omega = \sqrt{k/m}$ , with no $g$ in it. Gravity relocates home; it does not change the stiffness of the pull toward home. The companion error is a sign slip: $-kx$ is the force the spring exerts, while the force you apply to hold it stretched is $+kx$ — a Newton's-third-law pair, not the same force.

Worked examples

1 · Weighing a spring (confidence builder)

A 2.0 kg mass hangs at rest from a vertical spring, stretching it 4.0 cm. Find the spring constant and the energy stored in the spring. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. At rest, the net force is zero, so the spring's upward pull balances the weight: $kx_0 = mg$ .

Step 2. Solve for the stiffness: $k = mg/x_0 = (2.0)(9.8)/0.040 = \boxed{490\ \mathrm{N/m}}$ .

Step 3. Energy stored at that stretch, from (2): $U = \tfrac12 kx_0^2 = \tfrac12 (490)(0.040)^2 = \boxed{0.39\ \mathrm{J}}$ . This is precisely how a spring scale works — and how you measure $k$ in the lab: hang a known mass, read the static stretch.

2 · Release and fly (exam level)

A 0.50 kg block on a frictionless table is attached to a spring of constant $k = 200\ \mathrm{N/m}$ , pulled 10 cm from natural length, and released from rest. Find (a) the spring force and the block's acceleration at the moment of release, (b) the block's speed as it passes through equilibrium, and (c) the displacement at which kinetic and potential energy are equal.

Step 1. Force at release, from (1): $|F| = kA = 200 \times 0.10 = 20\ \mathrm{N}$ , directed back toward equilibrium.

Step 2. Acceleration at release: $a = F/m = 20/0.50 = 40\ \mathrm{m/s^2}$ — about $4g$ , and this is the largest acceleration anywhere in the motion, since the force shrinks as the block approaches home.

Step 3. Energy bookkeeping for (b): at release, all energy is elastic, $E = \tfrac12 kA^2 = \tfrac12(200)(0.10)^2 = 1.0\ \mathrm{J}$ . At equilibrium the spring is relaxed, so all 1.0 J is kinetic.

Step 4. Solve for the speed: $\tfrac12 mv^2 = 1.0\ \mathrm{J} \Rightarrow v = \sqrt{2(1.0)/0.50} = \boxed{2.0\ \mathrm{m/s}}$ .

Step 5. For (c), set $\tfrac12 kx^2 = \tfrac12 E$ : that gives $x^2 = A^2/2$ , so $x = A/\sqrt{2} \approx \boxed{7.1\ \mathrm{cm}}$ — the energy split is not at half the amplitude, because the potential grows as the square of the displacement. Cross-check Step 4: $v = A\sqrt{k/m} = 0.10\sqrt{400} = 2.0\ \mathrm{m/s}$ ✓ — the same answer written as amplitude × angular frequency, exactly as equation (5) predicts.

3 · Collapse the network first (combining springs)

Two springs of constant $k = 300\ \mathrm{N/m}$ each are joined end to end, and that chain is mounted side by side with a third identical spring. A 0.50 kg block rides on the combination. Find the effective spring constant, the oscillation frequency, and the static sag if the assembly hangs vertically. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Series pair first, from (6): $1/k_s = 1/300 + 1/300 \Rightarrow k_s = 150\ \mathrm{N/m}$ — softer than either member, as series must be.

Step 2. That chain in parallel with the third spring, from (7): $k_{\text{eff}} = 150 + 300 = \boxed{450\ \mathrm{N/m}}$ .

Step 3. Frequency from (5): $\omega = \sqrt{k_{\text{eff}}/m} = \sqrt{450/0.50} = 30\ \mathrm{rad/s}$ , i.e. $f = \omega/2\pi \approx \boxed{4.8\ \mathrm{Hz}}$ , period $T \approx 0.21\ \mathrm{s}$ .

Step 4. Static sag: $x_0 = mg/k_{\text{eff}} = (0.50)(9.8)/450 \approx \boxed{11\ \mathrm{mm}}$ — and by the pitfall above, the block oscillates about this sagged point at the same 4.8 Hz. The order of operations is the whole lesson: reduce the network to one $k_{\text{eff}}$ before touching dynamics.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Royal Society, London

Hooke was the Royal Society's Curator of Experiments and perhaps the most inventive experimentalist of his century: he coined the word "cell" from his microscope observations, surveyed London for its rebuilding after the Great Fire, and quarrelled bitterly with Newton over credit for gravitation. He found his spring law around 1660 but, protective of priority, published it in 1676 as the scrambled anagram ceiiinosssttuv, revealing the solution only in his 1678 lecture De Potentia Restitutiva: "ut tensio, sic vis" — as the extension, so the force. Fittingly for a man who feuded with Newton, no authenticated portrait of him survives.

History

1660

— Robert Hooke discovers that a spring's force is proportional to its extension.

1676

— Hooke stakes his claim as the anagram ceiiinosssttuv, hiding the law in plain sight.

1678

— De Potentia Restitutiva unscrambles it: "ut tensio, sic vis," with experiments on springs, coils, and watch balances.

1807

— Thomas Young introduces the elastic modulus, recasting Hooke's law as a material property independent of a spring's shape.

1821

— Navier and Cauchy build the full tensor theory of elasticity; Hooke's law becomes the linear stress–strain relation of continuum mechanics.

1986

— Binnig, Quate and Gerber's atomic force microscope turns a Hookean cantilever into a piconewton force sensor that images individual atoms.

Going further

Texts. Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 10 · Taylor, Classical Mechanics, Ch. 5 · Morin, Introduction to Classical Mechanics, Ch. 4 · Landau & Lifshitz Vol. 1 §21 for small oscillations · Goldstein, Classical Mechanics, 3rd ed., Ch. 6. All key results can be traced back to first principles: $U = \tfrac12 kx^2$ follows from $F = -\mathrm{d}U/\mathrm{d}x$ , and $\mathrm{d}E/\mathrm{d}t = 0$ follows from the equation of motion; the SHM solution $\omega = \sqrt{k/m}$ , $T = 2\pi\sqrt{m/k}$ is confirmed by direct substitution into $m\ddot x = -kx$ ; the universal-limit result $k_{\text{eff}} = U''(x_0)$ comes from Taylor expansion about the minimum; and the series/parallel rules follow from equal-force and equal-stretch arguments respectively. The worked-example numbers are: $k = 490$ N/m, $U = 0.39$ J; $F = 20$ N, $a = 40\ \mathrm{m/s^2}$ , $v = 2.0$ m/s, $x = A/\sqrt{2} \approx 7.1$ cm; $k_{\text{eff}} = 450$ N/m, $\omega = 30$ rad/s, $f \approx 4.8$ Hz, $x_0 \approx 11$ mm.

Builds on: Newton's second law · Position, displacement, and reference frames · Vectors and coordinate systems · Velocity and acceleration. Leads to: Simple harmonic motion ( $\omega = \sqrt{k/m}$ ) · Conservative forces and potential energy · Small oscillations about equilibrium · Damped and driven oscillators · Normal force.

Does the minus sign in F = −kx mean the force is negative?

No — it means the force opposes the displacement, whichever way that happens to point. The sign of $F$ flips with the sign of $x$ : stretch the spring ( $x > 0$ ) and the force is negative (pulling back toward the origin); compress it ( $x < 0$ ) and the force comes out positive (pushing back toward the origin). The minus sign is a statement about direction, not magnitude: force and displacement always disagree in sign, which is exactly what "restoring" means. If the law were $F = +kx$ the force would run away from equilibrium and the slightest nudge would grow exponentially — the minus sign is the difference between a system that oscillates and one that explodes.

Does the same formula really cover compression, or is that a separate law?

Same formula, same $k$ , no separate law — provided you treat $x$ as a signed quantity measured from natural length. Compression just means $x < 0$ , and $F = -kx$ then automatically points in the positive direction: the squeezed spring pushes outward. One physical caveat: a coil spring can only be compressed until its coils touch (and a rubber band or bungee cord can't be "compressed" at all — it simply goes slack and exerts zero force). Within those geometric limits, an ideal spring is perfectly symmetric: stretch and compression of equal size store equal energy $\tfrac12 kx^2$ and fire back forces of equal magnitude.

If I cut a spring in half, is each half still a spring with the same k?

Each half is a spring — but with double the stiffness, $2k$ , not the same $k$ . The cleanest way to see it: the original spring is its two halves connected in series, so $1/k = 1/k_{\text{half}} + 1/k_{\text{half}}$ , which forces $k_{\text{half}} = 2k$ . Physically, under a given force every coil stretches the same tiny amount; half as many coils means half the total stretch for the same force — i.e. a stiffer spring. The general rule is $k \propto 1/L$ for a given construction (same reason a short wire is stiffer than a long one, $k = EA/L$ ). Cutting a spring is the classic trick question precisely because intuition says "smaller spring, weaker spring" and the physics says the opposite.

Why doesn't gravity change the oscillation frequency of a vertical spring?

Because gravity is a constant force, and constant forces can't change the restoring slope — they only relocate the point of zero net force. Hanging the mass shifts equilibrium down by $x_0 = mg/k$ ; write the equation of motion about that new point and the constant $mg$ cancels the static spring force exactly, leaving $m\ddot y = -ky$ . The frequency is still $\omega = \sqrt{k/m}$ , with no $g$ anywhere in it. A pendulum, by contrast, does care about $g$ — there gravity supplies the restoring force itself rather than a constant offset. Comparing the two cases is a quick litmus test of whether you've understood what sets $\omega$ : the slope of the net force, not the forces' absolute sizes.

Is the spring constant a property of the material, like density?

No — $k$ is a property of the object: its material and its geometry. A steel wire of cross-section $A$ and length $L$ acts as a spring with $k = EA/L$ , where $E$ is Young's modulus — that $E$ (about $200$ GPa for steel) is the true material property. Double the length and $k$ halves; double the cross-section and $k$ doubles; coil the same wire into a helix and $k$ drops by orders of magnitude because the wire now deforms by twisting rather than stretching. That is why "the spring constant of steel" is a meaningless phrase while "the Young's modulus of steel" is tabulated in every handbook: Hooke's law at the object level ( $F = -kx$ ) is the shadow of Hooke's law at the material level (stress $= E \times$ strain).

Why is the stored energy ½kx² and not just force × distance = kx²?

Because the force isn't constant during the stretch — it grows from $0$ at natural length to $kx$ at the end. "Force × distance" only works for constant forces; here you must integrate, $W = \int_0^x kx'\,\mathrm{d}x'$ , which is the area of a triangle under the force–displacement line: $\tfrac12 \times kx \times x = \tfrac12 kx^2$ . Equivalently, the average force during the stretch is $\tfrac12 kx$ (halfway between $0$ and $kx$ , legitimate because the growth is linear), and average force × distance gives the same $\tfrac12 kx^2$ . Forgetting the half is the single most common spring-energy error — if your answer is exactly twice the expected one, this is where to look.

If Hooke's law is "just an approximation," why do we call it a law at all?

Partly history, partly earned respect. It is not fundamental in the sense Newton's laws are — it fails at large stretch and says nothing about why materials resist deformation. But it is the controlled leading-order limit of essentially every stable system: Taylor-expand any smooth potential about a minimum and Hooke's law is what survives, with $k = U''(x_0)$ . That gives it a validity guarantee most "laws" would envy — it is always right, provided the displacement is small enough, and the theory itself tells you the correction terms ( $-\beta x^3 + \cdots$ ) that signal the boundary. Modern physics is full of such "effective laws," and Hooke's is the prototype: not a statement about springs, but a theorem about equilibria.

Real springs have mass — does that break F = −kx and ω = √(k/m)?

The force law survives, but the frequency formula picks up a correction. When the spring itself has mass $m_s$ , its coils near the block move almost as fast as the block while coils near the wall barely move; integrating that velocity profile shows the spring contributes an effective extra mass of $m_s/3$ . The oscillation frequency becomes $\omega = \sqrt{k/(m + m_s/3)}$ . For a lab spring of 60 g driving a 100 g mass, ignoring the correction makes your predicted period about 10% short — easily visible with a stopwatch, and a favourite source of "my experiment disagrees with theory" panic. The ideal-spring formulas assume $m_s \ll m$ ; the $m_s/3$ rule is the first patch when that fails.