Spring-mass system — horizontal, vertical: Visual walkthrough

We will build, in order:

What "displacement" even means, and where zero is.
Why the spring pushes back (the restoring force).
How that force becomes a rule about acceleration (Newton).
Why that rule forces a wiggle (the shape of the motion).
Where the number $ω$ hides inside the rule.
Turning $ω$ into a period $T$ .
Hanging the spring: does gravity break everything? (No — one more picture shows why.)
The edge cases: zero displacement, heavy mass, stiff spring.

Step 1 — Set the stage and name "zero"

WHAT. A block of mass $m$ sits on a frictionless table, tied to a spring. When nothing pushes it, the spring is at its natural length — not stretched, not squashed. We plant a ruler there and call that point $x = 0$ .

WHY. Every measurement needs a home base. We measure how far the block is from home and call that number $x$ (in metres). Positive $x$ means "pushed right," negative $x$ means "pushed left." Without fixing zero, "displacement" is a word with no number.

PICTURE. The lavender block sits on the mint line. The ruler's zero (the coral tick) is exactly where the relaxed spring ends.

Step 2 — The spring always points you home ( $F = - k x$ )

WHAT. Pull the block right ( $x > 0$ ): the spring stretches and tugs it left. Push it left ( $x < 0$ ): the spring squashes and shoves it right. Either way the force points toward $x = 0$ . This is Hooke's Law: $F = opposite way - stiffness k how far x$

WHY. Read the equation as a sentence. $k$ (newtons per metre) says "how many newtons of push per metre of stretch" — a stiff spring has big $k$ . The $x$ says the force grows the farther you go. The minus sign is the whole personality: force and displacement always point opposite ways, so the block is forever herded back to zero.

PICTURE. Two snapshots. On the right, red spring force arrow points left (back home). On the left, red arrow points right (back home). The force arrow never points away from zero.

Step 3 — Turn force into a rule about acceleration (Newton)

WHAT. Newton's second law says force equals mass times acceleration. Acceleration is how fast the velocity changes; we write it $\overset{x}{¨}$ (the two dots mean "rate of change, twice" — velocity is the first change, acceleration the second). $inertia m acceleration \overset{x}{¨} = spring force - k x$

WHY. We have a force (Step 2) but we want to know how the block moves. Newton's law is the bridge from "what force acts" to "how motion changes." Why the double-dot notation? Because the spring force depends on position $x$ , but Newton controls acceleration — the same $x$ appears on both sides, and that self-reference is exactly what makes an oscillation.

PICTURE. Left box: the spring force $- k x$ (coral). Arrow through Newton's law. Right box: the acceleration $\overset{x}{¨}$ (lavender), pointing the same way as the force but scaled by $1/ m$ .

Divide by $m$ and tidy up: $\overset{x}{¨} + \frac{k}{m} x = 0$

$\overset{x}{¨}$ — acceleration.
$\frac{k}{m}$ — a single positive number combining stiffness and inertia. Big when spring is stiff, small when mass is heavy. Remember this bundle — it is the star of Step 5.

Step 4 — Why this rule forces a wiggle

WHAT. The rule $\overset{x}{¨} = - \frac{k}{m} x$ says: acceleration is always the flip of position. At the far right, position is big-positive, so acceleration is big-negative (yanked hard left). The block slows, stops, reverses, races through the middle (where force is zero but speed is greatest), overshoots to the left, gets yanked back — forever. Push-back plus inertia's overshoot = endless oscillation.

WHY. We need to see that this equation cannot describe anything but a smooth back-and-forth. The clue: the function whose second rate-of-change is its own negative is the cosine. So we guess a cosine wave and check it fits.

PICTURE. A cosine curve of $x$ against time. At the crests (far from home) the little acceleration arrows point steeply back to zero; at the centre crossings the arrows vanish and the speed arrows are longest.

Step 5 — Where the number $ω$ hides

WHAT. Write the guess $x (t) = A cos (ω t + ϕ)$ and feed it back into the rule.

$A$ — amplitude (metres): how far the biggest swing reaches.
$ω$ — angular frequency (radians/second): how fast the cosine cycles.
$ϕ$ — phase: where in the cycle we started (which crest at $t = 0$ ).

Differentiating a cosine twice brings down two factors of $ω$ and flips the sign: $\overset{x}{¨} = - ω^{2} A cos (ω t + ϕ) = - ω^{2} x$

WHY. We compare this to our physics rule $\overset{x}{¨} = - \frac{k}{m} x$ . Two expressions for the same $\overset{x}{¨}$ must match term for term: $from the wiggle ω^{2} = from the physics \frac{k}{m}$

So $ω$ is not a free choice — the spring and mass dictate it: $ω = \frac{k}{m}$

PICTURE. The bundle $k / m$ (from the equation) flowing into the wiggle-rate $ω^{2}$ of the cosine — a labelled matching of the two sides.

Step 6 — From $ω$ to the period $T$

WHAT. A full cosine cycle takes an angle of $2 π$ radians. Since $ω$ is radians per second, the time for one round trip is $T = \frac{2 π}{ω} = 2 π \frac{m}{k}$

WHY. $ω$ answers "how many radians per second"; the period $T$ answers "how many seconds per full cycle." They are reciprocals scaled by the $2 π$ radians in one turn. That is why we divide $2 π$ by $ω$ .

PICTURE. One full cosine cycle marked off; the horizontal span from crest to next crest is labelled $T$ , and beside it the algebra $T = 2 π / ω = 2 π m / k$ .

Step 7 — Hang it up: does gravity break the picture?

WHAT. Now the spring hangs and gravity pulls the mass down with force $m g$ . Let $y$ point down from the natural length. Gravity first stretches the spring to a resting point $y_{0}$ where the pulls balance: $k y_{0} = m g \Rightarrow y_{0} = \frac{m g}{k}$

$g$ — gravity's pull per kilogram ( $\approx 9.8 m/s^{2}$ ).
$y_{0}$ — static extension: how far gravity alone stretches the spring.

Displace to a general $y$ ; the net downward force is spring-up minus... let's do it carefully with the shifted coordinate $x = y - y_{0}$ (distance from the new home): $m \overset{x}{¨} = gravity down m g - spring up k (x + y_{0}) = = 0 by balance (m g - k y_{0}) - k x = - k x$

WHY. We ask this because gravity looks like it should slow the bounce. The algebra shows the constant $m g$ is entirely swallowed by the constant $k y_{0}$ — they cancel. What survives is the same $- k x$ from Step 2. Gravity moves the home base down by $y_{0}$ ; it never touches the restoring term. So $T = 2 π m / k$ unchanged.

PICTURE. Left: horizontal spring with home at natural length. Right: hanging spring with home dropped by $y_{0}$ (butter arrow), but the same coral restoring arrow about the new home. The cancellation $m g - k y_{0} = 0$ is shown.

Step 8 — The edge cases (never leave a stone unturned)

WHAT & WHY & PICTURE, all three limits at once:

Zero displacement ( $x = 0$ ). Force $= - k \cdot 0 = 0$ : no push. If also at rest, it stays — a legal (boring) solution, amplitude $A = 0$ . If moving through, this is the point of maximum speed (Step 4's centre crossing). Nothing breaks.
Heavier mass ( $m \to$ large). $T = 2 π m / k$ grows: slower and slower. Inertia resists the spring. In the limit of huge $m$ , the bounce becomes glacial.
Stiffer spring ( $k \to$ large). $T$ shrinks: faster and faster. In the limit $k \to \infty$ the spring is rigid, $T \to 0$ — no oscillation, it just holds.
Zero stiffness ( $k \to 0$ ) or free mass. $T \to \infty$ : no restoring force, no oscillation. The block, once pushed, drifts forever. This is the boundary where SHM stops existing — the picture degenerates into straight-line drift.

The one-picture summary

Everything on one canvas: the block and its restoring force → Newton's rule → the cosine it forces → the matched $ω$ → the period → and gravity's harmless shift.

Recall Feynman retelling of the whole walkthrough

Put a block on a smooth table tied to a spring and mark where it rests — call that home. Pull it away and the spring, being polite but firm, always tugs it back toward home, and the farther you pull the harder it tugs (that's $F = - k x$ ). Newton says a force makes the block speed up in that direction, but the block has weight-of-motion (inertia), so it overshoots home, stretches the other side, gets tugged back again — a never-ending back-and-forth. That back-and-forth is exactly a cosine wave, and the only thing that sets its rhythm is the tug-per-metre ( $k$ ) divided by the laziness of the mass ( $m$ ): stiff and light means fast, floppy and heavy means slow. Turn that rhythm into seconds-per-bounce and you get $T = 2 π m / k$ . Finally hang the whole thing from the ceiling: gravity drags home base downward a little, but once you find that new home, the spring behaves exactly as before — the bounce is the same. Gravity moves the house; it doesn't change the heartbeat.

Recall Quick self-check

Which quantity, increased, makes the bounce slower? ::: The mass $m$ — it sits under the root in $T = 2 π m / k$ . Where does the block move fastest? ::: At equilibrium $x = 0$ , where the spring force is zero and all energy is kinetic. Why doesn't gravity change the vertical period? ::: It's constant, so $m g - k y_{0} = 0$ cancels it; only the restoring $- k x$ about the new home survives. What is $ω$ in terms of $k$ and $m$ ? ::: $ω = k / m$ .

Connections

Spring-mass System — horizontal, vertical — the parent result this page derives visually.
Simple Harmonic Motion — the cosine motion of Step 4.
Hooke's Law — the $F = - k x$ of Step 2.
Energy in SHM — the fast-at-centre idea in Steps 4 and 8.
Springs in Series and Parallel — changes $k$ , hence Step 8's limits.
Simple Pendulum — contrast: its period does depend on $g$ .
Damped Oscillations — what Step 4's endless wiggle becomes with friction.