Spring forces

An ideal spring has a length it sits at when nothing pulls on it: the relaxed length. Stretch it past that length and it pulls back. Compress it shorter and it pushes out. Either way, the spring force on whatever's attached points back toward the relaxed configuration.

The size depends on how far the spring is from its relaxed length, not on which way. Pull the end out by $0.10$ m, you get the same magnitude as pushing it in by $0.10$ m. The direction flips; the size doesn't.

The magnitude of the spring force is proportional to how far the spring is from its relaxed length:

$$|\vec F_s| = k\,|x|.$$

The constant $k$ is the spring constant, a property of the spring itself, with units of $\text{N/m}$. A spring with $k = 200$ N/m exerts $20$ N when stretched by $0.10$ m, $40$ N when stretched by $0.20$ m, and so on. Double the stretch, double the force.

Watch the units. $k$ is in $\text{N/m}$, not in newtons: it's force per unit displacement, not the force itself. A spring with $k = 250$ stretched by zero exerts zero, not $250$ N.

Force depends on $k$ and $|x|$. A stiff spring barely stretched can match a soft spring stretched far. With $k_A = 50$ N/m and $x_A = 0.20$ m, the force is $10$ N. With $k_B = 200$ N/m and $x_B = 0.05$ m, also $10$ N. Same product, same force.

The spring force on a block always points back toward the relaxed length. Stretch the spring and it pulls the block in. Compress it and it pushes the block out. Direction is set by where the block is, not by which way it's moving.

That last part is the trap. A block oscillating on a spring spends half its time moving one way and half the other, but the spring force does not track velocity. It always points from the block toward equilibrium. Take a block at $x = -0.08$ m moving in the $-x$ direction (away from equilibrium): the spring force points in $+x$, opposite the motion. A moment later, after the block has slowed, stopped, and reversed, the same block at the same position $x = -0.08$ m feels exactly the same spring force in $+x$, now parallel to the motion. Same position, same force, very different velocity.

Compactly: $F_s = -k x$. The minus sign is a direction marker, like the minus in $v = -3$ m/s. It means "toward equilibrium," not "no force" or "small force."

A block of mass $m$ hangs at rest from a vertical spring attached to a ceiling. The spring is stretched by $x_{eq}$ from its relaxed length. Find the spring constant.

Step 1: forces. Two act on the block: gravity ($mg$, down) and the spring force ($k x_{eq}$, up). The block is at rest, so they balance.

Step 2: equilibrium equation, with up positive:

$$k\, x_{eq} - m g = 0.$$

Step 3: solve.

$$k = \dfrac{m g}{x_{eq}}.$$

This is how spring constants are measured in lab. Hang a known mass, measure the stretch with a ruler, divide.

Unit check: $\dfrac{m g}{x_{eq}}$ has units $\dfrac{\text{kg} \cdot \text{m/s}^2}{\text{m}} = \dfrac{\text{N}}{\text{m}}$, correct for $k$.

Pull a block on a spring out and let it go. It accelerates back toward equilibrium, picks up speed, overshoots, slows down on the far side, stops, then reverses. Each "stop" is a turning point. At each one, the block's speed is, for one instant, zero.

The trap: zero speed is not zero force, and not zero acceleration. At a turning point, the block is at the maximum displacement of the oscillation, $|x| = x_{max}$. The spring force is at its largest, $|\vec F_s| = k\, x_{max}$. By Newton's second law, so is the acceleration:

$$|\vec a| = \dfrac{k\, x_{max}}{m}.$$

Why this is easy to miss: in everyday life, motion and force feel coupled. You push something, it moves; you stop pushing, it stops. So when a block stops, the reflex is to assume nothing's pushing it. At a spring's turning point, the opposite is true. The spring is pushing the block harder than at any other instant in the cycle. The block stopped only because it was decelerating the whole way out and ran out of speed exactly when the spring force has built up to its peak.

A block of mass $m$ is attached to a horizontal spring with spring constant $k$, on a frictionless surface. A student pulls the block out by $d$ from the relaxed length, holds it there, and releases it from rest. Find the block's acceleration at the instant of release.

Step 1: net force at $x = d$. The vertical forces (gravity, normal) balance and don't affect horizontal motion. The only horizontal force is the spring, with magnitude $|\vec F_s| = k\, d$, pointing back toward equilibrium.

Step 2: Newton's second law.

$$|\vec a| = \dfrac{|\vec F_{net}|}{m} = \dfrac{k\, d}{m}.$$

The block has speed zero (just released from rest) and acceleration $kd/m$ toward equilibrium. Zero velocity, nonzero acceleration, same instant. That's the turning-point situation from §5.

If you stopped at $|\vec F_s| = kd$ and called that the acceleration, you'd be reporting newtons (a force) instead of m/s$^2$ (an acceleration). $F = ma$ says: to recover acceleration, divide force by mass.

Three failure modes show up over and over on spring problems. Each has a family of distractors built around it on the AP exam.