Springs and simple harmonic motion

TL;DR:

• Hooke's law: a spring pulls back with a force F = −kx, proportional to how far it is stretched.

• A mass on a spring swings with period T = 2π√(m/k). The amplitude does not change the period.

• Energy swaps between spring energy ½kx² and kinetic energy ½mv², and the total stays constant.

Pull a mass hanging from a spring down a few centimetres and let go. It bobs up and down again and again, taking the same time for each trip. That steady rhythm is called simple harmonic motion, and it turns up in clocks, guitar strings, car suspensions and the atoms in a solid.

Hooke's law: the pull gets stronger as you stretch

For most springs, within a limited range, the restoring force is proportional to the stretch: F = −kx. Here x is the displacement from the relaxed position, and k is the spring constant in newtons per metre. A stiffer spring has a larger k.

The minus sign says the force points back toward the equilibrium. Stretch the spring and it pulls in. Squash it and it pushes out. That always-pointing-home behaviour is what makes the motion repeat.

Why the motion repeats with a fixed period

Newton's second law gives ma = −kx, so the acceleration is proportional to the displacement and opposite to it. Any system obeying that rule oscillates, and solving it gives a sine wave with an angular frequency ω = √(k/m).

The period is the time for one full cycle: T = 2π/ω = 2π√(m/k). A larger mass makes it slower, because it is harder to accelerate. A stiffer spring makes it faster. The amplitude does not appear in the formula, so a big swing and a small swing take the same time.

Worked example: a 0.50 kg mass on a 200 N/m spring

A 0.50 kg block sits on a smooth horizontal surface, attached to a spring with k = 200 N/m. You pull it 5.0 cm (0.050 m) and release it. Find the period, frequency, maximum force, maximum speed and total energy.

1. Period: T = 2π√(m/k) = 2π√(0.50/200) = 2π√0.0025 = 2π × 0.05 = 0.314 s.

2. Frequency: f = 1/T = 1/0.314 ≈ 3.18 Hz.

3. Maximum force, at the ends: F = kA = 200 × 0.050 = 10 N.

4. Total energy, all stored in the spring at the ends: E = ½kA² = ½ × 200 × 0.0025 = 0.25 J.

5. Maximum speed, in the middle where all energy is kinetic: ½mv² = 0.25, so v² = 0.5 / 0.5 = 1, and v = 1.0 m/s.

Check the speed a second way. With ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s, the maximum speed is ωA = 20 × 0.050 = 1.0 m/s. Both routes agree. Also T = 2π/ω = 2π/20 = 0.314 s, matching step 1.

Energy trading back and forth

At each end of the swing the block stops for an instant, so all 0.25 J is stored in the spring. In the middle the spring is relaxed and all 0.25 J is kinetic. In between the energy is split, but the sum ½kx² + ½mv² never changes, because the only force doing work is the conservative spring force.

Halfway out at x = 0.025 m, the spring holds ½ × 200 × 0.025² = 0.0625 J. The kinetic energy is the rest, 0.25 − 0.0625 = 0.1875 J, giving v = √(2 × 0.1875 / 0.50) = √0.75 ≈ 0.87 m/s.

What if the spring hangs vertically?

Gravity only shifts the resting position. The 0.50 kg mass stretches the spring by mg/k = 0.50 × 9.8 / 200 = 0.0245 m at rest. Oscillations around that new equilibrium have the same period as before, 0.314 s. Gravity does not change T.

Springs can fly: A spring pulled past its limit can snap or slip off and whip back. Wear eye protection for demonstrations, never stretch a spring toward your face, and use a mass and a spring that match so nothing is overloaded.

How to check your answer

• Units. √(kg ÷ (N/m)) = √(kg·m/N) = √(s²), since N = kg·m/s². That is seconds.

• Direction. More mass should lengthen the period. A stiffer spring should shorten it.

• Sanity. Quadrupling the mass doubles the period. Here m = 2.0 kg gives T = 0.628 s, twice 0.314.

• Energy. Total energy at the ends must equal total energy at the middle.

Mistakes that cost marks

• Using the amplitude in the period formula. It does not belong there.

• Forgetting to convert centimetres to metres before using k in N/m.

• Mixing up k (spring stiffness) with ω or with a generic constant.

• Using T = 2π√(L/g) for a spring. That formula is for pendulums.

• Forgetting that the speed is zero at the ends and largest in the middle.

Where you will use it next

The same maths describes a pendulum at small angles, sound vibrations and the vibrations behind how waves carry energy. To see how a constant force sets a motion, compare why everything falls at the same rate.

Frequently asked questions

Why does the oscillation eventually die out?

Friction and air drag take energy out each cycle, so the amplitude shrinks. The ideal formula assumes no losses.

Does Hooke's law always hold?

No. Stretch a spring far enough and it deforms or breaks, and the force is no longer proportional to the stretch.