Physics I: Mechanics › Oscillations & gravitation › Mass–spring

T = 2π√(m/k)Say it: “the period of a mass on a spring equals two pi times the square root of mass over k”

The mass–spring period

Why a mass on a spring always takes the same time per swing — no matter how far you pull it.

Notation on this page: T is the period (seconds per oscillation), m the oscillating mass, k the spring constant. The companion form is ω = √(k/m).

Before this lesson: Hooke's law, SHM position

Where it comes from

Pull a mass on a spring and let go. Two facts from the last two lessons collide here. Hooke's law says the force is F = −kx, and Newton's second law says F = ma. Put them together:

ma = −kx  ⇒  a = −(k/m) xthe acceleration is minus a constant times x — the SHM fingerprint

But SHM position taught us that simple harmonic motion means a = −ω²x. The two expressions for a must be the same — so ω² = k/m, and since T = 2π/ω:

T = 2π√(m/k)Say it: “two pi root m over k”
Before reading on: you double the hanging mass. Does the period double? And if you swap in a spring twice as stiff, does the period halve? Guess the scaling, then check it against the square root.

Neither scales linearly — the square root tames both. Double m → period grows by √2 ≈ 1.41×. Double k → period shrinks by 1/√2 ≈ 0.71×. Heavier is slower, stiffer is faster, and the square root keeps both changes modest.

Derivation

The whole derivation is one comparison: Newton's second law applied to a Hooke's-law spring must match the SHM acceleration pattern. Watch the last step — it is where the period drops out.

ma
=
−kx
Step 1 — Newton meets Hooke. The only horizontal force on the mass is the spring's, F = −kx. So ma = −kx.
a
=
−(k/m) x
Step 2 — isolate a. Divide by m. The acceleration is minus (k/m) times the displacement.
a
=
−ω²x
Step 3 — the SHM pattern. Simple harmonic motion is defined by a = −ω²x (verified on the SHM position page). Same shape — so the constants must match.
ω
=
√(k/m),   T = 2π/ω = 2π√(m/k)
Step 4 — read off the timing. ω² = k/m, so ω = √(k/m). One full cycle is 2π radians: T = 2π/ω. ∎

Units check (free verification): k/m has units (N/m)/kg = (kg·m/s²)/(m·kg) = 1/s². Square root: 1/s. So ω is per-second ✓ and T = 2π/ω is in seconds ✓.

How to use it

The procedure, every time:

  1. Confirm SHM. A mass on a Hooke's-law spring, small oscillations — then the formula applies.
  2. Compute ω = √(k/m) first — it is the natural intermediate. Then T = 2π/ω or f = ω/(2π).
  3. Solve backwards when needed. Measured T? k = 4π²m/T². Unknown mass? m = kT²/(4π²).
  4. Think in ratios. m ×4 → T ×2. k ×9 → T ÷3. The square root halves every scaling.
  5. Ignore the amplitude. It isn't in the formula — a bigger pull doesn't change the timing (while Hooke's law holds).

The rearranged forms

ω = √(k/m)angular frequency — the form the derivation gives you first
f = (1/2π)√(k/m)ordinary frequency in Hz — oscillations per second
k = 4π²m / T²measure the period, weigh the mass, read off the stiffness
Common mistake: writing T = 2π√(k/m) — k on top. That says stiffer springs oscillate slower, which is backwards (and the units come out 1/s instead of s). Mass belongs on top: inertia slows things down.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: m = 0.50 kg, k = 200 N/m

  1. Angular frequency first. ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s.
  2. Period. T = 2π/ω = 2π/20 = π/10 ≈ 0.314 s.
  3. Frequency. f = 1/T ≈ 3.18 Hz — about three swings per second.
  4. Sanity check. A stiff spring (200 N/m) with a light mass (0.5 kg) should be quick — a third of a second per swing feels right ✓
Common mistake: T = 2π√(k/m) = 2π√400 = 40π ≈ 126 s — a two-minute “period” for a bouncy spring. Absurd answers are the formula telling you k and m are swapped.
Your turn — m = 0.20 kg, k = 80 N/m. Find ω, T, f.

Answer: ω = 20 rad/s, T = 0.314 s, f = 3.18 Hz. k/m = 80/0.20 = 400, so ω = 20 rad/s — the same ratio as Example 1 gives the same timing. T = 2π/20 ≈ 0.314 s, f ≈ 3.18 Hz.

Example 2 — working backwards: T = 0.80 s measured, m = 0.25 kg. Find k.

  1. Rearrange first. T = 2π√(m/k) → T² = 4π²m/k → k = 4π²m/T².
  2. Substitute. k = 4π² × 0.25 / 0.80² = π² / 0.64 ≈ 15.4 N/m.
  3. Sanity check. A slow 0.8 s period with a light mass means a soft spring — 15 N/m is soft ✓
Common mistake: forgetting to square T (writing k = 4π²m/T). Squaring matters: T = 0.8 vs T² = 0.64 changes k by 25%.
Your turn — T = 1.0 s, m = 0.40 kg. Find k.

Answer: ≈ 15.8 N/m. k = 4π² × 0.40/1.0² = 1.6π² ≈ 15.8 N/m. A one-second period is slow, so the spring is soft — consistent.

Example 3 — scaling: the mass quadruples. What happens to T?

  1. Use ratios, not numbers. T ∝ √m, so T′/T = √(m′/m) = √4 = 2.
  2. The period doubles. With the Example 1 numbers: m = 2.0 kg → ω = √(200/2.0) = 10 rad/s → T = 2π/10 ≈ 0.628 s (was 0.314 s) ✓
Common mistake: “four times the mass, four times the period.” The square root halves every scaling — ×4 mass is ×2 period, ×9 stiffness is ÷3 period.
Your turn — k is multiplied by 9 (same mass). What happens to T?

Answer: T shrinks to one third. T ∝ 1/√k, so T′/T = 1/√9 = 1/3. Nine times stiffer, three times faster.

Before reading on: pull the Example 1 mass back 5 cm vs. 10 cm and release. Which swing finishes one period first — or is it a tie?

Example 4 — judgment call: does amplitude matter?

  1. Check the formula. T = 2π√(m/k) — no A anywhere.
  2. It's a tie: both give T ≈ 0.314 s. The 10 cm pull reaches a higher vmax (Aω = 2.0 m/s vs 1.0 m/s) and covers double the distance in the same time.
  3. Why it works: bigger pull → bigger restoring force (F = −kx) → bigger acceleration — the extra distance is exactly compensated by extra speed. (This isochronism is what makes spring clocks possible.)
Common mistake: “farther to travel, so longer period.” True for a car at fixed speed — false here, because the “engine” (restoring force) gets stronger with distance.
Your turn — amplitude is tripled. What happens to T? To vmax?

Answer: T unchanged; vmax triples. T has no A in it. vmax = Aω, so tripling A triples the top speed — the mass covers triple the distance in the same time.

Memorization tips

  • Chant it: “two pi root m over k.” Mass on top (inertia slows), k below (stiffness quickens).
  • The absurdity check: if your T comes out in minutes for a hand-sized spring, k and m are swapped. T = 2π√(k/m) has units of 1/s — instantly wrong.
  • Square-root scalings: ×4 mass → ×2 T; ×4 stiffness → ÷2 T. Halve every scaling in your head.
  • Derive, don't memorize: ma = −kx → a = −(k/m)x → match a = −ω²x → ω = √(k/m) → T = 2π/ω. Four lines, thirty seconds.
  • No A, no problem: amplitude independence is the most-tested fact on this page — “which completes a period first?” is always a tie.
  • k from the clock: k = 4π²m/T² — timing an oscillation is the standard lab route to an unknown stiffness.

Final challenge

Five mixed questions — computations, scalings, and the traps, all in one. Score 5/5 and the mass–spring period is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Does the amplitude affect the period of a mass on a spring?

No — T = 2π√(m/k) has no amplitude in it. Pull the mass twice as far and it moves faster, covering the longer round trip in exactly the same time. (This holds while the spring obeys Hooke's law.)

Does a mass-spring oscillator work hanging vertically?

Yes, with the same period. Gravity shifts the equilibrium point down by mg/k but doesn't change the stiffness, so oscillations about the new equilibrium still have T = 2π√(m/k).

What if the spring itself has mass?

The formula assumes a massless spring. A real spring contributes about one-third of its mass to the moving mass: use m + mspring/3 in the formula for better accuracy.

How are T, f, and ω related for the mass-spring?

ω = √(k/m), f = ω/2π = (1/2π)√(k/m), and T = 1/f = 2π√(m/k). They are three faces of the same timing: angular frequency, cycles per second, seconds per cycle.

Why is m on top and k on the bottom under the square root?

Mass is inertia — it resists changes in motion, so more mass means a slower (longer-period) oscillation. Stiffness k is restoring strength — more k means a snappier return, so a shorter period.

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