Physics I: Mechanics › Oscillations & gravitation › Simple pendulum

T = 2π√(L/g)Say it: “the period of a simple pendulum equals two pi times the square root of length over g”

The simple pendulum

Galileo's great discovery in one line: the swing's timing ignores the bob's mass — only the string's length and local gravity matter.

Notation on this page: L is the pendulum length (pivot to the bob's centre of mass), g the local gravitational acceleration, θ the swing angle from vertical.

Before this lesson: SHM position, Mass–spring

Where it comes from

Displace a pendulum bob by an angle θ and gravity pulls it back — but only the tangential component does the restoring work. Resolving the weight mg along the swing direction:

Ft = −mg sin θthe restoring force along the arc — minus because it opposes the displacement

For small angles, sin θ ≈ θ (in radians). And the bob's position along its arc is s = Lθ, so θ = s/L:

F ≈ −mg(s/L) = −(mg/L) sHooke's law in disguise — with an effective spring constant keff = mg/L

It is a mass on a “spring” of stiffness mg/L. Borrowing T = 2π√(m/k) with k → mg/L, the m cancels:

T = 2π√(L/g)Say it: “two pi root L over g”
Before reading on: two pendulums, same length — one with a 100 g bob, one with a 1 kg bob. Which has the longer period? And what if you double the string length instead?

The bobs tie: mass cancelled out of the formula (Galileo's isochronism discovery, ~1588). Doubling L multiplies T by √2 ≈ 1.41 — longer string, lazier swing.

Derivation

We derive the period from the tangential restoring force, with the small-angle approximation doing the crucial linearising step. The mass cancellation in Step 4 is the famous part.

Ft
=
−mg sin θ
Step 1 — the restoring force. Weight mg points straight down; its component along the swing is mg sin θ, opposing the displacement — hence the minus.
sin θ
≈
θ   (θ in radians, θ ≲ 15°)
Step 2 — small angles. For small θ, sin θ ≈ θ. This is what turns the pendulum into SHM — without it the motion isn't sinusoidal.
F
≈
−mgθ = −(mg/L) s
Step 3 — Hooke's form. Arc position s = Lθ, so θ = s/L. The force is −(mg/L) s — Hooke's law with keff = mg/L.
T
=
2π√(m/keff) = 2π√(m/(mg/L)) = 2π√(L/g)
Step 4 — the mass cancels. Borrow the mass–spring result T = 2π√(m/k): m/m = 1 vanishes. Heavier bob → stronger pull, but proportionally more inertia. ∎

Why must θ be in radians? Both s = Lθ and sin θ ≈ θ are radian-only relationships. In degrees the arc formula needs a π/180 factor and the approximation fails.

How to use it

The procedure, every time:

  1. Measure L to the centre of mass. Pivot to the middle of the bob — not to the top of the bob, not the knot.
  2. Check the angle. Release from θ ≲ 15°. Bigger swings need the full (non-sinusoidal) treatment.
  3. Use local g. 9.8 m/s² on Earth; the local value anywhere else.
  4. Ignore the mass. It cancelled — a cannonball and a marble on equal strings keep identical time.
  5. Solve backwards for g. Measured T and L give g = 4π²L/T² — the classic absolute-gravimetry experiment.

The rearranged forms

L = gT² / (4π²)design a pendulum for a target period — clockmaker's equation
g = 4π²L / T²measure g from a swinging string — no scale needed
Common mistake: using the string length alone for L when the bob is large — or the total drop from the ceiling. L is pivot → centre of mass; a 5 cm bob on a 95 cm string is L = 1.00 m, not 0.95 m.

Worked examples

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

Example 1 — the seconds pendulum: L = 1.00 m, g = 9.8 m/s²

  1. Substitute. T = 2π√(L/g) = 2π√(1.00/9.8).
  2. Evaluate. √(1/9.8) ≈ 0.3194; T ≈ 2π × 0.3194 ≈ 2.01 s.
  3. Why “seconds pendulum”? Half a period is ~1.00 s — each half-swing ticks one second. Clockmakers used exactly this length for centuries.
Common mistake: answering T ≈ 1.00 s — forgetting the full period is there and back. The tick you hear is half a period.
Your turn — L = 0.50 m, g = 9.8 m/s². Find T.

Answer: ≈ 1.42 s. T = 2π√(0.50/9.8) = 2π × 0.2259 ≈ 1.42 s. Half the length does not halve the period — the square root gives 1/√2 ≈ 0.707×.

Example 2 — measuring g: T = 2.01 s, L = 1.00 m

  1. Rearrange first. T = 2π√(L/g) → g = 4π²L/T².
  2. Substitute. g = 4π² × 1.00 / 2.01² = 39.48/4.04 ≈ 9.77 m/s².
  3. Sanity check. Within 0.3% of 9.8 — a string, a stopwatch, and the formula recover gravity ✓
Common mistake: squaring only T's number but forgetting the square applies to the measured value including its error — a 1% timing error becomes a 2% g error. Time many swings and divide.
Your turn — you want T = 1.00 s exactly. What L (g = 9.8)?

Answer: ≈ 0.248 m. L = gT²/(4π²) = 9.8 × 1.00/39.48 ≈ 0.248 m — about 25 cm. (Check Example 3: a 0.25 m pendulum indeed ticks ~1 s.)

Example 3 — the metronome pendulum: L = 0.25 m

  1. Substitute. T = 2π√(0.25/9.8).
  2. Evaluate. √(0.25/9.8) = √0.02551 ≈ 0.1597; T ≈ 2π × 0.1597 ≈ 1.00 s.
  3. Notice. Quarter the length (1.00 m → 0.25 m) halves the period (2.01 s → 1.00 s) — the square root in action ✓
Common mistake: “quarter the length, quarter the period.” Square-root scaling again: ×1/4 length → ×1/2 period.
Your turn — L is quadrupled. What happens to T?

Answer: T doubles. T ∝ √L, so ×4 length → √4 = ×2 period. A 4 m pendulum would swing with T ≈ 4.01 s.

Before reading on: an astronaut takes the 1.00 m pendulum to the Moon (g = 1.62 m/s²). Faster, slower, or the same — and by roughly what factor?

Example 4 — judgment call: the pendulum on the Moon

  1. Substitute local g. T = 2π√(1.00/1.62).
  2. Evaluate. √(1/1.62) ≈ 0.7857; T ≈ 2π × 0.7857 ≈ 4.94 s.
  3. The lesson. Weaker gravity → weaker restoring pull → lazier swing: 4.94/2.01 ≈ 2.46× slower — and √(9.8/1.62) ≈ 2.46 ✓. g sits on the bottom: less g, more T.
Common mistake: putting g on top (“more gravity, more period”). Gravity drives the restoring force here — stronger g means a snappier return and a shorter period.
Your turn — the 1.00 m pendulum on Mars (g = 3.71 m/s²). T?

Answer: ≈ 3.26 s. T = 2π√(1.00/3.71) = 2π × 0.5192 ≈ 3.26 s — between Earth's 2.01 s and the Moon's 4.94 s, as Mars's gravity is between the two.

Memorization tips

  • Chant it: “two pi root L over g.” Length on top (longer = lazier), g below (stronger gravity = snappier).
  • The mass eraser: mass cancels — the single most-tested fact. If a question gives you the bob's mass, it's decoration (or a trap).
  • Seconds pendulum anchor: 1.00 m ↔ 2.01 s on Earth. Halve the period? Quarter the length. Double it? Quadruple the length.
  • Small-angle gate: θ ≲ 15°. Past that, sin θ ≠ θ and the formula drifts — the motion stops being SHM.
  • Derive via the fake spring: F ≈ −(mg/L)s, so keff = mg/L; T = 2π√(m/keff) = 2π√(L/g). Thirty seconds, no memorisation needed.
  • Radians or ruin: s = Lθ and sin θ ≈ θ both demand radians.

Final challenge

Five mixed questions — computations, the mass trap, and the small-angle limit, all in one. Score 5/5 and the pendulum 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

Why doesn't the pendulum's mass affect its period?

A heavier bob feels a larger restoring force (mg sin θ) but has proportionally more inertia — the mass cancels in a = F/m. Galileo discovered this: mass drops out of the period entirely.

What counts as a ‘small angle’ for the pendulum formula?

Below about 15° the sin θ ≈ θ approximation is good to ~1%. At 60° the true period is about 7% longer than the formula predicts — fine for clocks, not for precision work.

Is L the string length or something else?

L is measured from the pivot to the bob's centre of mass. For a small dense bob on a light string, that's the string length plus the bob's radius.

Can I use the pendulum formula on the Moon?

Yes — just use the local g. On the Moon (g = 1.62 m/s²) a 1.00 m pendulum has T = 2π√(1/1.62) = 4.94 s, nearly 2.5 times slower than on Earth.

Why does a longer pendulum swing more slowly?

The restoring acceleration is a = −(g/L)s: a longer L weakens the restoring pull per unit displacement, so the bob accelerates more lazily — exactly like a softer spring.

More from the codex

Live

Physics I formula sheet

All the mechanics formulas — kinematics, forces, energy — printable and quiz-ready.

Open sheet → Live

Formula Sheet Builder

Mix and match any sections into your own printable sheet.

Open tool → Live

Prompt Simulator

Practice prompt engineering with deterministic scoring.

Open tool →