Simple Pendulum Period

Equivalent forms

\omega = \sqrt{g/L}

f = \frac{1}{2\pi}\sqrt{g/L}

Mass-independent timing — the heart of every mechanical clock for 300 years.

Symbol	Name	SI unit	Dimension	Range
$T$	Periodoutput Time for one complete swing back and forth	s	T	0 – 20
$L$	Pendulum Length Distance from pivot to center of bob	m	L	0.05 – 5
$g$	Gravitational Acceleration Local gravitational acceleration	m/s²	LT^-2	0.1 – 25

Unit systems

Symbol	SI	CGS	Imperial
$T$	s	s	s
$L$	m	cm	ft
$g$	m/s²	cm/s²	ft/s²

Where it holds

Small-angle approximation

(\theta

<

\sim 15^{\circ})

, rigid massless rod, point bob. For larger swings or extended bodies, use the physical-pendulum or elliptic-integral formula.

Dimensional analysis

T \to [T]

\sqrt{L/g} \to \sqrt{[L]/[LT^{-2}]} = \surd [T^{2}] = [T]

Both sides are seconds.

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Discovery

Christiaan Huygens · 1656

Galileo noted pendulum isochronism in 1602; Huygens derived T = 2π√(L/g) and built the first pendulum clock, revolutionizing timekeeping accuracy from minutes to seconds per day.

Try this

How long should a grandfather-clock pendulum be to tick exactly once per second?

Solve T = 2π√(L/g) for L given T = 2 s (one full swing per 2 seconds = 1 tick per second). Use g = 9.8 m/s².

Research status: stable

Real-world applications

Pendulum clocks (1656-1930s, dominant timekeeping)
Foucault pendulum demonstrations of Earth's rotation
Seismographs (long-period pendulum-based)
Local g measurements via gravimetry

Common misconceptions

Period does NOT depend on bob mass — only length and g
Amplitude affects period weakly (only $\sim 0.4$ % at $15^{\circ})$
Foucault's pendulum is a 3D effect — the precession proves Earth rotates

Experimental verification

Huygens's clock achieved

\pm 10\,\mathrm{s/day}

in 1656. Modern timing of seconds pendulums

(L = 0.994\,\mathrm{m}

on Earth) matches

T = 2\,\mathrm{s}

within microseconds.

Derivation

Torque about pivot:

\tau = -mgL \sin \theta

.

For small

\theta

:

\sin \theta \approx \theta

, so

I\cdot d^{2}\theta /dt^{2} \approx -mgL\theta

with

I = mL^{2}

.

This gives

d^{2}\theta /dt^{2} = -(g/L)\theta \to SHM

with

\omega = \sqrt{g/L}

, hence

T = 2\pi /\omega = 2\pi \sqrt{L/g}

.

Limiting cases

L \to 0

⟶

T \to 0

Tiny pendulums oscillate infinitely fast — but real pendulums fail below a few millimeters.

L \to \infty

⟶

T \to \infty

Foucault's 67 m pendulum has

T \approx 16.4\,\mathrm{s}

.

\theta _\max

>

15^{\circ}

⟶

T \to T\cdot (1 + \theta ^{2}/16

+ ...)Small-angle approximation breaks down; period grows slightly with amplitude.

What if…

What if L quadruples?

Period doubles $(T \propto \sqrt{L})$ . A 4 m pendulum has $T = 2\,\mathrm{s} \times \sqrt{4} = 4\,\mathrm{s}$ .

What if you take the clock to the Moon?

g_Moon $\approx 1.62$ . T_Moon $= T$ _Earth $\times \sqrt{9.8/1.62} \approx 2.46\times$ longer — the clock runs slow.

What if amplitude is

60^{\circ}

?

Small-angle T under-predicts by about 7%. Use elliptic integral or the series $T \approx 2\pi \sqrt{L/g}\cdot (1 + \theta ^{2}/16$ + ...).

1

Grandfather clock pendulum

Given ·

T:: 2
g:: 9.8

Find · L

Steps

Rearrange: $L = g(T/2\pi )^{2}$
$L = 9.8 \times (2 / 6.2832)^{2} = 9.8 \times 0.1013$
$L \approx 0.994\,\mathrm{m}$ — used since the 17th century as the 'seconds pendulum' standard

Result ·

L \approx 0.994\,\mathrm{m}

— the classic 'seconds pendulum'

2

Foucault's pendulum (Paris Pantheon)

Given ·

L:: 67
g:: 9.8

Find · T

Steps

$T = 2\pi \times \sqrt{67 / 9.8} = 2\pi \times \sqrt{6.84}$
$T = 2\pi \times 2.614$
$T \approx 16.42\,\mathrm{s}$ — slow majestic swings reveal Earth's spin

Result ·

T \approx 16.42\,\mathrm{s}

simple harmonic motion periodHooke's Law→Gravitational Potential Energy→