The simple pendulum: what sets its period

Simple pendulum

Does a heavier bob or a wider swing make a pendulum slower? Only the length does, nearly.

Change the length, the swing angle and the mass of the bob. For small swings T = 2π√(L/g): quarter the length and the period halves, while the mass does nothing. Swing wider and the true period creeps above the formula.

High school
Restoring force mg sin θ
Small-angle approximation
T = 2π√(L/g)
Seconds pendulum

L

1.50

\theta_0\,(^\circ)

small angle

m

0.5

t

0.00

T = 2\pi\sqrt{\dfrac{L}{g}}

$T_0 = 2\pi\sqrt{L/g}$: 2.46
$T_{\text{true}}$: 2.49
$T_{\text{true}} / T_0$: 1.012
$\theta\,(^\circ)$: 25.0
$F = mg\sin\theta$: 2.07

Formulaunderestimates

Guided tour

1A one-metre pendulum
A $1\,\text{m}$ pendulum swung through $10^\circ$ : $T=2\pi\sqrt{L/g}\approx2.01\,\text{s}$ . Press play and count: one full swing out and back takes about two seconds.
2A quarter of the length
Shorten the string to $0.25\,\text{m}$ . The period halves to $1.00\,\text{s}$ : $T$ grows with $\sqrt L$ , so a quarter of the length gives half the period.
3A wide swing runs slow
Back to $1\,\text{m}$ , now swung through $60^\circ$ . The formula still says $2.01\,\text{s}$ , but the true period is about $2.15\,\text{s}$ , roughly $7\%$ longer. $T=2\pi\sqrt{L/g}$ rests on $\sin\theta\approx\theta$ , which only holds for small angles.
4The seconds pendulum
Set $L=0.99\,\text{m}$ , a small swing and a bob four times heavier. The period is $2.00\,\text{s}$ , one second each way, and the mass changed nothing: a heavier bob feels more pull but is harder to move by the same factor. Clockmakers used exactly this length.

Explanation

1
Where the restoring force comes from
Split gravity along the string and along the arc. The string cancels the first part; the second, $mg\sin\theta$ , pulls the bob back toward the bottom. Along the arc the displacement is $s=L\theta$ , so
$F=-mg\sin\theta\approx-mg\theta=-\frac{mg}{L}\,s$
For small $\theta$ this has the form $F=-ks$ with $k=mg/L$ : the pendulum moves like a mass on a spring.
2
The period formula
Put $k=mg/L$ into the spring formula $T=2\pi\sqrt{m/k}$ :
$T=2\pi\sqrt{\frac Lg}$
The mass cancels and the amplitude never appeared. Measure $T$ and $L$ and you can work out $g$ , which is how $g$ was measured for two centuries.
3
When the angle is not small
Since $\sin\theta<\theta$ , the real restoring force is weaker than the approximation, so a wide swing takes longer. The correction starts as
$T\approx2\pi\sqrt{\frac Lg}\left(1+\frac{\theta_0^2}{16}+\cdots\right)$
Below about $10^\circ$ the error is under $0.2\%$ , which is why textbooks ask for small swings.

Check yourself

Q01

A pendulum has a period of $2\,\text{s}$ . Its length is made four times longer. The new period is

Q02

The same pendulum is taken to the Moon, where $g$ is about $\tfrac16$ of Earth's. Its period

Q03

Which change makes a pendulum swing more slowly (longer period)?

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