A pendulum has no spring, yet it swings in nearly perfect simple harmonic motion. Gravity, the same pull that makes things fall, provides the restoring force.
Why it works
Pull the bob aside by an angle $\theta$. Gravity $mg$ splits into a part along the string (balanced by the tension) and a part along the arc, $mg\sin\theta$, back toward the bottom. For small angles, $\sin\theta \approx \theta = \dfrac{x}{L}$, where $x$ is the distance along the arc. So the restoring force is about $$F = -\frac{mg}{L}x,$$ the same form as a spring with "$k$" $= \dfrac{mg}{L}$. Put that into $T = 2\pi\sqrt{m/k}$: $$T = 2\pi\sqrt{\frac{m}{mg/L}} = 2\pi\sqrt{\frac{L}{g}}.$$ The mass cancels: a heavier bob is pulled harder but is harder to speed up, in the same proportion.
Key idea
For small swings, a simple pendulum of length $L$ (pivot to the center of the bob) has period $T = 2\pi\sqrt{\dfrac{L}{g}}$. It depends on the length and on $g$ only, not on the mass or (for small angles) the amplitude. Four times the length doubles the period.
Worked examples
Worked example: Period of a pendulum
Find the period of a $1.2\ \text{m}$ pendulum swinging through small angles.
- $T = 2\pi\sqrt{\dfrac{1.2\ \text{m}}{9.8\ \text{m/s}^2}} = 2.2\ \text{s}$.
Answer: $2.2\ \text{s}$
Worked example: A seconds pendulum
A clock pendulum should take exactly $1.0\ \text{s}$ for each swing one way, so its period is $2.0\ \text{s}$. How long must it be?
- Square: $T^2 = 4\pi^2\dfrac{L}{g}$, so $L = \dfrac{gT^2}{4\pi^2}$.
- $L = \dfrac{(9.8)(2.0)^2}{4\pi^2} = 0.99\ \text{m}$: about a meter.
Answer: $0.99\ \text{m}$
Worked example: On the Moon
The same $0.99\ \text{m}$ pendulum is taken to the Moon, where $g = 1.62\ \text{m/s}^2$. Find its period.
- $T = 2\pi\sqrt{\dfrac{0.99}{1.62}} = 4.9\ \text{s}$. Weaker gravity pulls the bob back more gently, so it swings more slowly.
Answer: $4.9\ \text{s}$
Common mistake
Thinking a heavier bob swings more slowly. A heavier bob does feel a bigger restoring pull, but it has more inertia in exactly the same proportion, so its motion is unchanged: there is no $m$ in $T = 2\pi\sqrt{L/g}$. To slow a pendulum, make it longer. Check: in the visual above, the heavy and light bobs stay in step.
"Small" swings means up to about $15^\circ$: there the formula is off by less than half a percent. Big swings take a little longer, because $\sin\theta$ grows more slowly than $\theta$. You can see this in the Pendulums and Resonance lab of the Science Lab.