Pendulum Period: T = 2π√(L/g)

For a simple pendulum oscillating at small angles, the period depends only on the length $L$ and gravity $g$ — not on the mass of the bob, and not on the amplitude (as long as it's small):

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

This astonishing result was discovered by Galileo around 1582, watching a chandelier swing in the Pisa Cathedral. He timed it against his pulse and noticed the period stayed the same as the chandelier's swings died down. Eventually it led to the pendulum clock — the most accurate timekeeping device for the next 300 years.

Step-by-Step Solution

Given: Pendulum length $L$, gravity $g = 9.81\;\text{m/s}^{2}$, small-angle approximation.

Find: The period for $L = 1\;\text{m}$ and $L = 4\;\text{m}$, and demonstrate the $\sqrt{L}$ scaling.

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Step 1 — Apply the formula at $L = 1\;\text{m}$.

$T_1 = 2\pi\sqrt{\dfrac{1}{9.81}} = 2\pi \times 0.3193 \approx 2.006\;\text{s}$

A 1-meter pendulum has a period of about 2 seconds — close enough to be the basis of "seconds pendulums" used in old clocks. The exact length for a 2-second period is $L = g/\pi^{2} \approx 0.9936\;\text{m}$.

Step 2 — Apply the formula at $L = 4\;\text{m}$.

$T_2 = 2\pi\sqrt{\dfrac{4}{9.81}} = 2\pi \times 0.6386 \approx 4.013\;\text{s}$

The 4-meter pendulum has a period twice as long as the 1-meter pendulum — quadrupling the length doubles the period.

Step 3 — Verify the $\sqrt{L}$ scaling.

$\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{L_2/g}}{2\pi\sqrt{L_1/g}} = \sqrt{\dfrac{L_2}{L_1}} = \sqrt{4} = 2 \;\;\checkmark$

Step 4 — Length needed for a 1-second period.

Solve $T = 2\pi\sqrt{L/g}$ for $L$:

$L = \dfrac{g\,T^{2}}{4\pi^{2}}$

For $T = 1\;\text{s}$:

$L = \dfrac{9.81 \times 1}{4\pi^{2}} = \dfrac{9.81}{39.48} \approx 0.2484\;\text{m}$

A roughly 25-cm pendulum has a 1-second period.

Step 5 — What does NOT affect the period?

• Mass of the bob. A bowling ball and a marble on equal-length strings swing at the exact same period.
• Amplitude (small angles). Releasing from 5° vs 10° gives the same period — this is the famous isochronism of small oscillations.
• (At large angles, this breaks down — the period grows slightly with amplitude. But for small swings, you can ignore that effect.)

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Answer:

$\boxed{\;T = 2\pi\sqrt{\dfrac{L}{g}}\;}$

• $L = 1\;\text{m}$ → $T \approx 2.006\;\text{s}$
• $L = 2\;\text{m}$ → $T \approx 2.836\;\text{s}$
• $L = 4\;\text{m}$ → $T \approx 4.013\;\text{s}$

Period scales as $\sqrt L$ — not linearly. Quadrupling the length only doubles the period. To double the period from 2 s to 4 s, you have to quadruple the length from 1 m to 4 m.

Try It

• Adjust the length with the slider.
• Watch the pendulum swing faster or slower.
• Try doubling the length — the period only grows by $\sqrt{2} \approx 1.414$.
• Adjust the amplitude — notice the period doesn't change (small-angle approximation).
• The HUD shows the live period.