Imagine tying a small weight to a string, holding the other end fixed, and giving it a gentle push. It swings back and forth — that’s a simple pendulum. The question is: what determines how fast it swings? Does a heavier bob swing faster? Does a longer string make it slower?
Let’s start with what you already feel. If you hold a short string (say 20 cm) and swing it, the bob zips back and forth quickly. If you use a long string (say 1 m), the swing is noticeably slower. So length matters — longer means slower.
What about the weight? Try a light plastic bob and a heavy metal one of the same size, on the same string. You’ll find they swing at the same speed. That’s surprising — heavier things don’t fall faster, and here they don’t swing faster either. So mass does not affect the period (the time for one complete back-and-forth swing).
What about how hard you push? If you give a big push, the bob swings wider, but does it take more time? For small swings (small angles, say less than about 15°), the period is almost the same regardless of amplitude. That’s the key: for small oscillations, the pendulum is isochronous — its period is independent of amplitude.
Note
This is only true for small angles. If you pull the bob to 60° and let go, the period becomes noticeably longer. In most exam problems, you assume “small oscillations” (usually < 10°).
The Precise Statement
For a simple pendulum of length L (measured from pivot to centre of bob), swinging with small amplitude in a uniform gravitational field g, the time periodT (time for one complete oscillation) is:
T=2πgL
That’s it. No mass term. No amplitude term (for small angles).
T=2πgL
Why does this formula make sense?
L in numerator: longer string → larger T (slower swing). Doubling L multiplies T by 2≈1.4.
g in denominator: stronger gravity (larger g) → smaller T (faster swing). On the Moon (g≈1.6 m/s²), the same pendulum swings much slower.
2π: comes from the mathematics of simple harmonic motion — the pendulum’s motion is approximately sinusoidal for small angles.
Tip
To remember: the formula is identical to that of a mass on a spring (T=2πm/k), but here the “restoring force per unit displacement” is mg/L, so the effective “k” is mg/L, giving T=2πL/g.
Common exam pitfalls
Don’t confuse L with amplitude.L is the string length, not how far you pull it. …
A simple pendulum's tangential restoring force gives a non-linear equation that linearises under the small-angle approximation to SHM, with T=2πl/g. …
Step 1. Setup. A simple pendulum is a bob of mass m on a light, inextensible string of length l. At any displaced angle θ from the vertical, two forces act: weight mg (down) and tension T (along the string). Resolving weight gives a normal component mgcosθ (along the string, balanced by tension) and a tangential component mgsinθ, which always points back towards equilibrium -- the restoring force.
Step 2. Equation of motion. Applying Newton's second law tangentially, with arc-length s=lθ so s¨=lθ¨: mlθ¨=−mgsinθ, i.e. θ¨=−(g/l)sinθ -- a non-linear equation because of sinθ.
Step 3. Small-angle approximation. For angular amplitudes up to about 10°, sinθ≈θ (in radians), which linearises the equation to θ¨=−(g/l)θ, the standard SHM form.
Step 4. Time period. Comparing gives ω2=g/l, so ω=g/l, and T=2π/ω=2πl/g. …