Swap the spring for a piece of string with a weight on the end, nudge it sideways, and it swings back and forth like a grandfather clock. For small swings this is simple harmonic motion — and this time gravity is back in charge.
📘 What you need to know
A simple pendulum is a small heavy bob on a light, inextensible string fixed to a point above
Give it a small sideways pull and release, and it swings with simple harmonic motion
The period for small swings is T = 2π√(L/g)
L is measured from the pivot to the centre of the bob
Unlike a mass–spring, the period does depend on gravity — a pendulum runs slower on the Moon
The formula only holds for small angles (θ < 10°); the mass of the bob and the amplitude don’t affect the period
What Is a Simple Pendulum?
A simple pendulum is about as basic as an oscillator gets. It has just two parts: a bob — a small, dense weight we treat as a single point — hanging from a string we assume is weightless and can’t stretch. The top of the string is tied to a fixed pivot, and the bob swings from side to side beneath it. The string stays taut (in tension) the whole time.
The length L runs from the pivot all the way down to the centre of the bob, and the bob swings through a small angle θ either side of equilibrium.
The Time Period Formula
For small swings, the time for one complete there-and-back swing is:
Period of a simple pendulumT = 2π√(L/g)
where T is the period in seconds, L is the string length (pivot to centre of bob) in metres, and g is the gravitational field strength in N kg⁻¹. A longer string swings more slowly (longer T), which is why a tall grandfather clock ticks at a stately pace.
Here’s the neat contrast with the mass–spring system. There, gravity was nowhere in the formula. Here, g sits right inside the root — so the period genuinely changes with gravity. Take the same pendulum to the Moon, where g is about six times weaker, and each swing takes noticeably longer. Notice too that the mass of the bob doesn’t appear at all: a heavy bob and a light one on equal strings keep the same time.
Why Only Small Angles?
What pulls the bob back to the middle is the part of its weight that acts along the arc of the swing. If the string makes an angle θ with the vertical, that restoring part of the weight is mg sin θ, always pointing back towards equilibrium.
The weight mg (red) always points down. Only its component along the swing, mg sin θ (teal), pulls the bob back towards equilibrium.
For SHM we need the restoring force to be proportional to displacement. That only works when the angle is small enough that sin θ is almost equal to θ itself (measured in radians):
Small-angle approximation
sin θ ≈ θ (for θ < 10°)
Keep the swings small — under about 10° from the vertical — and the pendulum behaves as a textbook simple harmonic oscillator, which is exactly when T = 2π√(L/g) is valid.
🧭 Working out a pendulum period
Check the angle is small (under ~10°) so the formula actually applies
Convert the length to metres and make sure you’re using g = 9.81 N kg⁻¹ on Earth
Substitute into T = 2π√(L/g) — divide L by g first, then square-root, then × 2π
For angular frequency, combine with T = 2π/ω to get ω = √(g/L)
On another planet or the Moon, change g — the length stays the same, but the period doesn’t
Quick recap: a small-swing pendulum does SHM with T = 2π√(L/g) — longer string is slower, the bob’s mass doesn’t matter, but gravity very much does.
WE 1
A pendulum of length 80.0 cm swings with a maximum angle of 8° from the vertical.
(a) Calculate the period of the swing.
(b) Determine the angular frequency of the oscillation.
Part (a)
The 8° swing is under 10°, so the formula applies. L = 80.0 cm = 0.80 m
T = 2π√(L/g) = 2π√(0.80 / 9.81)T = 2π × 0.2856T ≈ 1.8 sPart (b)ω = √(g/L) = √(9.81 / 0.80)ω ≈ 3.5 rad s⁻¹You’d get the same ω from ω = 2π/T — the two routes always agree.
WE 2
A pendulum has a period of 2.0 s on Earth (g = 9.81 N kg⁻¹).
(a) Calculate the length of the pendulum.
(b) Find its period on the Moon, where g = 1.6 N kg⁻¹.
Part (a)
Rearranging T = 2π√(L/g) gives L = g(T ÷ 2π)²
L = 9.81 × (2.0 ÷ 2π)² = 9.81 × 0.1013L ≈ 0.99 mPart (b)
Same length, but g is now 1.6 N kg⁻¹
T = 2π√(0.994 / 1.6)T ≈ 5.0 sWeaker gravity means a lazier swing — the clock runs badly slow on the Moon.
💡 Top tips
Measure L to the centre of the bob, not to the top of it
Inside the root it’s L over g; a quick check — a longer string should give a bigger T
The bob’s mass never enters the formula, so don’t go looking for it in the numbers
If a question moves the pendulum to the Moon or another planet, only g changes — reuse the same L
Watch the “small angle” condition: beyond about 10° the formula stops being accurate
⚠ Common mistakes
Flipping the fraction to √(g/L) when finding the period (that ratio is for ω, not T)
Trying to bring the bob’s mass into the calculation — it isn’t in the formula
Assuming the period is the same everywhere; it changes wherever g is different
Using a large swing angle and still trusting the small-angle formula
Measuring the length to the top of the bob instead of its centre
Up next: we stop timing the swings and start following the energy — Energy Changes in Simple Harmonic Motion, where kinetic and potential energy trade places as the object oscillates.
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Book a free meeting and let’s work through the tricky bits together.
