Simple pendulum — small angle approximation, T = 2π√(L - g) derivation: Question bank

The symbols this page uses — read first

Before the traps, let's pin down every letter so nothing surprises you. Look at the picture: it shows the whole setup and every symbol anchored to something you can point at.

Figure — Simple pendulum — small angle approximation, T = 2π√(L - g) derivation

Definition The five symbols

$m$ — the mass of the bob, the little weight at the end (in kilograms). "How much stuff" is swinging.
$L$ — the length of the pendulum: the straight-line distance from the pivot (where it hangs) down to the bob's centre of mass, not to the top of the bob (in metres). This is the radius of the circular arc the bob sweeps.
$g$ — the magnitude of gravitational acceleration, $g \equiv ∣ g ∣ \approx 9.8 m/s^{2}$ on Earth. Always positive; it measures how strongly gravity pulls.
$θ$ — the angular displacement from vertical: the angle (in radians) between the string and the straight-down rest position. $θ = 0$ is hanging straight down; $θ > 0$ is pulled to one side.
$θ_{0}$ — the amplitude: the largest value $θ$ reaches, i.e. the angle you release the bob from. It labels "how big the swing is".

The result these traps orbit around

Everything below tests one formula, so here is the whole story in one breath, with a picture of the swinging geometry.

True or false — justify

A pendulum's period depends on the mass of the bob.

False — in

m L \ddot{θ} = - m g sin θ

the mass

m

divides out of both sides, so

T = 2 π L / g

contains no

m

. Gravity pulls a heavy bob harder but its inertia resists just as much more, exactly cancelling.

Doubling the string length doubles the period.

False —

T \propto L

, so doubling

L

multiplies

T

2 \approx 1.41

. You must quadruple

L

to double

T

The formula

T = 2 π L / g

is exact for all swing angles.

False — it comes from

sin θ \approx θ

, which is only an approximation. The true period grows with amplitude:

T \approx 2 π L / g (1 + θ_{0}^{2} /16 + \dots)

, where

θ_{0}

is the amplitude (max angle).

A pendulum swings faster on the Moon than on Earth.

False — the Moon's smaller

g

appears under the root as

L / g

, so smaller

g

means larger

T

: slower, lazier swings.

Tension in the string does no work in speeding the bob along its arc.

True — tension points along the string, which is always perpendicular to the arc (the direction of motion), so its component along the motion is zero. Only gravity's tangential component drives the swing.

At the lowest point of the swing the net force on the bob is zero.

False — the tangential force is zero there (equilibrium of the angular motion), but the bob moves in a circle, so tension exceeds

m g

to provide the upward centripetal force. Net force points up, toward the pivot.

Isochronism means the period is the same no matter how heavy the bob is.

False — isochronism specifically means the period is independent of amplitude (for small swings). Mass-independence is a separate fact; don't merge the two.

If you replace the string with a stiff rod of the same length,

T = 2 π L / g

still applies unchanged.

False — a rod has its own mass distributed along its length, so it becomes a physical pendulum whose period depends on its moment of inertia, not the simple-pendulum formula.

Spot the error

"A student writes

sin (10°) \approx 10

, so the small-angle rule is broken."

The rule is

sin θ \approx θ

with

θ

in radians.

10°

0.1745

rad, and

sin (10°) = 0.1736

— excellent agreement. The error is comparing

sin

to the degree number.

"Since

T = 2 π L / g

, a longer pendulum has a bigger

ω

and swings more times per second."

Backwards.

ω = g / L

, so larger

L

gives smaller

ω

— fewer cycles per second and a longer period. Long pendulums are slow.

"The restoring force is

F = - m g θ

, so it's linear and this is always SHM."

The true tangential force is

F = - m g sin θ

, which is not linear. It only reduces to

- m g θ

after the small-angle approximation, so it's SHM only for small swings.

L

is the length of the string, measured from the pivot to where you tied the knot at the top of the bob."

L

is the distance from the pivot to the bob's centre of mass, not to its top. For a large bob this difference matters.

"On a fast-spinning merry-go-round, a hanging pendulum's period is still

2 π L / g

because gravity is unchanged."

In a rotating (non-inertial) frame the effective gravity combines

g

with the outward centrifugal effect, changing the restoring acceleration. Use effective

g_{eff}

, not plain

g

"The minus sign in

- m g sin θ

is just a convention and can be dropped."

It cannot — the minus encodes that the force points back toward

θ = 0

. Without it the equation would describe runaway growth, not oscillation. See Restoring Force and Equilibrium.

"Increasing amplitude increases the speed at the bottom, so the pendulum must complete each cycle faster."

The bob does move faster at the bottom for larger amplitude, but it also travels a longer arc. For small angles these balance (isochronism); for large angles the longer path wins and the period actually increases.

Why questions

Why does gravity get split into components at all, instead of using

m g

directly?

Because only the part of gravity along the arc (tangential) changes the bob's speed; the radial part is cancelled by tension. Splitting isolates the force that actually drives the oscillation.

Why must

θ

be in radians for

sin θ \approx θ

to hold?

The Taylor series

sin θ = θ - θ^{3} /6 + \dots

is derived assuming radian measure, where the derivative of

sin

cos

cleanly. In degrees an extra factor of

π /180

appears and

sin θ \approx θ

fails. See Taylor Series and Small-Angle Approximations.

Why does the mass cancel, physically rather than algebraically?

Gravity's pull scales with mass (

m g

), and resistance to acceleration (inertia) also scales with mass. The two mass factors are the same

m

, so the acceleration they produce is mass-free — the same reason all objects fall at the same rate.

Why is

T = 2 π / ω

and not, say,

π / ω

4 π / ω

One complete cycle corresponds to the phase advancing by

2 π

radians, and

ω

is the phase rate (radians per second). Time for one cycle is total phase over rate:

2 π / ω

. See Angular Frequency and Period.

Why does the pendulum equation look just like the mass-spring equation?

Both have the form

\overset{q}{¨} = - ω^{2} q

: a restoring "acceleration" proportional to displacement. For the spring

ω^{2} = k / m

; for the pendulum

ω^{2} = g / L

. Same maths, same SHM. See Mass-Spring System.

Why is timing 10 swings and dividing by 10 better than timing one swing to find

g

Your reaction-time error (say

0.2

s) is fixed per measurement. Spread over 10 periods it contributes only

0.02

s per period — a tenfold reduction in fractional error. See Measuring g with a Pendulum.

Why does a larger amplitude make the true period longer, not shorter?

For large

θ

sin θ < θ

, so the real restoring force is weaker than the linear approximation predicts. A weaker restoring pull means slower return and a longer period.

Edge cases

What happens to the period as the amplitude approaches

180°

(bob nearly balanced straight up)?

The period tends to infinity — near the top the restoring force

sin θ \to 0

, so the bob lingers arbitrarily long. The simple formula is hopelessly wrong here.

If you released the bob from exactly

θ = 0

with zero speed, what motion results?

None — it's already at equilibrium with no restoring force and no velocity, so it stays at rest. This is the degenerate zero-amplitude "oscillation".

What is the period of a pendulum in free fall (an elevator with the cable cut)?

Effectively infinite / undefined — in free fall the effective

g

is zero, there's no restoring force, and the bob just floats. The formula gives

T = 2 π L /0 \to \infty

Take the limit

L \to 0

: what does the formula predict, and is it physical?

T \to 0

: infinitely fast swinging. Mathematically consistent, but physically the point-mass idealization breaks down long before, so treat it as a limiting trend, not a real device.

g

were to double (a hypothetical stronger-gravity planet), how does the period change?

T \propto 1/ g

, so doubling

g

divides the period by

2 \approx 1.41

— the pendulum swings noticeably faster.

What happens to the energy of an ideal (frictionless) pendulum over many swings?

It is conserved — kinetic and potential energy trade back and forth but their sum stays constant, so the amplitude never decays. Real pendulums lose energy to air and friction. See Energy in Oscillations.

Active recall

Recall Which two "independences" of the period do people most often confuse?

Mass-independence (any bob, same $T$ ) versus amplitude-independence / isochronism (small swings of any size, same $T$ ). They are separate facts with separate reasons.

Recall Name the single approximation that turns the pendulum into SHM.

$sin θ \approx θ$ (radians), valid only for small angles — everything downstream depends on it.

Connections

Simple Harmonic Motion — the pattern every trap comes back to
Restoring Force and Equilibrium — source of the minus sign
Taylor Series and Small-Angle Approximations — why radians, why approximate
Angular Frequency and Period — the $T = 2 π / ω$ relation
Energy in Oscillations — the energy edge case
Mass-Spring System — analogous SHM
Measuring g with a Pendulum — the timing-many-swings trick