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

Before anything, let us fix the three symbols we lean on, in plain words:

Cell	Scenario class	What is unknown	Trick / danger
A	Forward: given $L, g$ → find $T$	$T$	direct substitution
B	Backward: given $T, g$ → find $L$	$L$	invert & square
C	Measure gravity: given $L, T$ → find $g$	$g$	$g = 4 π^{2} L / T^{2}$
D	Ratio / scaling: how does $T$ change if $L$ or $g$ changes?	ratio $T_{2} / T_{1}$	no numbers needed
E	Different planet/Moon (small $g$ )	$T$ or comparison	$g$ is the only change
F	Degenerate / limiting inputs ( $L \to 0$ , $g \to 0$ , $L \to \infty$ )	behaviour	conceptual limits
G	Many-swings timing (real lab)	$T$ then $g$	divide total time by count
H	Large-amplitude correction (formula breaks)	true $T$	$T \approx 2 π L / g (1 + θ_{0}^{2} /16)$
I	Exam twist: units, radians vs degrees, "half a swing"	trap-aware answer	read the wording

Cell A — Forward: find the period

Worked example Example 1 — a 2.5 m pendulum on Earth

(Cell A) A bob hangs on a $L = 2.5 m$ string, $g = 9.8 m/s^{2}$ . Find $T$ .

Forecast: A 1 m pendulum was $\approx 2$ s. This one is $2.5 \times$ longer — but $T \propto L$ , so guess a bit over $3$ s, not $5$ s.

Write the formula: $T = 2 π L / g$ . Why this step? $L$ and $g$ are both known, $T$ is the only unknown — pure substitution.
Divide inside the root: $L / g = 2.5/9.8 = 0.2551$ . Why this step? Do the division before the square root to keep the number tidy.
Square root: $0.2551 = 0.5051$ . Why this step? The formula wraps the ratio in a square root, so we must undo that root to get the number that $2 π$ multiplies.
Multiply by $2 π$ : $T = 6.2832 \times 0.5051 = 3.17 s$ . Why this step? The $2 π$ out front converts the "root of the ratio" into a full period — it is the leftover factor from one complete cycle of phase (see Angular Frequency and Period).

Verify: Units inside root: $m \div (m/s^{2}) = s^{2}$ , and $s^{2} = s$ ✓. Our forecast of "a bit over 3 s" lands right. $✓$

Cell B — Backward: find the length

Worked example Example 2 — length of a "seconds pendulum"

(Cell B) A seconds pendulum completes one full period in $T = 2.0 s$ . What length gives this on Earth?

Forecast: Example 1 in the parent showed $L = 1.0 m$ gives $T \approx 2.0 s$ , so we expect $L$ very close to $1$ m.

Start from $T = 2 π L / g$ and isolate the root: $\frac{T}{2 π} = \frac{L}{g}$ . Why this step? We solve algebraically first — fewer rounding slips than plugging numbers early.
Square both sides: $(\frac{T}{2 π})^{2} = \frac{L}{g}$ . Why this step? The square root is the wrapper hiding $L$ ; squaring is the inverse operation that peels it off.
Multiply by $g$ : $L = g (\frac{T}{2 π})^{2}$ . Why this step? $L$ is now divided by $g$ ; multiplying both sides by $g$ leaves $L$ alone on one side — the goal.
Plug in: $L = 9.8 (\frac{2.0}{6.2832})^{2} = 9.8 \times (0.3183)^{2} = 9.8 \times 0.1013 = 0.993 m$ . Why this step? Only now, with $L$ fully isolated, do we substitute numbers — a clean single evaluation with no algebra left to slip on.

Verify: Feed $L = 0.993$ back into Cell A: $T = 2 π 0.993/9.8 = 2.00 s$ ✓. Matches our forecast of "just under 1 m." $✓$

Cell C — Measuring $g$

Worked example Example 3 — find

$g$ from one clean period (Cell C) A pendulum of $L = 0.80 m$ is found to have period $T = 1.80 s$ . What is $g$ here?

Forecast: Earth's $g$ is $\approx 9.8$ . If this comes out wildly off, we made an error.

Square the formula: $T^{2} = 4 π^{2} \frac{L}{g}$ . Why this step? $g$ is trapped under a square root; squaring frees it into a plain fraction.
Rearrange for $g$ : $g = \frac{4 π ^{2} L}{T ^{2}}$ . Why this step? Multiply both sides by $g$ , divide by $T^{2}$ — standard algebra to isolate the target.
Plug numbers: $g = \frac{4 π ^{2} ( 0.80 )}{( 1.80 ) ^{2}} = \frac{31.58}{3.24} = 9.75 m/s^{2}$ . Why this step? With $g$ alone on the left, substitution is now a single safe evaluation; we compute the top ( $4 π^{2} \times 0.80$ ) and bottom ( $1.8 0^{2}$ ) separately to avoid slips.

Verify: $9.75$ is within $0.5%$ of $9.8$ — a believable Earth value ✓. Units: $\frac{m}{s ^{2}} = m/s^{2}$ ✓. See Measuring g with a Pendulum. $✓$

Cell D — Ratios and scaling (no calculator needed)

Here the whole point is that you never plug in $g$ — it cancels. This is the fastest kind of problem once you spot it.

Look at the figure below. The horizontal axis is length $L$ ; the vertical axis is period $T$ . Notice the yellow dashed markers at $L = 1$ m and the red ones at $L = 4$ m: the length jumps by $4 \times$ but the height (period) only doubles. The whole curve bends over — that flattening is exactly what "square root" looks like as a picture.

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

Cell E — Another world: the Moon

We now compare the same pendulum in two places. To keep the two apart we attach a small subscript: a letter written low and small to tag which world a quantity belongs to. Here $g_{E}$ and $T_{E}$ mean gravity and period on Earth (subscript $E$ ), while $g_{M}$ and $T_{M}$ mean the same quantities on the Moon (subscript $M$ ).

Worked example Example 5 — same pendulum, Earth vs Moon

(Cell E) A pendulum has $T_{E} = 2.00 s$ on Earth ( $g_{E} = 9.8 m/s^{2}$ ). On the Moon $g_{M} = 1.62 m/s^{2}$ . Find its period there, $T_{M}$ .

Forecast: Weaker gravity = lazier, slower swing = bigger $T$ . Expect a couple of times longer.

Only $g$ changes, $L$ is fixed, so take the ratio: $\frac{T _{M}}{T _{E}} = \frac{g _{E}}{g _{M}}$ . Why this step? $g$ is downstairs inside the root, so a smaller $g$ makes a larger $T$ — and taking the ratio kills the unknown $L$ .
Compute: $9.8/1.62 = 6.049 = 2.460$ . Why this step? This bare number is the factor by which the Moon stretches the period; note $g_{E}$ sits on top so weaker Moon gravity ( $g_{M}$ small) makes the ratio bigger than 1, confirming a slower swing.
So $T_{M} = 2.460 \times 2.00 = 4.92 s$ . Why this step? The ratio from Step 2 tells us how many times longer than Earth's period the Moon's is; multiplying it by the known Earth period $T_{E} = 2.00$ s converts that pure factor into an actual time in seconds — the quantity asked for.

Verify: Longer period on the Moon — matches the "slow lazy swing" prediction ✓. Sanity: $6 \approx 2.45$ , so $\approx 4.9$ s ✓. $✓$

Cell F — Degenerate & limiting inputs

You must never be surprised by an extreme. These are conceptual, but the formula answers each cleanly.

Worked example Example 6 — the limits

$L \to 0$ , $g \to 0$ , $L \to \infty$ (Cell F) Predict the period behaviour in three extreme cases from $T = 2 π L / g$ .

Forecast: Think about what each ratio does inside the root before reading on.

$L \to 0$ (string shrinks to nothing): $T = 2 π 0/ g = 0$ . Why this makes sense: a zero-length pendulum has nowhere to swing — the period collapses to zero. Physically it "buzzes" infinitely fast.
$L \to \infty$ (endlessly long string): $L / g \to \infty$ , so $T \to \infty$ . Why this makes sense: an enormously long pendulum swings ever more slowly; in the limit it never returns.
$g \to 0$ (deep space, no gravity): $T = 2 π L /0 \to \infty$ . Why this makes sense: with no restoring pull, there is nothing to bring the bob back — it never oscillates. Compare Restoring Force and Equilibrium: no restoring force ⇒ no period.

Verify: Try $L = 1 0^{- 6}$ m, $g = 9.8$ : $T = 2 π 1 0^{- 6} /9.8 = 2.01 \times 1 0^{- 3} s$ , tiny ✓. Try $g = 1 0^{- 6}$ , $L = 1$ : $T = 6.28 \times 1 0^{3} s$ , huge ✓. Both trend to their predicted extremes. $✓$

Cell G — Real-lab timing of many swings

Worked example Example 7 — 20 swings, then find

$g$ (Cell G) A student times $20$ complete swings of a $L = 1.20 m$ pendulum: total $43.9 s$ . Find $T$ , then $g$ .

Forecast: One swing should be roughly $2.2$ s (a bit longer than the 1 m ≈ 2 s pendulum), and $g$ should land near $9.8$ .

Period from the count: $T = \frac{total time}{number of swings} = \frac{43.9}{20} = 2.195 s$ . Why this step? Timing many swings and dividing spreads your reaction-time error over 20 periods, so each $T$ is far more accurate — the key lab trick.
Use $g = \frac{4 π ^{2} L}{T ^{2}}$ (from Cell C). Why this step? Same rearranged formula; $g$ is what the experiment is really after.
Compute: $g = \frac{4 π ^{2} ( 1.20 )}{( 2.195 ) ^{2}} = \frac{47.37}{4.818} = 9.83 m/s^{2}$ . Why this step? We evaluate top ( $4 π^{2} \times 1.20$ ) and bottom ( $2.19 5^{2}$ ) separately, then divide — keeping the two operations apart guards against calculator-order mistakes.

Verify: $9.83 \approx 9.8$ , an excellent Earth measurement ✓. Forecast of " $T \approx 2.2$ s" confirmed. $✓$

Cell H — When the swing is too big (formula breaks)

The formula $T = 2 π L / g$ is an approximation valid for small angles (that came from $sin θ \approx θ$ in the parent — see Taylor Series and Small-Angle Approximations). For a wide swing you need the first correction:

Worked example Example 8 — how wrong is the simple formula at

$θ_{0} = 30°$ ? (Cell H) A $L = 1.00 m$ pendulum swings with amplitude $θ_{0} = 30°$ . Find the small-angle period and the corrected period, and the percent difference.

Forecast: The correction is small — expect the true period to be a hair longer, around $1$ – $2%$ .

First the small-angle period — call it $T_{0}$ , meaning the period the simple formula predicts before any large-swing correction: $T_{0} = 2 π 1.00/9.8 = 2.007 s$ . Why this step? This $T_{0}$ is the baseline the correction factor multiplies; we need it before we can nudge it.
Convert amplitude to radians: $θ_{0} = 30° = 30 \times \frac{π}{180} = 0.5236 rad$ . Why this step? The correction term $θ_{0}^{2} /16$ is a Taylor coefficient — it only works in radians, never degrees.
Correction factor: $1 + \frac{( 0.5236 ) ^{2}}{16} = 1 + \frac{0.2742}{16} = 1 + 0.01714 = 1.01714$ . Why this step? Squaring the amplitude and dividing by 16 gives the fractional extra period; adding 1 turns it into a multiplier we can apply to $T_{0}$ .
Corrected period: $T = T_{0} \times 1.01714 = 2.007 \times 1.01714 = 2.041 s$ . Why this step? Multiplying the baseline $T_{0}$ by the factor bigger than 1 stretches it — the wide swing genuinely takes longer.
Percent difference: $\frac{2.041 - 2.007}{2.007} \times 100% = 1.71%$ . Why this step? Comparing the corrected value to $T_{0}$ as a percentage tells us exactly how much error the simple formula carries at $30°$ — the practical takeaway.

Verify: The parent note stated "at $θ_{0} = 30°$ the error is $\sim 1.7%$ " — we got $1.71%$ ✓. The corrected period is indeed slightly larger than the small-angle $T_{0}$ ✓, and both sit close to $2$ s as expected for a 1 m pendulum. $✓$

Study the figure below. The horizontal axis is the amplitude $θ_{0}$ in degrees; the vertical axis is how many percent longer the true period is than the simple formula. At small angles (yellow arrow) the curve hugs zero — the simple formula is excellent. By $30°$ (red marker) it has climbed to about $1.7%$ , matching Example 8. The message: the error grows as amplitude squared, slowly at first then faster.

Cell I — Exam twists (read the wording!)

Worked example Example 9 — "half a swing" and a degrees trap

(Cell I) A pendulum takes $0.90 s$ to travel from one extreme to the other extreme (that is, half a full cycle). Its amplitude is stated as $8°$ . On Earth, find $L$ .

Forecast: "Extreme to extreme" is only half a period, so the full period is $1.80$ s, not $0.90$ s. And $8°$ is small — the simple formula is fine, no correction needed.

Identify the real period: one full cycle is extreme → other extreme → back = two half-swings, so $T = 2 \times 0.90 = 1.80 s$ . Why this step? A "swing" from side to side is half a period; forgetting this halves or doubles every later answer. This is the single most common exam trap.
Check the angle: $8° = 0.1396 rad$ , tiny — small-angle formula valid, ignore the correction. Why this step? We must justify using $T = 2 π L / g$ before trusting it (Cell H showed it fails for big swings).
Solve for $L$ (as in Cell B): $L = g (\frac{T}{2 π})^{2} = 9.8 (\frac{1.80}{6.2832})^{2}$ . Why this step? $L$ is the unknown; we reuse the Cell B rearrangement that isolates it.
Compute: $L = 9.8 \times (0.28648)^{2} = 9.8 \times 0.08207 = 0.804 m$ . Why this step? With the correct full period $T = 1.80$ s substituted, this single evaluation gives the true length; using the trap value $0.90$ s here would have poisoned the answer.

Verify: Feed back: $T = 2 π 0.804/9.8 = 1.800 s$ , a full cycle ✓, and half of that is $0.90$ s as stated ✓. Had we wrongly used $T = 0.90$ s we would have got $L = 0.201$ m — four times too small. $✓$

Active recall

Recall Match each cell to its move

Given $L, g$ , want $T$ ::: substitute directly into $T = 2 π L / g$ (Cell A).
Given $T, g$ , want $L$ ::: $L = g (T /2 π)^{2}$ (Cell B).
Given $L, T$ , want $g$ ::: $g = 4 π^{2} L / T^{2}$ (Cell C).
Length made 9× bigger, period does what? ::: multiplied by $9 = 3$ (Cell D).
Weaker gravity (Moon), period does what? ::: gets bigger, since $g$ is under the root's denominator (Cell E).
"One swing side to side" is what fraction of a period? ::: one half (Cell I).

Connections

Parent: the derivation
Simple Harmonic Motion
Restoring Force and Equilibrium
Taylor Series and Small-Angle Approximations
Angular Frequency and Period
Energy in Oscillations
Mass-Spring System
Measuring g with a Pendulum

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

The scenario matrix