Simple pendulum

Every grandfather clock, every metronome, and the earliest reliable timekeepers ever built all lean on one small, elegant fact: a bob swinging on a string keeps remarkably steady time. This is the simple pendulum — the third strand of the "Measurements and Experimentation" chapter, after you have already learnt to measure length and mass precisely. Here you learn how time is measured, using nothing more exotic than a string, a bob, and a stopwatch. The syllabus keeps things focused: know what time period and frequency mean, use the formula $T = 2\pi\sqrt{l/g}$ directly in numericals (its derivation is not required), and understand why plotting $l$ against $T^2$ gives a straight line whose slope hands you the local value of $g$ . This article expands on that ground with a fuller picture of the physics, a graph-based derivation of $g$ , and extra worked practice — but stays exactly within this scope.

The core insight

A simple pendulum's time period depends only on its length and on $g$ , following $T = 2\pi\sqrt{l/g}$ , and this lets a length-versus- $T^2$ graph be used to measure $g$ experimentally.

The core idea

A simple pendulum is an idealised system: a small, heavy bob hung from one end of a light, inextensible string, with the other end fixed, free to swing back and forth in a vertical plane. "Light" means the string's mass is negligible compared to the bob's, and "small" means the bob can be treated as a single point of mass — these idealisations are what make the formula below clean and simple. One complete oscillation (or vibration) is the bob's journey from its starting position, through the mean position to one extreme, back through the mean position to the other extreme, and back again to the start. Two useful quantities describe this repeating motion: The time period ( $T$ ) is the time taken for one complete oscillation, in seconds. The frequency ( $f$ ) is the number of complete oscillations in one second, in hertz (Hz). They are reciprocals of each other:

f = \frac{1}{T}

where: $f$ is frequency in Hz and $T$ is time period in seconds. For small angles of swing (a few degrees), the time period depends only on the pendulum's length $l$ and the local acceleration due to gravity $g$ :

T = 2\pi\sqrt{\frac{l}{g}}

where: $T$ is the time period (s), $l$ is the length of the pendulum measured from the point of suspension to the centre of the bob (m), and $g$ is the acceleration due to gravity (m/s²). Notice what is missing from the right side: the mass of the bob, and (for small swings) the amplitude. A heavy bob and a light bob of the same length, released from the same small angle, keep exactly the same time. Squaring the formula turns it into something you can test on a graph:

T^2 = \frac{4\pi^2}{g}\,l

This is a straight line through the origin when $T^2$ (y-axis) is plotted against $l$ (x-axis), with slope $\dfrac{4\pi^2}{g}$ . A student measures $T$ for several different lengths $l$ , plots the points, draws the best-fit line, and reads off its slope — turning a swinging string into a gravity-measuring instrument.

Figure

A straight line graph of T squared on the vertical axis against length l on the horizontal axis passing through the origin, with a right triangle marked on the line showing rise over run equal to the slope four pi squared over g

Derivation AGetting g from the slope of the l–T² graph

The goal is to start from the given relation $T = 2\pi\sqrt{l/g}$ , turn it into the straight-line form actually plotted in the laboratory, and show precisely how the measured slope of that line hands you the local acceleration due to gravity — the calculation the syllabus expects you to carry out even though the pendulum formula itself is not derived.

Step 1. Begin with the given formula and remove the square root by squaring both sides. Starting from $T = 2\pi\sqrt{l/g}$ , squaring gives $T^2 = (2\pi)^2 \times \dfrac{l}{g} = \dfrac{4\pi^2}{g}\,l$ . This single algebraic step converts an equation with a square root — awkward to plot directly — into one that is linear in $l$ and $T^2$ .

Step 2. Recognise the linear form. Compare $T^2 = \dfrac{4\pi^2}{g}\,l$ with the general equation of a straight line through the origin, $y = mx$ . Here $y$ is $T^2$ , $x$ is $l$ , and the constant multiplying $x$ — the slope $m$ — is $\dfrac{4\pi^2}{g}$ . Because $g$ is treated as constant at a given place on Earth, this slope is a fixed number, and the line must pass through the origin (when $l = 0$ , $T^2 = 0$ ).

Step 3. Connect the algebra to the experiment. In the lab, a student measures the time period $T$ for several different pendulum lengths $l$ , squares each $T$ , and plots $T^2$ against $l$ . The points lie on a straight line through the origin, exactly as Step 2 predicts. Measuring the slope of the best-fit line — rise in $T^2$ divided by run in $l$ — gives a numerical value for $\dfrac{4\pi^2}{g}$ .

Step 4. Rearrange to isolate $g$ . Since slope $= \dfrac{4\pi^2}{g}$ , cross-multiplying gives $g \times \text{slope} = 4\pi^2$ , so

g = \frac{4\pi^2}{\text{slope}}

This is the working rule: measure the slope experimentally, then divide $4\pi^2$ by it to obtain $g$ . The Worked example below applies this exact rule to real numbers.

Worked example

Question: A simple pendulum has a length of 1.0 m. Find its time period and frequency. This same pendulum is one of several lengths used to plot a $T^2$ (y-axis) against $l$ (x-axis) graph, which comes out as a straight line through the origin with slope $4.02\text{ s}^2/\text{m}$ . Find the value of $g$ this experiment gives. (Take $\pi = 3.14$ .)

Step 1. Substitute into the time period formula for the 1.0 m pendulum.

T = 2\pi\sqrt{\frac{l}{g}} = 2 \times 3.14 \times \sqrt{\frac{1.0}{9.8}}

Step 2. Evaluate the square root and complete the calculation.

\sqrt{\frac{1.0}{9.8}} = \sqrt{0.102} \approx 0.319 \quad\Rightarrow\quad T = 2 \times 3.14 \times 0.319 \approx 2.0\text{ s}

Step 3. Find the frequency from $f = 1/T$ .

f = \frac{1}{2.0} = 0.5\text{ Hz}

Step 4. Now use the measured graph slope, applying $g = 4\pi^2/\text{slope}$ from the derivation above.

g = \frac{4\pi^2}{\text{slope}} = \frac{4 \times (3.14)^2}{4.02} = \frac{39.44}{4.02} \approx 9.8\text{ m/s}^2

Answer: The time period is approximately $\mathbf{2.0\text{ s}}$ , the frequency is $\mathbf{0.5\text{ Hz}}$ , and the slope of $4.02\text{ s}^2/\text{m}$ on the $l$ – $T^2$ graph gives back $\mathbf{g \approx 9.8\text{ m/s}^2}$ , matching the standard value — a check that both parts of the question are physically consistent.

Competitive stretch

If the same pendulum were taken to a place where the measured slope of the $l$ – $T^2$ graph is $4.20\text{ s}^2/\text{m}$ , what value of $g$ would this indicate?

Step 1. Apply $g = 4\pi^2/\text{slope}$ directly: $g = 39.44 / 4.20 \approx 9.39\text{ m/s}^2$ .

Answer: This location has a lower effective $g \approx 9.39\text{ m/s}^2$ — a larger slope on the $l$ – $T^2$ graph always signals a smaller $g$ , since slope and $g$ are inversely related.

Common traps

Mixing up which axis is which — depending on how a question phrases the graph ( $l$ vs $T^2$ or $T^2$ vs $l$ ), the slope formula flips between $\dfrac{g}{4\pi^2}$ and $\dfrac{4\pi^2}{g}$ ; always check the axes stated in the question first.
Believing a heavier bob swings faster or slower — the formula shows $T$ depends only on $l$ and $g$ , never on the bob's mass.
Miscalculating $4\pi^2$ as $4\pi$ , a common arithmetic slip when finding $g$ from the slope.
Trying to derive $T = 2\pi\sqrt{l/g}$ itself from first principles — the syllabus gives this formula directly; only the graph-to- $g$ working (as shown above) is expected of you.

Ask the tutor: Want more practice turning an $l$ – $T^2$ graph slope into a value of $g$ , or a quick refresher on rearranging $T = 2\pi\sqrt{l/g}$ ? Ask the AI tutor for guided problems on this topic.

Why does the mass of the bob not affect the time period at all?

It seems surprising until you notice that in the formula $T = 2\pi\sqrt{l/g}$ , mass simply does not appear — only length and gravity do. Physically, a heavier bob needs a larger force to accelerate it through the swing, but gravity also pulls it down with proportionally more force, and the two effects exactly cancel. So a lead bob and a wooden bob of the same size and length, released from the same small angle, complete each swing in identical time.

Does the amplitude of the swing change the time period?

Only if the amplitude is large. The formula $T = 2\pi\sqrt{l/g}$ is derived under the assumption of a small swing angle (typically under about 10 degrees), and within that range the time period is independent of amplitude — a small swing and a slightly larger small swing take the same time. For large angles, this approximation breaks down and the true time period increases slightly with amplitude, but this lies outside the ICSE Class 9 syllabus, which only expects the small-angle case.

Why do we plot l against T squared instead of l against T directly?

Because $T = 2\pi\sqrt{l/g}$ is not a straight-line relation between $l$ and $T$ — it involves a square root, so a graph of $T$ against $l$ would be a curve, which is hard to read a precise slope from. Squaring both sides gives $T^2 = (4\pi^2/g)\,l$ , a genuine straight line through the origin. Straight lines are far easier to draw accurately and to extract a reliable slope from, which is exactly why the experiment is designed around $T^2$ rather than $T$ .

Simple pendulum

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads