PhysicsICSE Class 9 · Measurements and Experimentation

Simple Pendulum: Time Period, Frequency and the l–T² Graph (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

A simple pendulum swings with time period T = 2π√(l/g), where l is the effective length (suspension point to centre of the bob). T depends on l and g, not on the mass of the bob or a small amplitude. A graph of l against T² is a straight line through the origin with slope g/4π².

Understand Simple Pendulum: Time Period, Frequency and the l–T² Graph

The simple pendulum. A small, heavy bob hung from a rigid support by a light, inextensible thread. Pull it a little to one side and release: it oscillates to and fro.

**Effective length $l$ : from the point of suspension to the centre of the bob** = length of thread + radius of the bob.
Oscillation: one complete to-and-fro motion (A to B and back to A).
**Time period $T$ **: time for one oscillation. To reduce error, time $20$ oscillations and divide: $T = \frac{time for 20 oscillations}{20}$ .
**Frequency $f$ **: oscillations per second, $f = \frac{1}{T}$ , in hertz (Hz).
Amplitude: the largest displacement from the mean position (kept small, less than about $1 0^{\circ}$ ).

The formula (given without derivation):

$T = 2 π \frac{l}{g}$

What it tells us:

$T \propto l$ : four times the length doubles the period.
$T \propto \frac{1}{g}$ : where $g$ is less, the pendulum is slower. On a planet with $g$ one-sixth, $T$ becomes $6 \approx 2.45$ times as large.
$T$ does not depend on the mass of the bob, nor (for small swings) on the amplitude.

Seconds pendulum. One with $T = 2$ s (one second each way). Its length is $l = \frac{g T ^{2}}{4 π ^{2}}$ , about $0.99$ m for $g = 9.8$ m s⁻².

**The $l$ – $T^{2}$ graph.** Squaring: $T^{2} = \frac{4 π ^{2}}{g} l$ , so a graph of $l$ (y-axis) against $T^{2}$ (x-axis) is a straight line through the origin. Its slope is

$slope = \frac{l}{T ^{2}} = \frac{g}{4 π ^{2}}, g = 4 π^{2} \times slope .$

A graph of $T$ against $l$ is a curve; of $T$ against $l$ a straight line.

Sources of error in timing. Starting and stopping the stopwatch late, counting oscillations wrongly, using a large amplitude, a stretching thread, and measuring $l$ to the top of the bob instead of its centre.

Pendulum clocks. If a clock's pendulum lengthens (heat), $T$ rises and the clock runs slow; if it shortens, the clock runs fast.

Simple Pendulum: Time Period, Frequency and the l–T² Graph: formulas and key facts

Time period: $T = 2 π \frac{l}{g}$; No derivation; small amplitude.
Experimental period and frequency: $T = \frac{t _{20}}{20}, f = \frac{1}{T}$; f in Hz.
Length from period: $l = \frac{g T ^{2}}{4 π ^{2}}$; Seconds pendulum: T = 2 s.
Slope of the l–T² graph: $\frac{l}{T ^{2}} = \frac{g}{4 π ^{2}}$; g = 4π² × slope.
Scaling: $\frac{T _{2}}{T _{1}} = \frac{l _{2}}{l _{1}} = \frac{g _{1}}{g _{2}}$; Ratio method for comparison questions.

How to solve Simple Pendulum: Time Period, Frequency and the l–T² Graph questions

Decide whether to use the formula directly or the ratio form.

1Convert l to metres and find the effective length (thread + radius of bob).
2Find T from the time of 20 oscillations if given.
3Use $T = 2 π l / g$ directly, or the ratio $T_{2} / T_{1} = l_{2} / l_{1}$ for two pendulums.
4For graphs, find the slope as $Δ l /Δ (T^{2})$ and use $g = 4 π^{2} \times$ slope.
5State the value of $π$ and $g$ you used.

Simple Pendulum: Time Period, Frequency and the l–T² Graph in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A pendulum clock in summer

Suppose a wall clock has a seconds pendulum (T = 2 s) and in the heat the rod lengthens by 0.2%.

Does the clock gain or lose, and about how much in a day?

Try it first, then show the solutionSolution

New $T = T 1.002 \approx 1.001 T$ , so each swing takes slightly longer.
Fewer swings are counted: loss $\approx 86400 \times (1 - \frac{1}{1.001}) \approx 86$ s per day.

Answer: It loses about 86 seconds a day.

Memory hook: Longer pendulum, slower clock.

Tricks to remember Simple Pendulum: Time Period, Frequency and the l–T² Graph

Short memory aids. Say each one aloud once, then use it on a question the same day.

Root of the length
T goes as √l: 4× length gives 2× period, 9× length gives 3×.
Use it when: Ratio questions.
Mass and amplitude do nothing
T depends on l and g only (small swings).
Use it when: True or false and reason questions.
Slope is g over 4π²
l on y, T² on x. g = 4π² × slope ≈ 39.4 × slope.
Use it when: Graph questions.

Simple Pendulum: Time Period, Frequency and the l–T² Graph: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Frequency from timing · 1 mark · MCQBoard level

A pendulum completes 20 oscillations in 40 s. Its frequency is:
(a) 0.5 Hz
(b) 2 Hz
(c) 2 s
(d) 0.5 s

Try it first, then show the solutionSolution

$T = \frac{40}{20} = 2$ s; $f = \frac{1}{2} = 0.5$ Hz.

Answer: (a) 0.5 Hz

Example 2: Time period of a given pendulum · 2 marks · Short answerBoard level

A bob of radius 1.5 cm hangs from a thread of length 98.5 cm. Find the effective length and the time period ( $g = 9.8$ m s⁻², $π = 3.14$ ).

Try it first, then show the solutionSolution

$l = 98.5 + 1.5 = 100$ cm $= 1$ m. (1)
$T = 2 π \frac{1}{9.8} = 2 \times 3.14 \times 0.3194 \approx 2.0$ s. (1)

Answer: l = 1 m; T ≈ 2.0 s

Example 3: Pendulum on the Moon · 3 marks · Short answerHarder board level

A seconds pendulum is taken to the Moon, where $g$ is one-sixth of its value on Earth. Find its time period there, and the length of a pendulum that would still be a seconds pendulum on the Moon (Earth value $l = 0.99$ m).

Try it first, then show the solutionSolution

$T = 26 \approx 4.9$ s. (1)
A seconds pendulum needs $T = 2$ s; $l \propto g$ for fixed $T$ . (1)
$l_{M oo n} = \frac{0.99}{6} \approx 0.165$ m. (1)

Answer: About 4.9 s; about 0.165 m

Example 4: g from a graph · 3 marks · Short answerHarder board level

Students plot $l$ against $T^{2}$ and get a straight line through the origin. At $l = 0.80$ m the time period is $T = 1.8$ s. Find the slope of the graph and the value of $g$ ( $π = 3.14$ ).

Try it first, then show the solutionSolution

$T^{2} = 1. 8^{2} = 3.24$ s². (½)
Slope $= \frac{0.80}{3.24} = 0.247$ m s⁻². (1)
$g = 4 π^{2} \times slope = 4 \times 3.1 4^{2} \times 0.247 = 39.44 \times 0.2469 \approx 9.74$ m s⁻². (1½)

Answer: Slope ≈ 0.247 m s⁻²; g ≈ 9.74 m s⁻²

Example 5: A clock that loses time · 4 marks · Case-basedHOTS / competency

A pendulum clock with a seconds pendulum is kept in a hot room, where the rod lengthens so that $l$ increases by 21%.
(a) By what factor does the time period change?
(b) After the clock has shown 1 hour, how much real time has passed?
(c) Is the clock fast or slow?
(d) The mass of the bob is also increased by 21%. What is the effect on the answer to (a)?

Try it first, then show the solutionSolution

(a) $\frac{T _{2}}{T _{1}} = 1.21 = 1.1$ . (1)
(b) Each swing takes 1.1 times as long, so the clock counts 1 hour when $1.1$ hours have passed (66 min). (1)
(c) Slow, because it counts fewer swings than there are seconds. (1)
(d) None: $T$ does not depend on the mass of the bob. (1)

Answer: (a) 1.1 times (b) 66 min (c) slow (d) no effect

Where students lose marks in Simple Pendulum: Time Period, Frequency and the l–T² Graph

The slips examiners see most often, and what to write instead.

Mistake: Length measured to the top of the bob
Effective length runs to the centre of the bob.
Wrong working
l = 98.5 cm
Right working
l = 98.5 + 1.5 = 100 cm
Do this: Add the radius every time.
Usually costs: The final answer and one method mark.
Mistake: Centimetres used in the formula with g in m s⁻²
Units must match.
Wrong working
T = 2π√(100/9.8)
Right working
T = 2π√(1/9.8) with l = 1 m
Do this: Convert l to metres first.
Usually costs: The whole numerical.
Mistake: Plotting T against l and calling it a straight line
T ∝ √l, so T against l is a curve; l against T² is straight.
Wrong working
Straight line of T vs l through origin.
Right working
l against T² gives the straight line.
Do this: Check which quantity is squared.
Usually costs: Graph marks.

Exam tips

Time at least 20 oscillations in practical-based answers, and start counting from the mean position.
Write "effective length = length of thread + radius of the bob".
State that T is independent of mass and (small) amplitude when asked for factors.
Quote the slope as g/4π² and then compute g.

Practice questions on Simple Pendulum: Time Period, Frequency and the l–T² Graph

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The time period of a simple pendulum (small swings) depends on:
(a) the mass of the bob
(b) the amplitude of the swing
(c) its effective length
(d) the material of the bob
Hint
Look at the formula.
Answer
(c) its effective length
Q2 · 1 mark · Very short answerBoard level
A pendulum has time period 0.8 s. Find its frequency.
Hint
f = 1/T.
Answer
$f = \frac{1}{0.8} = 1.25$ Hz.
Q3 · 2 marks · Short answerHarder board level
Find the length of a seconds pendulum for $g = 9.8$ m s⁻² and $π = 3.14$ .
Hint
Use $l = g T^{2} /4 π^{2}$ with $T = 2$ s.
Answer
$l = \frac{9.8 \times 4}{4 \times 3.1 4 ^{2}} = \frac{39.2}{39.44} \approx 0.994$ m.
Q4 · 3 marks · Short answerHarder board level
A student times 20 oscillations as 36 s. Find T and the effective length of the pendulum ( $g = 9.8$ m s⁻², $π = 3.14$ ).
Hint
Find T first.
Answer
$T = \frac{36}{20} = 1.8$ s; $l = \frac{9.8 \times 1. 8 ^{2}}{4 \times 3.1 4 ^{2}} = \frac{31.752}{39.44} \approx 0.805$ m.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A pendulum with a heavier bob has a longer time period.
Reason (R): The time period depends on the acceleration due to gravity.
(a) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(b) (A) is false and (R) is true.
(c) (A) is true and (R) is false.
(d) Both (A) and (R) are false.
Hint
Which quantities appear in the formula?
Answer
(b). T does not depend on the mass of the bob, so A is false; R is true but does not support A.
Q6 · 3 marks · Short answerHOTS / competency
Pendulum P has length $L$ and pendulum Q has length $4 L$ (same place). Both are released together from the same side at the same instant. After how many oscillations of P will both next be at the release position together? Give the reasoning.
Hint
Compare their periods first.
Answer
$T_{Q} = 2 T_{P}$ . P completes 2 oscillations in the time Q completes 1, so both are at the release position together after 2 oscillations of P (1 of Q).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.About what is the time period of a 1 m simple pendulum on the Earth?Show answer
About 2 s (T = 2π√(l/g)).
2.What is the effective length?Show answer
Distance from the suspension point to the centre of the bob.
3.Time period of a seconds pendulum?Show answer
2 s.
4.Slope of the l–T² graph?Show answer
g/4π².
5.Does T depend on the mass of the bob?Show answer
No.

Simple Pendulum: Time Period, Frequency and the l–T² Graph: frequently asked questions

What is the time period of a simple pendulum?

The time for one complete oscillation, T = 2π√(l/g) for small swings. Doubling the length multiplies T by √2.

Why do we measure 20 oscillations instead of one?

Reaction-time error in starting and stopping the stopwatch is spread over 20 oscillations, so the error in one period becomes 20 times smaller.

What does the slope of the l versus T² graph give?

g/(4π²); multiplying the slope by 4π² gives the value of g.