PhysicsICSE Class 10 · Conservation of Energy

Conservation of Energy in a Simple Pendulum: Energy at Extreme and Mean Positions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

In a swinging simple pendulum, the bob is momentarily at rest and highest at each extreme, so its energy is all potential. At the mean (lowest) position it moves fastest, so its energy is all kinetic. In between it is part of each, and without air resistance PE + KE stays constant.

At A and B the bob is highest with all PE; at O it is lowest and fastest with all KE, and PE + KE stays equal to mgh.

Illustration drawn for QuestionBuddy.

Understand Conservation of Energy in a Simple Pendulum

A simple pendulum is a small heavy bob hung from a rigid support by a light, inextensible string. Pulled aside and released, it swings to and fro about its mean position (the lowest point, where the string is vertical). The farthest points on either side are the extreme positions. The syllabus treats the pendulum qualitatively: you must describe the energy changes and explain them by the principle of conservation of energy.

**Taking the mean position as the reference level ( $U = 0$ there):**

Position	Height	Speed	PE	KE
Extreme (A or B)	Highest, $h$	Zero (momentarily at rest)	Maximum, $m g h$	Zero
Mean (O)	Lowest	Maximum	Zero	Maximum, $m g h$
Between	In between	In between	Part	Part

One swing, step by step (A → O → B).
1. At A the bob is at rest: all its mechanical energy is potential.
2. From A to O it falls: PE decreases and KE increases by the same amount.
3. At O all the energy is kinetic; the bob moves fastest.
4. From O to B it rises: KE changes back into PE.
5. At B it stops for an instant with all PE again, and B is at the same height as A (no air resistance).

At every point, $U + K = m g h$ , where $h$ is the height of the extreme position above the mean position. This is the principle of conservation of energy at work.

Why the swing dies away. In practice, air resistance on the bob and friction at the support convert a little mechanical energy into heat in each swing. So the bob rises slightly less each time: the amplitude decreases, and the pendulum finally stops at the mean position. Mechanical energy is lost, but total energy is still conserved: the 'lost' energy is now heat in the air and the support.

The time for one swing hardly changes as the amplitude slowly decreases, which is why pendulums were used in clocks. A clock's spring or battery feeds in just enough energy each swing to make up for the loss.

Useful results (simple numericals).

Speed at the mean position: $m g h = \frac{1}{2} m v^{2}$ , so $v = 2 g h$ , where $h$ is the vertical rise of the bob at the extreme (not the length of the string).
Height reached from speed $v$ at the mean position: $h = \frac{v ^{2}}{2 g}$ .
At a height $y$ above the mean position, the fraction of the energy that is kinetic is $\frac{h - y}{h}$ .
The mass cancels: bobs of different mass released from the same height pass the mean position at the same speed.

Conservation of Energy in a Simple Pendulum: formulas and key facts

Energy at any point: $U + K = m g h$; h = height of the extreme above the mean position; no air resistance.
Speed at the mean position: $v = 2 g h$; Independent of the mass of the bob.
Rise from the mean position: $h = \frac{v ^{2}}{2 g}$; v = speed at the lowest point.

How to solve Conservation of Energy in a Simple Pendulum questions

Pendulum questions are mostly descriptive; any numbers come from equating the PE at the extreme with the KE at the mean position.

1Take the mean (lowest) position as the reference level.
2At an extreme: speed zero, energy all PE ( $m g h$ ).
3At the mean position: energy all KE ( $\frac{1}{2} m v^{2}$ ).
4Equate them: $m g h = \frac{1}{2} m v^{2}$ , using the vertical height $h$ , not the string length.
5For energy loss, compare the PE at successive extremes: loss $= m g (h_{1} - h_{2})$ .
6In explanations, always name where the lost energy goes: heat due to air resistance and friction at the support.

Conservation of Energy in a Simple Pendulum 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 demolition wrecking ball

Suppose a 500 kg wrecking ball hangs from a crane and is pulled back until it is 1.25 m above its lowest point, then released. ( $g = 10 m s^{- 2}$ , ignore air resistance)

Find its speed and kinetic energy as it swings through the lowest point, where it strikes the wall.

Try it first, then show the solutionSolution

$v = 2 \times 10 \times 1.25 = 5$ m/s.
$K = m g h = 500 \times 10 \times 1.25 = 6250$ J.

Answer: 5 m/s; 6250 J, delivered at the lowest point.

Memory hook: Pull back higher, hit harder: the PE at the extreme becomes the KE at the bottom.

A pendulum in oil

Imagine two identical pendulums started from the same height, one swinging in air and one with its bob in a jar of thick oil.

Which stops sooner, and is energy conserved in each?

Try it first, then show the solutionSolution

Oil resists the bob far more than air, so it turns mechanical energy into heat much faster.
In both, the mechanical energy lost appears as heat, so total energy is conserved.

Answer: The one in oil stops much sooner; total energy is conserved in both.

Memory hook: More resistance, faster fade, same total energy.

Tricks to remember Conservation of Energy in a Simple Pendulum

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

Ends are Potential, Middle is Kinetic
E-P, M-K: at the Ends the energy is Potential; in the Middle it is Kinetic.
Use it when: MCQs on extreme and mean positions.
Stop at the top
The bob stops for an instant at the top of each swing, so KE is zero there.
Use it when: Velocity and KE at the extremes.
Heat steals the height
Each swing, a little energy turns to heat, so the bob reaches a little less height.
Use it when: Explaining the decreasing amplitude.
Height, not string
Use the vertical rise of the bob in mgh, never the string length.
Use it when: Pendulum numericals.

Conservation of Energy in a Simple Pendulum: solved examples

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

Example 1: Energy at the mean position · 1 mark · MCQBoard level

As the bob of a simple pendulum passes through its mean position, it has:
(a) maximum potential energy
(b) maximum kinetic energy and minimum potential energy
(c) zero kinetic energy
(d) equal kinetic and potential energy

Try it first, then show the solutionSolution

The mean position is the lowest point (least PE), and all the PE lost has become KE, so the speed is greatest.

Answer: (b) maximum kinetic energy and minimum potential energy

Example 2: Describing one swing · 3 marks · Short answerBoard level

A pendulum bob is released from an extreme position A. It swings through the mean position O to the other extreme B. Describe the energy changes from A to B, and state how the height of B compares with that of A if there is no air resistance.

Try it first, then show the solutionSolution

At A the bob is at rest: all its energy is PE. From A to O, PE changes into KE. (1)
At O the energy is all KE (maximum speed). From O to B, KE changes back into PE. (1)
At B the bob is momentarily at rest with all PE. With no air resistance, B is at the same height as A, since PE + KE stays constant. (1)

Answer: PE → KE → PE; B is at the same height as A.

Example 3: Speed at the lowest point · 2 marks · Short answerHarder board level

A pendulum bob of mass 0.2 kg is pulled aside until it is 0.45 m above its lowest position, and released. Find its speed and kinetic energy at the lowest position. ( $g = 10 m s^{- 2}$ , no air resistance)

Try it first, then show the solutionSolution

$v = 2 \times 10 \times 0.45 = 9 = 3$ m/s. (1)
$K = m g h = 0.2 \times 10 \times 0.45 = 0.9$ J. (1)

Answer: 3 m/s; 0.9 J

Example 4: How high does it swing? · 2 marks · Short answerHarder board level

A bob hanging at rest is given a sharp tap that sets it moving at 1.4 m/s from its lowest position. Through what vertical height does it swing up before stopping? ( $g = 10 m s^{- 2}$ , no air resistance)

Try it first, then show the solutionSolution

All the KE at the bottom becomes PE at the extreme: $\frac{1}{2} m v^{2} = m g h$ . (1)
$h = \frac{1. 4 ^{2}}{2 \times 10} = \frac{1.96}{20} = 0.098$ m $= 9.8$ cm. (1)

Answer: 9.8 cm

Example 5: Energy shares part-way · 3 marks · Short answerHOTS / competency

A bob is released from rest 0.2 m above its lowest position. When it is 0.05 m above the lowest position: (a) what fraction of its energy is kinetic? (b) what is its speed? ( $g = 10 m s^{- 2}$ , no air resistance)

Try it first, then show the solutionSolution

(a) It has fallen $0.2 - 0.05 = 0.15$ m out of 0.2 m, so $\frac{K}{total} = \frac{0.15}{0.2} = \frac{3}{4}$ . (1½)
(b) $v = 2 \times 10 \times 0.15 = 3 \approx 1.73$ m/s. (1½)

Answer: (a) Three quarters (b) About 1.73 m/s

Example 6: Case study: a playground swing · 4 marks · Case-basedHOTS / competency

A child sits on a park swing; together they have a mass of 25 kg. The swing is pulled back until the seat is 0.8 m above its lowest position, then released. ( $g = 10 m s^{- 2}$ )
(a) Find the energy at the extreme position.
(b) Find the speed at the lowest position, ignoring losses.
(c) After some swings, the seat rises only 0.6 m above the lowest position. How much mechanical energy has been lost, and where has it gone?
(d) How does a parent keep the swing going, in energy terms?

Try it first, then show the solutionSolution

(a) $U = 25 \times 10 \times 0.8 = 200$ J. (1)
(b) $v = 2 \times 10 \times 0.8 = 4$ m/s. (1)
(c) PE now $= 25 \times 10 \times 0.6 = 150$ J, so 50 J lost, as heat through air resistance and friction at the hooks. (1)
(d) Each push does work on the swing, adding mechanical energy to replace what is lost. (1)

Answer: (a) 200 J (b) 4 m/s (c) 50 J, to heat (d) Pushes add energy by doing work.

Where students lose marks in Conservation of Energy in a Simple Pendulum

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

Mistake: Putting maximum speed at the extremes
The bob stops for an instant at each extreme; it is fastest at the mean position.
Wrong working
'At the extreme the bob has maximum KE because it is farthest away.'
Right working
'At the extreme, KE is zero and PE is maximum.'
Do this: Link the extremes to 'stops and turns back'.
Usually costs: MCQ and definition marks.
Mistake: Saying the energy is destroyed when the pendulum stops
The mechanical energy changes into heat; total energy is conserved.
Wrong working
'Energy is lost, so the pendulum stops.'
Right working
'Mechanical energy is converted into heat by air resistance and friction at the support.'
Do this: Name the form the energy changes into.
Usually costs: The reasoning mark in 'why does it stop' questions.
Mistake: Using the string length or the wrong reference height
The PE that turns into KE depends on the vertical drop from the extreme to the mean position.
Wrong working
A 0.1 kg bob, 1.3 m above the floor at the extreme and 1.0 m at the mean: $K = 0.1 \times 10 \times 1.3 = 1.3$ J.
Right working
$K = 0.1 \times 10 \times (1.3 - 1.0) = 0.3$ J.
Do this: Measure heights from the mean position.
Usually costs: The method and answer marks.
Mistake: Saying the time period decreases as the swing dies down
As the amplitude falls, the time for one swing stays almost the same; only the size of the swing shrinks.
Wrong working
'The pendulum swings faster and faster until it stops.'
Right working
'The amplitude decreases; each swing takes about the same time.'
Do this: Separate amplitude (size of swing) from time period.
Usually costs: Explanation marks.

Exam tips

The syllabus keeps the pendulum qualitative: learn to describe PE and KE at the extremes, the mean position and in between, with the reason.
Always state the reference level (the mean position) before saying PE is zero there.
In 'why does it stop', name both causes (air resistance, friction at the support) and the form the energy takes (heat).
For any number work, use $m g h = \frac{1}{2} m v^{2}$ with $h$ in metres.

Practice questions on Conservation of Energy in a Simple Pendulum

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
At an extreme position of its swing, the velocity of a pendulum bob is:
(a) maximum
(b) zero
(c) half its maximum value
(d) equal to $2 g h$
Hint
It turns back there.
Answer
(b) zero
Q2 · 2 marks · Short answerHarder board level
A swinging pendulum slowly comes to rest. (a) Why? (b) Does this go against the principle of conservation of energy? Explain.
Hint
Name where the energy goes.
Answer
(a) Air resistance and friction at the support convert some mechanical energy into heat in every swing, so the amplitude keeps decreasing. (b) No: the mechanical energy lost appears as heat, so the total energy is unchanged.
Q3 · 3 marks · Short answerBoard level
Name the position(s) of a swinging bob at which its mechanical energy is (a) all PE, (b) all KE, (c) shared equally between PE and KE.
Hint
Take the lowest point as the reference.
Answer
(a) The two extreme positions. (b) The mean (lowest) position. (c) The two points at half the height of the extremes above the mean position.
Q4 · 2 marks · Short answerHarder board level
A bob of mass 0.05 kg is released from rest 12.8 cm above its lowest position. Find its speed and kinetic energy at the lowest position. ( $g = 10 m s^{- 2}$ , no air resistance)
Hint
Convert 12.8 cm to metres.
Answer
$v = 2 \times 10 \times 0.128 = 2.56 = 1.6$ m/s; $K = 0.05 \times 10 \times 0.128 = 0.064$ J.
Q5 · 2 marks · Short answerHarder board level
Two identical pendulums are started together from the same height. One swings in air; the other in a vacuum chamber. Compare their motion after several minutes, and explain using energy.
Hint
Which one loses energy to the surrounding air?
Answer
The one in the vacuum keeps nearly the same amplitude, as no air resistance converts its mechanical energy to heat (only slight friction at the support). The one in air swings with a steadily smaller amplitude.
Q6 · 2 marks · Short answerHarder board level
A pendulum bob is replaced by one of twice the mass, released from the same height. What happens to (a) its speed at the mean position, (b) its kinetic energy there? Explain.
Hint
Look at $v = 2 g h$ and $K = m g h$ .
Answer
(a) Unchanged, because $v = 2 g h$ does not depend on mass. (b) Doubled, because $K = m g h$ is proportional to mass.
Q7 · 3 marks · Short answerHOTS / competency
A pendulum is released with its bob 10 cm above the lowest position. In each swing (extreme to extreme) it loses 20% of the mechanical energy it had at the start of that swing. Find the height the bob reaches at the end of the second swing.
Hint
The height is proportional to the energy left.
Answer
After swing 1: $10 \times 0.8 = 8$ cm. After swing 2: $8 \times 0.8 = 6.4$ cm.

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.KE of a pendulum bob at an extreme position?Show answer
Zero.
2.Where is the speed of the bob greatest?Show answer
At the mean (lowest) position.
3.Speed at the mean position for a bob released 20 cm above it ( $g = 10$ m s⁻²)?Show answer
2 m/s.
4.What happens to the amplitude of a pendulum swinging in air?Show answer
It gradually decreases.
5.Does a pendulum coming to rest violate conservation of energy?Show answer
No; its mechanical energy becomes heat.

Conservation of Energy in a Simple Pendulum: frequently asked questions

Where is the potential energy of a pendulum maximum?

At the two extreme positions, where the bob is highest and momentarily at rest. All its mechanical energy is potential there.

Why does a pendulum not keep swinging forever?

Air resistance and friction at the support convert a little of its mechanical energy into heat on every swing, so each swing is smaller until it stops.

Do I need to solve pendulum numericals for ICSE Class 10?

The syllabus treats the pendulum qualitatively. Simple energy calculations, such as the speed at the lowest point from the height, use the same mgh = ½mv² as other conservation problems.