Drag the bob and let go: where you release it is the amplitude
Controls
Gravitational acceleration, g (m/s²)
Readings
- Angular displacement, θ
- 10.0°
- Bob speed, v
- 0.00m/s
- Formula period, T = 2π√(L/g)
- 2.007s
- Measured period
- measuring…
- Frequency, f
- 0.498Hz
- Kinetic energy, Eₖ
- 0.000J
- Potential energy, Eₚ
- 0.074J
How to use this simulation
- Move the length slider: four times the length doubles the measured period.
- Change the bob mass: the bob grows and shrinks, the period does not move.
- Raise the amplitude from 5° to 70°: at large swings the measured period pulls ahead of the formula.
- With damping at zero the total-energy bar stays flat; add damping and it slowly drains.
From the playground swing to the pendulum
Think about a playground swing. Someone gives you one push and you go forward, stop for a split second, and come back along the same arc, again and again. Here is the odd part: a gentle push or a harder one, the time for one forward-and-back trip is almost the same.
Now think of an old grandfather clock. Behind the glass a brass disc swings with a steady tick-tock. That disc is a pendulum, and the clock keeps time because every swing takes the same amount of time. For about three centuries, this one fact was how the world kept its clocks right.
In physics, the cleanest, idealised version of that swing or clock is called a simple pendulum: a string with a small heavy ball on the end, nothing more. With something that plain you can measure the strength of Earth's gravity, and even predict how the same pendulum would behave on the Moon.
Starting from zero: why does a pendulum come back?
When the pendulum hangs straight down and still, the string's pull and the bob's weight act along the same vertical line. Nothing pushes the bob sideways, so it can stay there. That is the equilibrium position.
Pull the bob to one side and things change. Its weight mg still points straight down, but the string now pulls at a slant. The part of the weight along the arc, mg·sinθ, pushes the bob back toward the middle. A force that always tries to return an object to equilibrium is called a restoring force.
Let go, and the bob speeds up as it falls toward the middle. It cannot stop there, because that is where it is moving fastest; its inertia carries it past to the other side. There the restoring force pulls the other way, the bob slows, stops and returns. That back-and-forth is an oscillation.
A handy picture: roll a marble inside a bowl. Push it up the side and let go; it rolls to the bottom, overshoots up the far side and rolls back. A pendulum bob is really rolling in an invisible bowl, and the string sets the shape of that bowl.
Words you need to know
These words turn up in every chapter and every exam paper. Each one is explained in plain words with an everyday example.
| Term | Plain meaning | Everyday example |
|---|---|---|
| Oscillation | The bob going from one end to the other and back to where it started | A swing going forward and coming back once |
| Time period (T) | Time for one complete oscillation, in seconds (s) | One tick-tock of a pendulum clock |
| Frequency (f) | Complete oscillations per second; f = 1/T, in hertz (Hz) | How many times a swing returns each second |
| Amplitude | The biggest displacement from equilibrium (angle or distance) | How high the swing rises |
| Effective length (L) | Distance from the point of support to the centre of the bob | String length plus the bob's radius |
| Bob | The small heavy object on the end of the string | The brass disc of a clock pendulum |
| Simple harmonic motion (SHM) | Motion whose acceleration is proportional to displacement and points back to equilibrium | A pendulum at small angles; a mass on a spring |
The time period formula of a simple pendulum
Here is the main result. For small amplitudes the period depends on only two things: the effective length L and the gravitational acceleration g. The mass drops out and so does the amplitude, and that is the most surprising thing about the formula.
T = 2π√(L/g)time period
f = 1/T = (1/2π)√(g/L)frequency
g = 4π²L / T²finding g
Where the formula comes from (the easy version)
Pull the bob through an angle θ and the restoring force is F = −mg·sinθ. For small angles measured in radians, sinθ is almost exactly θ, and the displacement along the arc is x = Lθ. So F ≈ −mg·(x/L).
Put that into Newton's second law F = ma and you get a = −(g/L)·x: the acceleration is proportional to the displacement and opposite to it. That is the definition of simple harmonic motion, a = −ω²x, which gives ω² = g/L.
Since T = 2π/ω, T = 2π√(L/g). Notice that the mass m cancelled from both sides. That cancellation is exactly why the period does not depend on mass.
a = −(g/L)·x ⇒ ω = √(g/L) ⇒ T = 2π√(L/g)
Law 1: the law of isochronism
For small amplitudes (roughly up to 10°) every swing takes the same time. A smaller swing covers a shorter path but moves more slowly, and the two effects balance. That is why a clock pendulum keeps good time even as its swing slowly shrinks.
Law 2: the law of length
With g fixed, T ∝ √L. Four times the length gives twice the period: a 1 m pendulum has T = 2.01 s, a 4 m pendulum has T = 4.01 s. A swing on long ropes moves lazily; one on short ropes is quick.
Law 3: the law of acceleration
With the length fixed, T ∝ 1/√g. Where gravity is weaker the pendulum swings more slowly. On the Moon g = 1.62 m/s², so a 1 m pendulum would take 4.94 s per swing, about 2.46 times as long as on Earth.
Law 4: the law of mass
With the same length and g, changing the bob's mass, size or material does not change the period. Whether your little brother sits on the swing or you do, on the same ropes you swing in the same rhythm.
The energy game: kinetic and potential energy trading places
At the very top of a swing you hang still for an instant and feel that flutter in your stomach. At that moment all your energy is potential energy. As you drop, height turns into speed; at the bottom you are fastest and all the energy is kinetic.
Without air resistance or friction, no energy is lost in this trade and the total mechanical energy stays the same. Watch the bars on the right of the simulation: Eₖ and Eₚ rise and fall, but the total E bar does not move.
Let's put numbers on it. A 0.5 kg bob on a 1 m string is pulled to 10°. It is then h = L(1 − cosθ) = 1.52 cm above the lowest point, so its potential energy is mgh = 0.074 J. All of it becomes kinetic energy at the bottom, so the top speed is v = √(2gh) = 0.55 m/s.
In real life, air drag and friction at the pivot steal a little energy on every swing, so a swing nobody pushes slowly comes to rest. That is damped oscillation. A pendulum clock hides a spring or a hanging weight that gives the pendulum a tiny kick each swing to replace what was lost.
Eₚ = mgL(1 − cosθ)potential energy
Eₖ = ½mv²kinetic energy
Eₖ + Eₚ = constantwithout damping
Try it yourself in the simulation
Seeing it happen sticks far better than memorising it. Run these little experiments in the simulation above, and before each one, guess what will happen.
- Take the length from 1 m to 2.5 m. Does the swing slow down? By roughly how much does the measured period grow?
- Take the mass from 0.1 kg to 2 kg. The bob gets bigger, but watch the measured period.
- Raise the amplitude from 10° to 70°. Do the measured period and the formula still agree?
- Switch to the Moon, Mars and Jupiter. Where does the pendulum swing fastest?
- Add damping. Watch the total energy bar and the θ graph shrink.
- Drag the bob with your mouse or finger to any angle and let go.
Solved problems, step by step
Let's do some numbers. Writing down what is given and what is asked before you start cuts mistakes in half. Take g = 9.8 m/s² unless a problem says otherwise.
Problem 1: period of a 1 m pendulum
Given L = 1 m and g = 9.8 m/s². T = 2π√(L/g) = 2 × 3.1416 × √(1/9.8) = 2.01 s. The frequency is f = 1/T = 0.50 Hz, a little under half a swing per second.
Problem 2: finding g in the lab
A 1 m pendulum makes 20 complete oscillations in 40.1 s. One oscillation takes T = 40.1/20 = 2.005 s.
Then g = 4π²L/T² = 4 × (3.1416)² × 1 / (2.005)² = 9.82 m/s². Timing many swings instead of one spreads your stopwatch reaction error over all of them, so it shrinks.
Problem 3: length of a seconds pendulum
A seconds pendulum has a period of 2 s, so each one-way swing takes 1 s. Put T = 2 into L = gT²/4π² and you get L = g/π² = 0.993 m, almost exactly one metre.
Problem 4: the same pendulum on the Moon
On Earth the 1 m pendulum has T = 2.01 s. On the Moon, g = 1.62 m/s², so T = 2π√(1/1.62) = 4.94 s. On Mars (g = 3.71 m/s²) the same pendulum has T = 3.26 s.
Problem 5: what length doubles the period?
Since T ∝ √L, doubling T means doubling √L, which means making L four times longer. Going from 1 m to 4 m takes T from 2.01 s to 4.01 s. Cut the length to a quarter (0.25 m) and T halves to 1.00 s.
Length vs period at a glance
Every number in this table comes from T = 2π√(L/g). Notice that when the length goes up four times (0.25 → 1 m), the period exactly doubles.
| Length L (m) | Period T (s), Earth | Frequency f (Hz) | Period T (s), Moon |
|---|---|---|---|
| 0.25 | 1.00 | 1.00 | 2.47 |
| 0.50 | 1.42 | 0.70 | 3.49 |
| 1.00 | 2.01 | 0.50 | 4.94 |
| 1.50 | 2.46 | 0.41 | 6.05 |
| 2.00 | 2.84 | 0.35 | 6.98 |
| 2.50 | 3.17 | 0.32 | 7.81 |
How wrong is the formula at large amplitudes?
T = 2π√(L/g) is an approximation, because we assumed sinθ ≈ θ. At bigger angles sinθ falls below θ, the restoring force is a bit weaker than the approximation says, and the pendulum takes a little longer. Here is the comparison with the exact result for a 1 m pendulum:
| Amplitude | Formula T (s) | Exact T (s) | Difference |
|---|---|---|---|
| 5° | 2.007 | 2.008 | 0.05% |
| 10° | 2.007 | 2.011 | 0.19% |
| 20° | 2.007 | 2.022 | 0.77% |
| 30° | 2.007 | 2.042 | 1.74% |
| 45° | 2.007 | 2.087 | 4.00% |
| 60° | 2.007 | 2.154 | 7.32% |
| 70° | 2.007 | 2.212 | 10.21% |
Mistakes almost everyone makes
These turn up again and again in exam scripts and lab vivas. Knowing them in advance means you can dodge them.
- "A heavier bob swings faster." No: at small amplitudes the period does not depend on mass.
- Measuring only the string. Effective length = string length + radius of the bob.
- Counting half a swing (one end to the other) as a full oscillation. A full one ends back where it started.
- "Double the length, double the period." No: the period grows by √2; you need four times the length to double it.
- Measuring g with a huge amplitude (30°–40°). The formula is no longer accurate there.
Pendulums in real life
Clocks: for centuries the pendulum clock was the most accurate clock there was. In hot weather a metal rod expands slightly, the period grows and the clock runs slow, so fine clocks used rods that barely change length with temperature. The nut under the bob lets you move it up or down to adjust the time.
Swings: push a swing in rhythm and it climbs higher each time. When your pushes match the swing's own natural frequency, that is resonance. Push at the wrong moment and you slow it down instead.
Foucault's pendulum: a very long, heavy pendulum left swinging for hours seems to turn its plane of swing slowly. The plane stays put; the Earth turns underneath it. It was the first direct demonstration that the Earth rotates.
Tall buildings: some skyscrapers hang a huge mass of hundreds of tonnes near the top, called a tuned mass damper. When wind or an earthquake sways the building, the mass swings out of step with it and reduces the sway.
Measuring g: geologists once carried precise pendulums from place to place to detect tiny differences in g, which could hint at dense ore bodies underground.
Exam corner: questions you will actually see
Pendulum questions follow a few familiar patterns. Here are the common ones with short model answers.
Definitions and short answers
Q: What is a seconds pendulum? A: A pendulum whose time period is exactly 2 seconds.
Q: What is the effective length of a pendulum? A: The distance from the point of suspension to the centre of gravity of the bob.
Reasoning questions
Q: Why does a pendulum clock lose time on a mountain top? A: g is smaller at altitude, and T ∝ 1/√g, so each swing takes longer and the clock runs slow.
Q: Why is the motion not SHM at large amplitudes? A: sinθ ≈ θ no longer holds, so the restoring force is no longer proportional to the displacement.
Numerical and graph questions
Numerical: find g from "L = 1 m, 20 oscillations in 40.1 s". Divide to get T, then use g = 4π²L/T², giving 9.82 m/s² as in Problem 2.
Graph: plot L on the x-axis against T² on the y-axis. The result is a straight line through the origin with slope 4π²/g, so g = 4π²/slope. Averaging over many lengths makes this more accurate than a single reading.
Quick revision summary
Read just this the night before the exam and the key ideas will come back.
- Simple pendulum = light inextensible string + point-like heavy bob + frictionless support.
- Restoring force F = −mg·sinθ; at small θ the motion is SHM.
- T = 2π√(L/g), f = 1/T, g = 4π²L/T².
- Four laws: isochronism, length (T ∝ √L), acceleration (T ∝ 1/√g), mass (no effect).
- Seconds pendulum: T = 2 s, L ≈ 0.993 m.
- All potential energy at the ends, all kinetic at the middle; the total is constant.
Frequently asked questions
What is a simple pendulum?
An idealised system: a heavy point-like bob hung from a fixed support by a light, inextensible string, swinging freely without friction. Real pendulums come close to it when the string is light and the swing is small.
What is the formula for the period of a simple pendulum?
T = 2π√(L/g), where L is the effective length and g the gravitational acceleration. It holds for small amplitudes, roughly below 10°.
Does mass affect the period of a pendulum?
No. A heavier bob feels a bigger gravitational pull but also needs a bigger force to accelerate. The two effects cancel exactly, so m does not appear in T = 2π√(L/g).
Why is a pendulum SHM only for small angles?
The restoring force is mg·sinθ. Only when sinθ ≈ θ is that force proportional to the displacement, which is the condition for SHM. At larger angles the period grows beyond 2π√(L/g).
What is a seconds pendulum?
A pendulum with a period of exactly 2 s. From L = g/π², its length is about 0.993 m where g = 9.8 m/s².
How do you find g using a simple pendulum?
Time many oscillations, divide to get T, and use g = 4π²L/T². For better accuracy, repeat for several lengths and take g from the slope of the L against T² graph.
Why does a pendulum clock run slow in summer?
Heat expands the pendulum rod, which increases the effective length. Since T ∝ √L the period grows, so every tick takes longer and the clock falls behind.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
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