Does a heavier bob or a bigger swing change the timing?
Competences: L1 fair testing · L2 clocks & swings · L4 force diagrams · L5 simulation
Four quick cards: where acceleration peaks, spring period ratios, components of a force, and small angles. Answer each in your head before you reveal it.
In the school physics lab, Tunde and Zainab each make a confident claim about what slows a pendulum down. Mrs Okafor sets them a fair test — and a puzzle about the okada.
The 3D story couldn't start on this device. Read the dialogue below instead.
Scene 1 · Tunde (holding a brass bob and a wooden bob): “This heavy one will swing slowly, steady like a loaded lorry. I'm sure.”
Scene 2 · Zainab: “And if I pull it back further, it must take longer — it has further to go.”
Scene 3 · Mrs Okafor: “Two claims, two fair tests. Change one thing at a time and time 20 swings.”
Scene 4 · Mrs Okafor (pointing at a pendulum about the length of a finger): “And here's a puzzle for later. This little pendulum keeps exactly the okada's time. Why that length?”
Scene 5 · Zainab (two pendulums of equal length, one brass, one wood, swinging side by side): “Tunde… they're together.”
Rank what you think changes a pendulum's period: bob mass, release angle, thread length, the planet you're on. Pin your idea before you test it.
Drag or tap each of the nine cards into a bin: Changes T, No change, or Only at large angles (changes only when the swing is wide).
Change one setting, press Release & time 20, and watch the stopwatch. Every time comes from the exact large-angle period of the pendulum model.
0.50 m, 10°: time 50 g, then 100 g.
100 g: time 5°, 10° and 30°.
10°: time 0.25 m, 0.50 m, 1.00 m.
0.50 m on the Moon: time again.
The 3D bench couldn't start on this device. Use the expected results and worked examples below.
| Test · value | t₂₀ / s | T / s | Changed? |
|---|---|---|---|
| Your timings appear here, one row per run. | |||
Variables tested: 0 of 4 (mass · angle · length · g)
| Test (10°, 0.50 m, Earth unless stated) | t / s (20 osc) | T / s | Changed? |
|---|---|---|---|
| Mass 50 g | 28.44 | 1.42 | — |
| Mass 100 g | 28.44 | 1.42 | ✗ no |
| Angle 5° | 28.40 | 1.42 | ✗ no |
| Angle 30° | 28.88 | 1.44 | slightly (≈ 1.7 % longer than at 5°) |
| Length 0.25 m | 20.11 | 1.01 | ✓ yes |
| Length 1.00 m | 40.22 | 2.01 | ✓ yes (quadrupling l doubles T) |
| Moon (g = 1.6), 0.50 m | 70.38 | 3.52 | ✓ yes |
Precautions (as written in a report): only one variable was changed at a time; the angle was kept small (≤ 10°) for the fair tests; the bob swung in one plane, not in a circle; timing started as the bob passed the rest mark.
Sources of error: reaction time at the start and stop (shared out over 20 oscillations); the support moved slightly, adding to the effective length.
Stand up and use your own arm as a pendulum.
Now we connect the bench to the formula WAEC, NECO and JAMB expect — and see exactly why WAEC says “small angle”.
Forces on the bob: the weight mg splits into mg cos θ along the thread (balanced by the tension) and mg sin θ along the arc, pulling the bob back towards the lowest point. That component is the restoring force.
The small-angle step: for small θ (in radians), sin θ ≈ θ ≈ x/l, so the restoring force is F ≈ −mg(x/l) and a = F/m = −(g/l)x. The mass cancels. This matches a = −ω²x with ω² = g/l, so T = 2π√(l/g). For large angles sin θ falls noticeably below θ, the force is no longer proportional to x, and the period grows.
| Simple pendulum | Loaded spring | |
|---|---|---|
| Period formula | T = 2π√(l/g) | T = 2π√(m/k) |
| Depends on mass? | no | yes, T ∝ √m |
| Depends on g? | yes, T ∝ 1/√g | no |
| Depends on small amplitude? | no | no |
| Restoring force from | weight component mg sin θ | spring tension kx |
Tap a release angle. The numbers come from the same pendulum models as the bench and the zoom.
Drag the angle and watch the components. Swap the 50 g bob for 100 g: every force arrow doubles, but the green acceleration arrow g sin θ stays the same. The small 3D view shows the real pendulum swinging.
The force zoom couldn't start on this device.
Try each on paper first, then check the working. Watch the square roots — most lost marks come from leaving one out.
Find the period of a simple pendulum 0.90 m long (g = 9.8 m s⁻²).
T = 2π√(0.90 ÷ 9.8)T = 2π√0.0918 = 2 × 3.142 × 0.303T = 1.90 sA pendulum 0.40 m long has period 1.27 s. The thread is lengthened to 0.90 m. Find the new period.
T₂/T₁ = √(0.90 ÷ 0.40)T₂/T₁ = √2.25 = 1.5T₂ = 1.5 × 1.27 = 1.90 sA pendulum has period 2.0 s on Earth. Find its period on the Moon, where g is one-sixth of Earth's.
T_M/T_E = √(g_E/g_M) = √6√6 = 2.45T_M = 2.0 × 2.45 = 4.9 sA “seconds pendulum” ticks once per second, so its period is 2.0 s. Find its length (g = 9.8 m s⁻²).
l = gT²/4π²l = 9.8 × 2.0² ÷ 39.5 = 39.2 ÷ 39.5l = 0.99 mRun Tunde's test on the bench (50 g and 100 g at 0.50 m), then find the pendulum that keeps the okada's 0.50 s bounce.
Story numbers are a teaching model, not real measurements.
The big idea of Lesson 4: T = 2π√(l/g) — only length and gravity count, for small swings. A heavier bob is not slower, and the okada's bounce keeps the time of a 6.2 cm pendulum.
Choose an option for instant feedback and a full explanation. Two review questions from Lessons 2 and 3 come at the end.
Rate your confidence from 1 to 4 on each goal of Lesson 4.
I can name the restoring force on a bob.
I can explain the small-angle condition.
I can use T = 2π√(l/g).
I can solve length and g ratio problems.
I can contrast the pendulum and the spring.
T ∝ √l (1), so T₂/T₁ = √(1/4) = 0.5: the period is halved (1).
Type your attempt first to unlock each model solution.
You need: two pieces of thread about 80 cm long, a light key, a heavy padlock, a broom handle or rod laid across two chairs, tape, a tape measure and a phone stopwatch.
| Pendulum | Length (m) | t₁ for 20 (s) | t₂ for 20 (s) | T (s) |
|---|---|---|---|---|
| Key | 0.60 | 31.2 | 31.0 | 1.56 |
| Padlock | 0.60 | 31.1 | 31.2 | 1.56 |
| Key (shortened) | 0.30 | 22.1 | 21.9 | 1.10 |
A good answer shows that the key and the padlock at the same length stay in step and have the same period within timing error (about 1.56 s), so the mass of the bob does not change T, which disproves Tunde's “heavier is steadier”; halving the length shortens the period by a factor of about √2 (1.56 ÷ 1.10 ≈ 1.4), so length does change it. A strong answer explains why mass cancels (bigger weight but also bigger inertia), notes that the angle was kept small and the same for a fair test, and names a source of error: the lengths were measured to the top of the objects rather than their centres, or a draught pushed the light key more than the padlock.
Define: (a) a simple pendulum; (b) a seconds pendulum; (c) the effective length of a simple pendulum.
(a) A small heavy bob hung from a fixed point by a light, inextensible thread, free to swing to and fro in a vertical plane. (1) (b) A simple pendulum that takes exactly 2 s for one complete oscillation (each swing from one end to the other takes 1 s). (1) (c) The distance from the point of suspension to the centre of mass (centre) of the bob. (1)
Mark scheme note: In (a) “light thread” and “heavy (small) bob” are both needed. In (b) “period 1 s” is a common error and scores zero. In (c) “the length of the thread” alone is not accepted.
(a) Name the two quantities on which the period of a simple pendulum depends. (b) Name two quantities that do not affect it (for small swings). (c) Why must the swing be small?
(a) The length l and the acceleration due to gravity g. (1) (b) The mass of the bob and the amplitude (angle) of a small swing; also accept the material or colour of the bob. (1) (c) Only for small angles is sin θ ≈ θ, so that the restoring force is proportional to displacement and the motion is SHM. (1)
Mark scheme note: Both answers are needed in (a) and in (b). “Small angles are safer” scores zero in (c).
A bob of mass m on a thread of length l is displaced through a small angle θ, its displacement along the arc being x. (a) Draw and label the forces on the bob. (b) Write the restoring force. (c) Using the small-angle approximation, show that a = −(g/l)x. (d) Hence show that the motion is SHM and that T = 2π√(l/g). (e) Explain why the mass of the bob does not appear.
(a) Weight mg vertically down and tension along the thread towards the support; mg resolved into mg cos θ along the thread and mg sin θ along the arc. (1) (b) F = −mg sin θ (towards the lowest point). (1) (c) For small θ in radians, sin θ ≈ θ = x/l, so F ≈ −mg(x/l) and a = F/m = −(g/l)x. (1) (d) g/l is a positive constant, so a ∝ −x: SHM with ω² = g/l; T = 2π/ω = 2π√(l/g). (1) (e) m appears in both the restoring force and in F = ma, so it cancels. (1)
Mark scheme note: The small-angle step (sin θ ≈ θ = x/l) must be stated for (c). A formula quoted without derivation earns only (d) if the comparison with a = −ω²x is shown.
A pendulum in a school hall is 1.21 m long. (Take g = 9.8 m s⁻².) Calculate: (a) its period; (b) the time for 50 oscillations; (c) the length needed for a period of 1.10 s; (d) the frequency of the 1.10 s pendulum.
(a) T = 2π√(1.21 ÷ 9.8) = 2π × 0.3514 = 2.21 s (1) (b) t = 50 × 2.208 = 110 s (1) (c) l = 9.8 × 1.21 ÷ 39.48 = 0.300 m (1) (d) f = 1 ÷ 1.10 = 0.909 Hz (1)
Mark scheme note: Error carried forward into (b). In (c) forgetting to square T gives 0.273 m and loses the mark. Units required.
A clock pendulum has a period of 2.40 s on Earth (g = 9.8 m s⁻²). (a) Calculate its length. (b) Calculate its period on the Moon, where g is one-sixth of Earth's. (c) What length would give a period of 2.40 s on the Moon? (d) Explain why a pendulum clock taken to the Moon without adjustment would show the wrong time, and say whether it would gain or lose.
(a) l = 9.8 × 5.76 ÷ 39.48 = 1.43 m (1) (b) T_M = 2.40 × 2.449 = 5.88 s (1) (c) For the same T, l ∝ g, so l_M = 1.43 ÷ 6 = 0.238 m (0.24 m). (1) (d) Each swing takes longer, so the clock ticks fewer times per real hour (1): it loses time (runs slow). (1)
Mark scheme note: In (b) multiplying by 6 (14.4 s) loses the mark. In (c) “l × 6” is the inverse error. In (d) “gains” scores zero for the last mark.
Zainab changed one variable at a time and timed 20 oscillations:
| Test | Length / m | Bob mass / g | Angle / ° | t (20 osc.) / s |
|---|---|---|---|---|
| 1 | 0.60 | 50 | 8 | 31.10 |
| 2 | 0.60 | 100 | 8 | 31.12 |
| 3 | 0.60 | 100 | 4 | 31.08 |
| 4 | 0.30 | 100 | 8 | 22.00 |
| 5 | 1.20 | 100 | 8 | 43.98 |
(a) Calculate T for tests 1, 4 and 5. (b) What do tests 1 and 2 show? (c) What do tests 2 and 3 show? (d) Use tests 4 and 5 to show that T ∝ √l.
(a) T₁ = 31.10 ÷ 20 = 1.555 s; T₄ = 22.00 ÷ 20 = 1.100 s; T₅ = 43.98 ÷ 20 = 2.199 s. (1) (b) Doubling the mass left T unchanged (within 0.02 s in 31 s), so T does not depend on the bob's mass. (1) (c) Halving the (small) angle left T unchanged, so for small swings T does not depend on amplitude. (1) (d) l₅/l₄ = 1.20 ÷ 0.30 = 4 (1); T₅/T₄ = 2.199 ÷ 1.100 = 2.0 = √4, so T ∝ √l. (1)
Mark scheme note: All three periods are needed in (a). In (b) and (c) the conclusion must name the variable. In (d) the ratio comparison is required; “T increases with l” alone earns no marks.
For each change, say whether the period of (i) a simple pendulum and (ii) a mass on a spring increases, decreases or stays the same: (a) the mass of the bob or load is doubled; (b) the system is taken to the Moon; (c) the amplitude is halved (still small).
(a) Pendulum: same; spring: increases (×√2). (1) (b) Pendulum: increases (×√6); spring: same. (1) (c) Both stay the same. (1)
Mark scheme note: Each part needs both systems right for its mark.
Mrs Okafor's short pendulum keeps time with the okada's bounce (T = 0.50 s). (Take g = 9.8 m s⁻².) (a) Calculate its length. (b) Show that it equals the okada's static sag, mg/k, with m = 200 kg and k = 3.16 × 10⁴ N m⁻¹.
(a) l = 9.8 × 0.25 ÷ 39.48 = 0.062 m (6.2 cm). (1) (b) mg/k = 200 × 9.8 ÷ 31600 = 0.062 m: the same, because the two periods are equal when l/g = m/k. (1)
Mark scheme note: In (b) the numerical check alone earns the mark; the l/g = m/k reason is credit-worthy but not required. Story numbers are a teaching model, not real measurements.
Timings: 0–5 Do Now · 5–15 Wonder · 15–45 Investigate · 45–50 movement break (swing one arm loosely, then holding a bag: the rate hardly changes; bend the elbow — a shorter “pendulum” — and feel it quicken) · 50–65 Make Sense · 65–75 Apply · 75–80 Reflect.
Competences: L1 fair testing (one variable at a time) · L2 clock pendulums and playground swings · L4 force diagrams (resolving mg) · L5 using the simulation.
Support: a ratio card “T ∝ √l: four times the length, twice the period”. Stretch: why does a grandfather clock lose time if its rod expands in the afternoon heat?
Bench note: the bench uses the exact large-angle period, so 30° runs about 1.7 % slower than 5°; at 5° and 10° the difference (0.04 s in 20 swings) is below what a hand stopwatch can resolve.