What is g in our own lab, and how good is our measurement?
Four quick cards: rearranging the pendulum formula, straight-line graphs, why we time many swings, and bob mass. Answer each in your head before you reveal it.
Zainab's lift-off rule says the crates fly when ω²A reaches g. But what is g in our own lab? Mrs Okafor hands Tunde a pendulum, and he slips twice.
The 3D story couldn't start on this device. Read the dialogue below instead.
Scene 1 · Mrs Okafor: "The lift-off rule needs g. Textbooks say 9.8. Let's measure our own, here in Ogbomoso."
Scene 2 · Tunde lays the ruler along the thread and stops at the top of the bob: "Eighty centimetres. Done."
Scene 3 · Zainab: "The bob's centre is lower. The pendulum swings from the cork to the centre of the bob."
Scene 4 · Tunde starts the stopwatch as the bob leaves his hand and counts "One!" Mrs Okafor stops him.
Scene 5 · Mrs Okafor: "Small slips like these are exactly where WAEC candidates lose marks. Let's do it the examiner's way and see how much each slip costs."
Which single slip do you think spoils the value of g most: wrong length, wrong count, wrong starting point, or a wide swing?
A precaution is something that was done to improve the result, written in the past tense. A source of error is something that spoils the result. Drag each card, or tap it and then tap a bin.
r = 2.0 cm. Per length: Measure, Release, Time 20 twice. Fault none for the table.
Cork edge to bob top, then add r.
Chalk the rest mark; about 10°, one plane.
Start at the mark counting "zero".
0.400, 0.550, 0.700, 0.850, 1.000 m.
T² against l; g = 4π² ÷ slope.
The 3D bench couldn't start on this device. Use the expected results below.
Set a fault, then time a length twice.
Fill t̄ (2 d.p.), T = t̄ ÷ 20 (3 d.p.), T² (3 d.p.).
| l | t₁ | t₂ | t̄ | T | T² |
|---|---|---|---|---|---|
| m | s | s | s | s | s² |
Fill a row of the table correctly and its point lands here.
| l / m | t₁ / s | t₂ / s | t̄ / s | T / s | T² / s² |
|---|---|---|---|---|---|
| 0.400 | 25.36 | 25.44 | 25.40 | 1.270 | 1.613 |
| 0.550 | 29.82 | 29.76 | 29.79 | 1.490 | 2.219 |
| 0.700 | 33.57 | 33.63 | 33.60 | 1.680 | 2.822 |
| 0.850 | 37.06 | 37.00 | 37.03 | 1.852 | 3.428 |
| 1.000 | 40.13 | 40.19 | 40.16 | 2.008 | 4.032 |
T² is worked from the unrounded T, so squaring the displayed T (2.220 at 0.550 m, 3.430 at 0.850 m) is also accepted (±0.002 s²). The line passes through the origin. Slope triangle on the line from (0.400, 1.613) to (1.000, 4.032):
Precautions (past tense):
Sources of error (with effect): length to the top of the bob → every l too small by r, slope unchanged but the line misses the origin · counting from "one" → only 19 oscillations timed, T too small by 19/20, g too large by (20/19)² ≈ 1.11 · wide angle → T slightly too large, g slightly too small · reaction time (random) · conical swing → shorter period, g too large.
Why does a graph of T² against l give g, and what changes when you can't reach the ceiling?
From formula to straight line: T² = (4π²/g) l has the form y = mx with y = T², x = l and slope 4π²/g, so g = 4π²/slope. The graph averages many readings and shows a constant error (an intercept) that a single reading would hide.
Ceiling variant: if the ceiling is out of reach, measure the height y of the bob's centre above the floor; then l = H − y and T² = (4π²/g)(H − y). The T²–y line slopes downwards (slope −4π²/g), and it cuts the horizontal (y) axis, where T² = 0, at y = H, the ceiling height.
| Feature | Precaution | Source of error |
|---|---|---|
| Definition | something that was done to improve the result | something that spoils the result |
| Tense | past tense, first person ("I ensured…") | a noun phrase ("reaction time…") |
| Example | I started timing as the bob passed the rest mark | reaction time in starting the stopwatch |
| Marks under "precautions"? | yes, if specific and in the past tense | no, it scores zero there |
Tap a length to see its period and its point on the T²–l line (g = 9.79 m s⁻²). Then switch on length to top.
Pick a fault and the pendulum replays it in slow motion while the mini graph shows how the points and g shift. Switch to the ceiling rig to see y measured from the floor.
The 3D zoom couldn't start on this device. The mini graph below still works.
Try each on paper first, then check every step.
The best-fit line of a T²–l graph passes through (0.400 m, 1.613 s²) and (1.000 m, 4.032 s²). Find g.
s = (4.032 − 1.613) ÷ (1.000 − 0.400)s = 2.419 ÷ 0.600 = 4.03 s² m⁻¹g = 39.48 ÷ 4.03 = 9.79 m s⁻²A thread is 0.780 m from the cork to the top of a bob of radius 2.0 cm. Twenty oscillations take 35.90 s. Find g.
l = 0.780 + 0.020 = 0.800 mT = 35.90 ÷ 20 = 1.795 s, so T² = 1.795² = 3.222 s²g = 39.48 × 0.800 ÷ 3.222g = 31.58 ÷ 3.222 = 9.80 m s⁻²In Example 2, Tunde forgets the radius and uses l = 0.780 m. Find his g and the percentage error.
g = 39.48 × 0.780 ÷ 3.222 = 9.56 m s⁻²9.802 − 9.557 = 0.245 m s⁻²0.245 ÷ 9.80 × 100 = 2.5 % (too low), the same as r/l = 0.020 ÷ 0.800 = 0.025With the bob's centre 1.80 m above the floor, T² = 4.028 s²; at 2.20 m, T² = 2.417 s². Find g and the ceiling height H.
s = (2.417 − 4.028) ÷ (2.20 − 1.80) = −1.611 ÷ 0.40 = −4.03 s² m⁻¹g = 39.48 ÷ 4.03 = 9.80 m s⁻²Δy = 4.028 ÷ 4.03 = 1.00 mH = 1.80 + 1.00 = 2.80 mTake the class value g = 9.79 m s⁻² and the okada's ω = 12.57 rad s⁻¹. Does our own g confirm Zainab's danger line?
1 · Our class graph has a slope of 4.03 s² m⁻¹. What is g?
2 · The crates leave the carrier when the carrier's greatest acceleration…
3 · With ω = 12.57 rad s⁻¹, the danger amplitude A = g ÷ ω² is about…
The big idea of Lesson 5: plot T² against l; g = 4π²/slope. Precautions are things you did; errors are things that went wrong. With our g = 9.79 m s⁻², the lift-off amplitude is A = g/ω² = 9.79 ÷ 158 = 0.062 m = 6.2 cm.
Choose an option for instant feedback and a full explanation. Two review questions from earlier lessons come at the end.
Rate your confidence from 1 to 4 on each goal of Lesson 5.
I can set up the pendulum with l measured to the bob's centre.
I can fill a composite table to consistent decimal places.
I can deduce g from a large slope triangle.
I can handle the ceiling variant.
I can write precautions in the past tense.
g = 4π²/s (1) = 39.48 ÷ 4.10 = 9.63 m s⁻² (1).
Type your attempt first to unlock each model solution.
You need: a stone or padlock, about 1.2 m of strong thread, a door frame or nail to hang it from, a tape measure, a phone stopwatch and a sheet of graph paper.
| l (m) | t₁ (s) | t₂ (s) | Mean t (s) | T (s) | T² (s²) |
|---|---|---|---|---|---|
| 0.40 | 25.5 | 25.2 | 25.4 | 1.27 | 1.61 |
| 0.70 | 33.4 | 33.7 | 33.6 | 1.68 | 2.82 |
| 1.00 | 40.0 | 40.3 | 40.2 | 2.01 | 4.04 |
Slope from two points read from the drawn line, e.g. (0.20, 0.81) and (1.00, 4.04): slope = 3.23 ÷ 0.80 = 4.04 s² m⁻¹, so g = 39.48 ÷ 4.04 = 9.8 m s⁻².
A good answer shows a complete table with consistent decimal places, T² against l close to a straight line through the origin, a slope from a large triangle on the line (not from one table row), and g within about ±0.3 m s⁻² of 9.8 m s⁻². A strong answer gives two or more precautions in the past tense ("I measured the length to the middle of the stone", "I ensured the stone swung in one plane") and one source of error with its effect (the stone is not a point mass, so its middle was estimated and each l may be out by a centimetre; the support moved slightly, adding to the effective length).
The practical. A candidate was given a retort stand with clamp and boss, a split cork, a light inextensible thread, a metal bob of diameter 3.0 cm, a metre rule and a stopwatch. She clamped the thread between the cork halves, set the length L of thread between the lower face of the cork and the top of the bob to 0.285 m, displaced the bob through a small angle and timed 20 complete oscillations twice. She repeated this for L = 0.435, 0.585, 0.735 and 0.885 m.
| L / m | t₁ / s | t₂ / s |
|---|---|---|
| 0.285 | 21.96 | 22.04 |
| 0.435 | 26.90 | 26.98 |
| 0.585 | 31.14 | 31.06 |
| 0.735 | 34.82 | 34.74 |
| 0.885 | 38.06 | 38.14 |
Copy the table and add rows for: the effective length l = L + r (3 d.p.); the mean time t̄ (2 d.p.); the period T = t̄/20 (3 d.p.); and T² (3 d.p.).
r = 3.0 ÷ 2 = 1.5 cm = 0.015 m. l: 0.300, 0.450, 0.600, 0.750, 0.900 m (1) · t̄: 22.00, 26.94, 31.10, 34.78, 38.10 s (1) · T: 1.100, 1.347, 1.555, 1.739, 1.905 s (1) · T²: 1.210, 1.814, 2.418, 3.024, 3.629 s² (1).
Mark scheme note: One mark per correct row; all five values must be correct to the stated decimal places. Using L instead of l (no radius added), or adding the diameter, loses the l row; error carried forward is allowed in later parts. Inconsistent decimal places within a row lose that mark.
Plot T² (vertical axis) against l (horizontal axis), starting both axes from zero. Draw the line of best fit.
Both axes labelled with quantity and unit, T²/s² vertical and l/m horizontal (1); a sensible linear scale using at least half the graph paper on each axis, e.g. 2 cm to 0.1 m and 2 cm to 0.5 s², not an awkward scale such as 3 cm to 0.1 (1); all five points correctly plotted, within half a small square (2: 1 mark for three or four correct); a single thin straight line of best fit passing through, or very close to, the origin with the points evenly spread about it (1); a title or clear origin and an overall neat graph with fine points (1).
Mark scheme note: Axes the wrong way round (l vertical) lose the axis mark and make the later slope the inverse; the examiner then allows only error carried forward. Joining the dots dot-to-dot, or a thick line, loses the line mark. A scale that cannot be read easily (multiples of 3 or 7) loses the scale mark.
Determine the slope s of your graph, showing the triangle you used.
A large right-angled triangle drawn on the line, using points read from the line (not from the table), e.g. (0.100 m, 0.403 s²) and (0.900 m, 3.628 s²) (1); ΔT² = 3.628 − 0.403 = 3.225 s² (1); Δl = 0.900 − 0.100 = 0.800 m (1); s = 3.225 ÷ 0.800 = 4.03 s² m⁻¹ (accept 3.98–4.08) (1).
Mark scheme note: A triangle smaller than half the length of the line loses the first mark. Using two table points that are not on the drawn line loses the first mark. Δl/ΔT² (0.248) is the inverse slope and loses the last mark. The unit s² m⁻¹ is required for the last mark.
(a) Show that the slope of the graph is 4π²/g. (b) Use your slope to determine g.
(a) T = 2π√(l/g), so T² = 4π²l/g = (4π²/g) l, which is y = mx with slope 4π²/g (1). (b) g = 4π²/s (1) = 39.48 ÷ 4.03 = 9.79 m s⁻² (accept 9.68–9.92 m s⁻² on error carried forward) (1).
Mark scheme note: In (b) the unit m s⁻² is required. g = s/4π² or g = 4π² × s loses the last two marks.
(a) State where your line cuts the T² axis. (b) Explain what this shows about the way the lengths were measured.
(a) At (or very near) the origin, T² ≈ 0 when l = 0 (1). (b) It shows that l was correctly measured to the centre of the bob (the radius was added), so no constant error was present in the lengths (1).
Mark scheme note: In (b), "the graph is correct" earns nothing; the link between a zero intercept and correct effective lengths is needed.
State three precautions that should have been taken to ensure accurate results.
Any three (1 each), in the past tense: I ensured the bob swung through a small angle (about 10° or less) · I started and stopped the stopwatch as the bob passed its rest position · I measured the length to the centre of the bob · I ensured the bob swung in one vertical plane and not in a circle · I ensured the cork was firmly clamped so the point of suspension did not move · I repeated each timing and used the mean · I avoided parallax when reading the metre rule · I avoided draughts by closing the windows.
Mark scheme note: Precautions written as commands ("ensure the angle is small") or in the future tense score zero. "I repeated the experiment" without saying what was repeated, and "I was careful", are not credited. Sources of error ("reaction time") offered as precautions score zero.
(a) Why is the T² against l graph preferred to working out g from a single reading? (b) A second candidate measured every length to the top of the bob. State and explain the effect on (i) his graph and (ii) his value of g. (c) A third candidate started her stopwatch while counting "one". State and explain the effect on her value of g.
(a) The graph averages many readings, so random errors partly cancel, and it exposes any constant error as an intercept instead of hiding it in the answer (1). (b)(i) Every l is too small by r, so every point shifts left by 0.015 m and the line cuts the T² axis above the origin (1). (ii) The slope is unchanged, so g from the slope is unaffected (1). (c) Only 19 oscillations were really timed but the time was divided by 20 (1), so T was too small and g = 4π²l/T² came out too large (by about 11 %) (1).
Mark scheme note: In (b)(ii) "g too small" scores zero, though it would be true for g calculated from a single reading. In (c) both the 19/20 reason and the direction "too large" are required for the two marks.
In another laboratory the support was out of reach, so a candidate measured the height y of the bob's centre above the floor. When y = 1.20 m, T² = 5.646 s²; when y = 1.80 m, T² = 3.226 s². (a) Calculate the slope of T² against y. (b) Hence find g. (c) Find the height H of the point of suspension above the floor.
(a) slope = (3.226 − 5.646) ÷ (1.80 − 1.20) = −2.420 ÷ 0.60 = −4.03 s² m⁻¹ (1). (b) g = 39.48 ÷ 4.033 = 9.79 m s⁻² (1). (c) The line cuts the horizontal axis (T² = 0) at y = H: from y = 1.20 m the extra height is 5.646 ÷ 4.033 = 1.40 m, so H = 1.20 + 1.40 = 2.60 m (1).
Mark scheme note: A negative value of g loses (b). In (c) error carried forward is allowed from (a).
Timetable (80 min): 0–5 Do Now · 5–15 Wonder · 15–45 Investigate · 45–50 movement break (the living T²–l graph) · 50–65 Make Sense · 65–75 Apply · 75–80 Reflect.
Competences: L1 (error detective), L3 (clamping and safe bench practice), L4 (composite table, graph), L6 (precision measurement as a job skill).
Note on g: the local value near Ogbomoso is about 9.78 m s⁻²; the bench uses 9.79 m s⁻² as a realistic class measurement (0.1 % high).
Support: a pre-ruled composite table. Stretch: the T²–r variant (fixed thread, bobs of different radius).