How can we put a number on a bounce?
Four quick cards on speed, units of time, reaction time and reciprocals. Answer each in your head before you reveal it.
Eggs keep cracking on the way from Hajia Ramatu's farm. Dayo says the back of his okada "jumps like a goat" — but only at some speeds. Before we argue about speed, we need to measure a bounce.
The 3D story is not available on this device. Read the dialogue below instead.
Scene 1 · Hajia Ramatu (at the farm gate, holding a cracked tray): “Third time this month, Zainab. Dayo swears he rode carefully.”
Scene 2 · Dayo (on the parked okada): “Slow, nothing breaks. Fast, nothing breaks. But somewhere in the middle, the back starts to jump like a goat.” The rear of the bike bobs gently as he shifts his weight.
Scene 3 · Tunde: “Simple. The faster you ride, the more it bounces. Ride slowly every time and finish.”
Scene 4 · Zainab presses down on the carrier and lets go; it bobs up and down several times. “Look — it bounces at its own rhythm even when the bike is standing still.”
Scene 5 · Mrs Okafor (in the physics lab, beside a pendulum and a loaded spring): “Before we argue about speed, let's measure that rhythm. Anything that repeats can be timed.”
Scene 6 · Mrs Okafor: “Count twenty bounces, not one. Then we'll know how long one takes, and how many happen in a second.”
If you had only a phone stopwatch, how would you measure how long one bounce of the okada takes — as accurately as possible?
Sort each card into its bin. Drag it, or tap it then tap a bin. After you check, every card shows why.
A 0.80 m pendulum, a 0.20 kg mass on a 24.5 N m⁻¹ spring, and Dayo's parked okada (200 kg). Time 20 oscillations of each lab rig, then try timing by hand.
Small displacement, let it settle into a steady swing.
Press as it passes, counting “zero”, not “one”.
Back to the mark, moving the same way.
T = t/20, f = 1/T, ω = 2π/T; take the mean.
The 3D bench is not available on this device. Use the expected table and the worked examples below.
Each timing fills a row. Work out f = 1/T yourself and type it (to 3 s.f.); the table checks it and then shows ω.
| Rig · n | t / s | T = t/n / s | f / Hz | ω / rad s⁻¹ |
|---|
Timings recorded: 0
| Rig | n | t / s | T / s | f / Hz | ω / rad s⁻¹ |
|---|---|---|---|---|---|
| Pendulum, l = 0.80 m | 20 | 35.90 | 1.80 | 0.557 | 3.50 |
| Spring, m = 0.20 kg, k = 24.5 N m⁻¹ | 20 | 11.35 | 0.568 | 1.76 | 11.1 |
| Parked okada, 200 kg | 20 | 10.0 | 0.500 | 2.00 | 12.6 |
The table checks your typed f within ±0.001 Hz, then shows ω to 3 s.f.: 1 ÷ 1.80 = 0.556 Hz and 2π ÷ 1.80 = 3.49 rad s⁻¹ from the rounded T, or 0.557 Hz and 3.50 rad s⁻¹ from the unrounded 1.795 s.
Why 20? A 0.2 s reaction error on one pendulum swing is (0.2 ÷ 1.80) × 100 = 11.1 %; spread over 20 swings it is (0.2 ÷ 36.0) × 100 = 0.56 %.
Precautions (past tense): I started and stopped the stopwatch as the bob passed the rest mark, where it moved fastest; I began counting at “zero”; I timed 20 oscillations and repeated the timing; I displaced the bob only slightly.
Sources of error: reaction time at the start and stop (shared out over 20 cycles); miscounting the oscillations; air resistance slowly reducing the amplitude (it does not noticeably change T).
Stand up. Let both arms hang and sway them to and fro in time with your teacher's clap. Count the swings for ten claps. Now the teacher doubles the clap rate: sway twice as often. You have just doubled your frequency and halved your period. Shake out, then sit down.
Now connect what you timed to the definitions and equations WAEC, NECO and JAMB expect.
Amplitude is measured from the middle to one end — not from end to end. A bob swinging between marks 12 cm apart has A = 6 cm.
| Kind | Repeats? | To and fro? | Fixed rest position? | Example |
|---|---|---|---|---|
| Oscillatory | Yes, in equal times | Yes | Yes | okada carrier, swing |
| Periodic only | Yes, in equal times | No (goes round) | No | ceiling fan, clock hand |
| Non-repeating | No | No | No | falling mango |
Tap a pin. The pendulum is our 0.80 m bench pendulum (T from the kit model). Watch the bob in the strip and the fraction of a cycle completed.
Release the bob and watch it leave ghost markers at A (0), O (¼), B (½), O (¾) and A (1). The ruler shows the amplitude O → A. The stopwatch face compares a 0.2 s reaction error on 1 swing with the same error on 20.
The 3D zoom is not available on this device.
Try each on paper first, then check the working.
Zainab counts 20 bounces of the loaded, parked okada in 10.0 s. Find T, f and ω.
T = t/n = 10.0 ÷ 20 = 0.50 sf = 1/T = 1 ÷ 0.50 = 2.0 Hzω = 2πf = 2 × 3.142 × 2.0 = 12.6 rad s⁻¹Story numbers are a teaching model, not real measurements.
A child on a compound swing makes 15 complete oscillations in 42.0 s. Find the period, frequency and angular frequency.
T = 42.0 ÷ 15 = 2.80 sf = 1 ÷ 2.80 = 0.357 Hzω = 2π ÷ 2.80 = 6.283 ÷ 2.80 = 2.24 rad s⁻¹A student's reaction error is 0.2 s. A pendulum's true period is 1.80 s. Compare the percentage error from timing one oscillation with that from timing 20.
error = (0.2 ÷ 1.80) × 100 = 11.1 %t = 20 × 1.80 = 36.0 s, then error = (0.2 ÷ 36.0) × 100 = 0.56 %ratio = 11.1 ÷ 0.56 = 20 — the error is shared out twenty times.A bob swings between two points 12.0 cm apart with frequency 0.50 Hz. Find the amplitude, the period and the distance the bob travels in 10 s.
A = 12.0 ÷ 2 = 6.0 cmT = 1/f = 1 ÷ 0.50 = 2.0 s, so N = 10 ÷ 2.0 = 5 complete oscillations.4A = 4 × 6.0 = 24 cm per oscillation, so s = 5 × 24 = 120 cm = 1.20 m.Hand-time the parked okada (200 kg, press and release) the way Tunde did — one bounce — and the way Zainab did — twenty bounces. Each press of your stopwatch carries a reaction delay of 0.1–0.3 s. Do each at least twice.
Story numbers are a teaching model, not real measurements.
The big idea of Lesson 1: an oscillation is measured by its period and frequency, f = 1/T; time many cycles to beat reaction time.
Choose an option for instant feedback and a full explanation.
Rate your confidence from 1 to 4 on each goal of Lesson 1.
I can tell oscillation from other periodic motion.
I can define A, T, f and ω with their units.
I can calculate T, f and ω from timing data.
I can explain why 20 oscillations are timed from the rest mark.
I can find the distance covered from the amplitude.
Close your notes and write down: periodic vs oscillation · one complete oscillation · A middle-to-end · T = t/n · f = 1/T · ω = 2πf · count from zero at the rest mark.
f = 600 ÷ 2.0 = 300 Hz (1); T = 1 ÷ 300 = 3.3 × 10⁻³ s (1).
Part A is a home investigation; Part B is a 30-mark written paper. Type your attempt first to unlock each model solution.
Sample results: at 1.00 m, t₁ = 40.3 s and t₂ = 40.0 s, mean 40.2 s, so T = 40.2 ÷ 20 = 2.01 s and f = 1 ÷ 2.01 = 0.498 Hz. At 0.50 m, mean 28.4 s, so T = 28.4 ÷ 20 = 1.42 s and f = 1 ÷ 1.42 = 0.704 Hz. Single-oscillation timings at 1.00 m: 1.8 s, 2.3 s, 2.0 s.
The shorter thread gave the shorter period and the higher frequency; only the length was changed, so it was a fair test. The three single timings scatter by several tenths of a second while the two 20-oscillation timings agree within about 0.3 s: the reaction-time error is shared out when many cycles are timed, just as Zainab timed 20 bounces of the okada rather than one. A source of error: the stone swung slightly in an ellipse, or a draught from the window nudged it.
(a) Periodic motion is motion that repeats itself in equal intervals of time. (1)
(b) An oscillation (vibration) is a periodic to-and-fro motion about a fixed rest (equilibrium) position. (1)
(c) Frequency is the number of complete oscillations made in one second; its unit is the hertz (Hz = s⁻¹). (1)
Mark scheme: (a) needs equal time intervals; (b) needs both to-and-fro and the fixed rest position — “moving up and down” alone scores zero; (c) needs “complete” oscillations per second.
(a) Amplitude is the greatest displacement of the bob from its rest position; on the sketch it is OA (or OB), not AB. (1)
(b) The period is the time taken for one complete oscillation (A → O → B → O → A). (1)
(c) ω = 2π/T (= 2πf), in radians per second (rad s⁻¹). (1)
Mark scheme: marking AB as the amplitude loses (a); in (c) both the relation and the unit are needed.
(a) The reaction-time error on pressing the stopwatch is about the same however long the timing is, so dividing by 20 makes it twenty times smaller as a fraction of T. (1)
(b) The bob moves fastest at the rest position, so the instant it passes the mark is sharp; at the ends it lingers. (1)
(c) Counting “one” at the start means only 19 oscillations are really timed when the count reaches 20, so T comes out too small. (1)
(d) Each count would be only half an oscillation, so the T obtained would be half the true period. (1)
Mark scheme: “to be accurate” without the reaction-time idea earns nothing; “because the bob stops there” in (b) is the misconception reversed and scores zero.
(a) One oscillation takes T = t/n, so the number of oscillations in one second is n ÷ t: f = n/t. (1) Since T = t/n, n/t = 1/T, so f = 1/T. (1)
(b) In one oscillation the body goes O → end (A), back to O (A), to the other end (A) and back to O (A): total 4A. (1) In time t there are t/T oscillations, so distance = (t/T) × 4A = 4Aft. (1)
(a) T = t/n = 27.0 ÷ 45 = 0.600 s (1)
(b) f = 1/T = 1 ÷ 0.600 = 1.67 Hz (1)
(c) ω = 2π/T = 6.283 ÷ 0.600 = 10.5 rad s⁻¹ (1)
(d) End to end = 18.0 − 12.0 = 6.0 cm, so A = 6.0 ÷ 2 = 3.0 cm (1)
(e) N = 6.0 ÷ 0.600 = 10 oscillations; s = 10 × 12.0 = 120 cm (1.20 m) (1)
Mark scheme: error carried forward allowed from (a); using 18.0 cm or 6.0 cm as A loses (d); units required in every part.
(a) 17.4 ÷ 10 = 1.74 s; 35.0 ÷ 20 = 1.75 s; 52.4 ÷ 30 = 1.75 s (1.747 s) (1)
(b) T = (1.74 + 1.75 + 1.747) ÷ 3 = 1.75 s (1)
(c) f = 1 ÷ 1.75 = 0.571 Hz (1); ω = 2π ÷ 1.75 = 3.59 rad s⁻¹ (1)
(d) 10 oscillations: (0.2 ÷ 17.4) × 100 = 1.1 %; 30 oscillations: (0.2 ÷ 52.4) × 100 = 0.38 % (1). The 30-oscillation timing is most reliable, because the same reaction error is the smallest fraction of the time measured. (1)
(a) Any two, in the past tense: I started and stopped the stopwatch as the bob passed the rest mark · I counted “zero” at the start · I timed 20 oscillations and repeated the timing · I displaced the bob only slightly · I ensured the bob swung in one plane. (2)
(b) Reaction time in starting and stopping the stopwatch (it can only be reduced, by timing many oscillations), or air resistance slowly reducing the amplitude. (1)
Mark scheme: precautions written as commands or in the future tense score zero; a precaution offered as an error scores zero.
(a) T = 10.0 ÷ 20 = 0.50 s (1)
(b) Tunde's reading includes a reaction error of a few tenths of a second, a large fraction (about 40 %) of a period of only 0.5 s; Zainab's error is shared over 20 bounces, so hers is far more reliable. (1)