Four quick cards on circular motion, the SHM condition, ω and radians. Answer each in your head before you reveal it.
1Wonder · The case · 10 min
The crates jump at the top
The crates on Dayo's carrier keep cracking. Zainab says they jump at the top of the bounce; Tunde says it is where the bike moves fastest. Mrs Okafor has an idea with a muddy wheel and the midday sun.
3D story · Ridge Road, Ogbomoso
The 3D story couldn't start on this device. Read the dialogue below instead.
Scene 1 of 5
Read the story as text or use the 3D player above
Scene 1 · Dayo (riding past the ridges at a moderate pace): “Hear how the crates rattle! The carrier is pumping up and down, but today they are staying on.”
Scene 2 · Zainab (watching from the roadside): “The crates stay on the carrier while it rises. They only jump when it is at the top and starting down.”
Scene 3 · Tunde:“They jump because that is where the bike is moving fastest.”
Scene 4 · Mrs Okafor (the okada on its centre stand; she spins the rear wheel slowly, a lump of red mud goes round, and the midday sun throws its shadow on the ground): “Watch the mud's shadow on the ground, not the mud. It moves exactly like the bounce.”
Scene 5 · Mrs Okafor:“At the top of the bounce the carrier is barely moving, but it is accelerating downwards harder than anywhere else. If that downward acceleration beats g, the carrier falls away faster than the crate can.”
1Wonder · Your predictions
The Wonder Wall
Where in the bounce is the carrier moving fastest, and where is its acceleration biggest? Same place or different? Pin your idea before we test it.
2Investigate · Activity 1
Where does it happen?
Drag each card into the bin where it is true for a body in SHM: at the centre (x = 0), at the ends (x = ±A), or nowhere.
At the centre
equilibrium, x = 0
At the ends
x = ±A
Nowhere
never true in SHM
2Investigate · Virtual practical · 25 min
The wheel-and-shadow bench: x, v and a
A wheel of radius A = 0.30 m turns once every T = 2.0 s (ω = π rad s⁻¹). Read the shadow's x, v and a at four angles.
01
Set A and T, spin
Watch the shadow slide to and fro.
02
Pause at 0°, 30°, 60°, 90°
Step 30° reads x, v and a.
03
Compare with the formulae
x = A sin ωt, v = ωA cos ωt, a = −ω²x.
04
Graphs on
Match each arrow to its trace.
2D wheel bench
The 2D bench couldn't draw on this device. The buttons, readout and table still work.
Step 30° records 0°–90° · drag the lump to turn
Record your angles
ωt / °
x / m
v / m s⁻¹
a / m s⁻²
a/x / s⁻²
Angles recorded: 0 of 4
Expected results & graphs open after you measure
A = 0.30 m, T = 2.0 s, so ω = 2π/T = π = 3.14 rad s⁻¹ and ω² = 9.87 s⁻².
ωt / °
x / m
v / m s⁻¹
a / m s⁻²
0
0
0.942
0
30
0.150
0.816
−1.48
60
0.260
0.471
−2.56
90
0.300
0
−2.96
Graphs: x–t is a sine curve; v–t is a cosine curve that peaks where x = 0; a–t is an upside-down copy of x–t. At every angle a/x = −9.87 s⁻² = −ω².
Practical analysis questions
Answer these five questions, then compare with the model answers.
At which angle is the shadow fastest, and why?
Why is a always opposite in sign to x?
Show that a/x is the same at 30° and at 60°.
What happens to vmax if T is halved?
And what happens to amax?
Model answers unlocks after you try
At 0° (and 180°), where the shadow passes the centre O. There v = ωA cos ωt has cos ωt = ±1, the largest possible, so the speed is ωA = 0.942 m s⁻¹.
The acceleration always points back to the centre: when the shadow is on the + side, a is towards −, and the other way round. That is the minus sign in a = −ω²x.
At 30°: −1.48 ÷ 0.150 = −9.87 s⁻². At 60°: −2.56 ÷ 0.260 = −9.87 s⁻². Both equal −ω² = −π² (using unrounded values).
Halving T doubles ω, so vmax = ωA doubles.
amax = ω²A, so doubling ω makes it four times as large.
45Movement break · 45 min
Movement break · 45 min · the shadow dance (5 minutes): stand up in pairs. One partner walks slowly round a small circle at a steady pace. The other stands by the wall and copies only the walker's side-to-side position, moving along the wall line. Notice where the copier has to move fastest and where they stop and turn. Swap roles once, then sit down.
3Make sense · 15 min
SHM is the shadow of a circle
Now connect what the shadow did to the equations WAEC, NECO and JAMB expect.
Explainer videoSHM as a Circle's Shadow13 min · narrated · captionsWatched ✓
The reference circle: a point moving round a circle of radius A at a steady angular speed ω casts a shadow on a diameter. That shadow moves in SHM with amplitude A and the same ω. Timing from the centre, after time t the radius has turned through ωt, so the shadow's displacement is x = A sin ωt. One turn of the circle is one complete oscillation, so they share the same period T = 2π/ω.
Equations
x = A sin ωt · v = ω√(A² − x²) · a = −ω²x
vmax = ωA (at x = 0) · amax = ω²A (at x = ±A)
Always work out ωt in radians before taking the sine (set your calculator to RAD).
The lift-off rule: the carrier can push the crate up, but it cannot pull it down. Only gravity pulls the crate down, so the crate can never accelerate downwards faster than g. At the top of the bounce the carrier accelerates downwards at ω²A. If ω²A > g, the carrier drops away faster than the crate can fall, and the crate leaves it. So the crates stay on only while A < g/ω².
Centre and ends compared
Quantity
At x = 0 (centre)
At x = ±A (ends)
Displacement
zero
greatest (A)
Speed
greatest (ωA)
zero
Acceleration
zero
greatest (ω²A), towards O
Restoring force
zero
greatest, towards O
Position Explorer
Tap a phase pin on the A = 0.30 m, T = 2.0 s circle. Read the shadow's x, v and a at that instant.
Tap a pin to read x, v and a at that angle.
3Make sense · 2D Zoom
Zoom in: circle, traces and the g line
The reference circle on the left drives three traces on the right: x, v and a against t, linked by one cursor. Switch to the okada (ω ≈ 12.6 rad s⁻¹) and drag A from 2 cm to 8 cm: the a peak crosses g at g/ω² = 6.2 cm.
2D circle zoom
The 2D zoom couldn't draw on this device.
x–v–a–amax (m/s²)–g/ω²–
4Apply · Worked examples · 10 min
Worked examples, step by step
Try each on paper first, then check the working. Calculator in radians.
EXAMPLE 1
Speed and acceleration at a point
A body moves in SHM with A = 0.10 m and ω = 5.0 rad s⁻¹. Find its speed and acceleration when x = 0.060 m, and its maximum speed.
Evaluate the speed:v = 5.0 × √0.0064 = 5.0 × 0.080 = 0.40 m s⁻¹
Acceleration:a = −ω²x = −25 × 0.060 = −1.5 m s⁻² (the minus sign means towards the centre)
Maximum speed:v_max = ωA = 5.0 × 0.10 = 0.50 m s⁻¹
v = 0.40 m s⁻¹ · a = 1.5 m s⁻² towards O · v_max = 0.50 m s⁻¹
EXAMPLE 2
Displacement at a time
The okada's carrier bounces with A = 5.0 cm and f = 2.0 Hz, timed from the centre. Find its displacement at t = 0.0625 s.
Angular frequency:ω = 2πf = 2π × 2.0 = 4π rad s⁻¹
Phase angle in radians:ωt = 4π × 0.0625 = π/4 = 0.785 rad
Displacement:x = A sin ωt = 5.0 × sin(π/4) = 5.0 × 0.707 = 3.5 cm
x ≈ 3.5 cm
Story numbers are a teaching model, not real measurements.
EXAMPLE 3
A gentle bounce
At f = 2.0 Hz the carrier bounces with A = 3.0 cm. Find vmax and amax. Do the crates stay on? (g = 9.8 m s⁻²)
Maximum speed:v_max = ωA = 12.6 × 0.030 = 0.38 m s⁻¹ (ω = 2π × 2.0 = 12.6 rad s⁻¹)
Maximum acceleration:a_max = ω²A = 158 × 0.030 = 4.7 m s⁻²
Compare with g:4.7 m s⁻² < 9.8 m s⁻², so the carrier never accelerates downwards faster than g.
v_max ≈ 0.38 m s⁻¹ · a_max ≈ 4.7 m s⁻² · crates stay on
EXAMPLE 4
The lift-off threshold
Find the smallest amplitude at which the crates leave the carrier at f = 2.0 Hz. (g = 9.8 m s⁻²)
Condition: lift-off starts when ω²A = g, so A = g/ω²
Square ω:ω² = 12.57² = 158 s⁻²
Threshold:A = g ÷ ω² = 9.8 ÷ 158 = 0.062 m = 6.2 cm
A ≥ 6.2 cm (the same as the okada's static sag mg/k)
4Apply · The big idea
Back to Ridge Road: was Tunde right?
Use the okada numbers in the zoom (ω ≈ 12.6 rad s⁻¹). Try two bounces, answer three questions, then give your verdict.
Do the crates fly, and where?
1 · With A = 3 cm, the crates…
2 · With A = 8 cm, the crates…
3 · Where in the bounce do they leave?
Story numbers are a teaching model, not real measurements.
The big idea of Lesson 3: in SHM, speed peaks at the centre (vmax = ωA) and acceleration peaks at the ends (amax = ω²A). The crates fly at the top of the bounce, where the carrier is still but accelerating downwards hardest. Lift-off needs A ≥ 6.2 cm, so something on Ridge Road must be pumping the bounce that big.
4Apply · Check yourself
Quick quiz: WAEC, NECO & JAMB questions
Choose an option for instant feedback and a full explanation. Two review questions from earlier lessons come at the end.
Score 0 / 12
5Reflect
How did you do?
Rate your confidence from 1 to 4 on each goal of Lesson 3.
1 Not yet2 With help3 On my own4 I could teach it
I can link SHM to a reference circle.
I can use x = A sin ωt in radians.
I can use v = ω√(A² − x²) and a = −ω²x.
I can say where v and a are zero and greatest.
I can apply the lift-off rule ω²A ≥ g.
Exit ticket
A body moves in SHM with A = 4.0 cm and ω = 10 rad s⁻¹. Find vmax and amax.
2 marks
Model answer unlocks after you try
v_max = ωA = 10 × 0.040 = 0.40 m s⁻¹ (1) and a_max = ω²A = 100 × 0.040 = 4.0 m s⁻² (1).
5Assignment
Assignment: Part A and Part B
A home investigation and a 30-mark written paper. Type your attempt first to unlock each model solution.
Part A · Home investigation: the shadow of a turning wheel
Where does the shadow move fastest, and where does it turn round?
You need: a bicycle (or a basin lid or round tray that can turn on a nail), a bottle cap and tape, a torch, a plain wall, a pencil and paper. Work with an adult; keep fingers away from the spokes.
Turn the bicycle upside down so the front wheel spins freely, or fix the lid on a nail so it can turn. Tape the bottle cap to the rim.
In a dim room, shine the torch from the side so the cap throws a shadow on the wall.
Turn the wheel slowly and as steadily as you can. Watch the shadow, not the cap.
Mark on the wall (with tape) the two ends of the shadow's path and its middle.
Turn the wheel through quarter turns, pausing each time, and note where the shadow is.
Describe where the shadow moves fastest and where it slows down and turns round.
Sample observations unlocks after you try
Cap on wheel
Shadow position
How the shadow moves
Level with the hub
Middle mark
Fastest
Top of wheel
Top end mark
Stopped, turning round
Level with the hub (other side)
Middle mark
Fastest, other direction
Bottom of wheel
Bottom end mark
Stopped, turning round
A good answer shows that the shadow's path is as long as the wheel's diameter, so its amplitude is the radius; the shadow is fastest at the middle and momentarily still at the ends, exactly like the bouncing carrier; and one turn of the wheel is one complete oscillation of the shadow, so they share the same period. A strong answer links this to the reference circle, explains that the shadow must accelerate hardest at the ends (it stops and reverses there), and names a source of error, for example the wheel was not turned at a steady rate, or the torch was so close that the shadow was enlarged unevenly.
Part B · Written paper (30 marks)
1. Definitions
3 marks
(a) Define simple harmonic motion in terms of acceleration. (b) What is meant by the reference circle for an SHM? (c) Define angular frequency, and give its unit.
Model solution unlocks after you try
(a) SHM is motion in which a body's acceleration is directly proportional to how far it is displaced from a fixed point and always points back towards that point. (1) (b) A circle of radius equal to the amplitude, round which a point moves at constant angular speed ω; the projection (shadow) of that point on a diameter moves with the SHM. (1) (c) Angular frequency is 2π times the frequency (ω = 2πf = 2π/T), the angle in radians swept per second by the point on the reference circle; unit rad s⁻¹. (1)
Mark scheme note: In (b) both "radius = amplitude" (or "point moving uniformly") and "projection on a diameter" are needed. In (c) the unit is required.
2. Where Things Happen
3 marks
Copy and complete the table for a body in SHM, writing zero or maximum in each box.
Displacement
Speed
Acceleration
At the equilibrium position
At an extreme position
Model solution unlocks after you try
Equilibrium: displacement zero, speed maximum, acceleration zero. Extreme: displacement maximum, speed zero, acceleration maximum. (1) for the displacement column, (1) for the speed column, (1) for the acceleration column.
Mark scheme note: Each column must have both entries right for its mark. Writing "maximum" for speed at the extreme position (the "fastest at the ends" error) loses the speed mark.
3. From the Circle to the Formulae
4 marks
A point P moves round a circle of radius A at constant angular speed ω, starting on the horizontal diameter at t = 0. N is the projection of P on the vertical diameter. (a) Show that the displacement of N from the centre is x = A sin ωt. (b) Given that the velocity of N is v = ωA cos ωt, show that v = ω√(A² − x²). (c) State the maximum speed and where it occurs.
Model solution unlocks after you try
(a) After time t the radius OP has turned through angle θ = ωt; the vertical component of OP is A sin θ, so x = A sin ωt. (1) (b) cos ωt = √(1 − sin² ωt) = √(1 − x²/A²) (1); so v = ωA√(1 − x²/A²) = ω√(A² − x²). (1) (c) vmax = ωA, when x = 0 (at the equilibrium position). (1)
Mark scheme note: In (a) the angle ωt must be identified. In (b) the identity sin² + cos² = 1 must be used or clearly implied; quoting the result without working earns no mark. In (c) the position is required.
4. A Bouncing Load
5 marks
A load on a spring oscillates in SHM with amplitude 0.12 m and period 0.80 s, timed from the equilibrium position. Calculate: (a) its angular frequency; (b) its maximum speed; (c) its maximum acceleration; (d) its speed when it is 0.060 m from equilibrium; (e) its displacement at t = 0.10 s.
Mark scheme note: Error carried forward from (a). In (e) a calculator in degree mode gives 0.0016 m and loses the mark. Units required.
5. The Lift-off Threshold
4 marks
The okada's rear bounces at its natural frequency f₀. (Take g = 9.8 m s⁻².) (a) Explain why a crate resting loosely on the carrier leaves it when ω²A > g. (b) With 200 kg on the carrier, ω = 12.57 rad s⁻¹. Find the smallest amplitude for lift-off. (c) With 250 kg, ω = 11.24 rad s⁻¹. Find the new threshold and say whether a 7.0 cm bounce would now throw the crates.
Model solution unlocks after you try
(a) At the top of the bounce the carrier accelerates downwards at ω²A; only gravity pulls the crate down, so it can only fall at g. If ω²A exceeds g the carrier moves down faster than the crate and they separate. (1) (b) A = g/ω² = 9.8 ÷ 158 = 0.062 m = 6.2 cm (1) (c) A = 9.8 ÷ 126.4 = 0.078 m = 7.8 cm (1); 7.0 cm is below 7.8 cm, so the crates would stay on (just). (1)
Mark scheme note: In (a) the comparison of the carrier's downward acceleration with g is essential; "the bike is going fast" scores zero. In (c) the conclusion must follow from the comparison, error carried forward allowed.
6. Practical: The Wheel-and-Shadow Bench
6 marks
On the 2D bench, the wheel had radius A = 0.20 m and period T = 2.5 s. Zainab paused the shadow at four angles θ = ωt. (a) Calculate ω. (b) Copy and complete the table (3 s.f.). (c) Show that a/x is the same at 30° and at 60°, and say what it equals. (d) State, with a reason, at which angle the shadow moved fastest.
θ / °
0
30
60
90
x / m
v / m s⁻¹
a / m s⁻²
Model solution unlocks after you try
(a) ω = 2π ÷ 2.5 = 2.51 rad s⁻¹ (ω² = 6.32 s⁻²). (1) (b) x = A sin θ: 0, 0.100, 0.173, 0.200 m (1); v = ωA cos θ: 0.503, 0.435, 0.251, 0 m s⁻¹ (1); a = −ω²x: 0, −0.632, −1.09, −1.26 m s⁻². (1) (c) −0.632 ÷ 0.100 = −6.32 s⁻² and −1.094 ÷ 0.1732 = −6.32 s⁻² (unrounded values); both equal −ω². (1) (d) At 0°, where x = 0 (equilibrium): the speed v = ωA cos θ is greatest when cos θ = 1. (1)
Mark scheme note: Each row of (b) needs all four entries (±1 in the last figure). Missing minus signs on a lose that row's mark. In (c) "−ω²" (or "minus omega squared") must be stated.
7. Sketching x, v and a
3 marks
On one time axis, describe (or sketch) the graphs of x, v and a against t for one cycle of SHM starting at equilibrium, and state the phase relationship of v and of a to x.
Model solution unlocks after you try
x is a sine curve starting at zero and rising. (1) v is a cosine curve starting at its maximum: it is a quarter of a cycle (90°) ahead of x. (1) a is an inverted sine curve, exactly opposite to x (half a cycle, 180°, out of phase), with peaks of size ω²A where x = ±A. (1)
Mark scheme note: Sketches must be correctly aligned in time. "a is the same as x" (missing the inversion) loses the third mark.
8. The Bridge: Correcting Tunde
2 marks
Tunde says the crates jump off "where the bike is moving fastest". Correct him in two sentences, using the ideas of this lesson.
Model solution unlocks after you try
The carrier is fastest at the middle of its bounce, where its acceleration is zero, so it is not throwing anything off there. (1) The crates leave at the top of the bounce, where the carrier is momentarily still but has its greatest downward acceleration ω²A; when that exceeds g the carrier drops away from under them, because only gravity pulls the crates down. (1)
Mark scheme note: Both "speed greatest at the middle" and "acceleration greatest at the top, compared with g" are needed for full marks.
TFor the teacher
Teacher notes
Timing (80 min): 0–5 Do Now · 5–15 Wonder · 15–45 Investigate · 45–50 movement break (shadow dance) · 50–65 Make sense · 65–75 Apply · 75–80 Reflect.
Competences: L1 challenge "fastest at the top"; L3 why bumps at speed throw loads; L4 the three time-graphs; L5 the 2D simulation.
Support: give a table of sine values in degrees (sin 30° = 0.500, sin 45° = 0.707, sin 60° = 0.866) before converting to radians. Stretch: show that vmax/amax = 1/ω.
Story model: okada rear m = 200 kg, k = 3.16 × 10⁴ N m⁻¹, f₀ = 2.00 Hz, ω₀ ≈ 12.6 rad s⁻¹; threshold g/ω² = 6.2 cm, equal to the static sag mg/k. These are constructed teaching numbers, not measurements.
Watch for: calculators left in degree mode; "a = 0 at the top because it stops"; and the idea that the carrier "throws" the crate. Stress that only gravity pulls the crate down — the carrier can only push up.