Four quick cards on energy formulas, the greatest speed in SHM, ω² for a spring and a practical precaution. Answer each in your head before you reveal it.
1Wonder · The case · 10 min
The shock absorber that weeps oil
Mallam Sule takes the rear shock off Dayo's okada. The spring is fine, he says, but the oil is leaking out. So what does oil have to do with a bounce?
3D story · Sule's shed
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 · Mallam Sule (unbolting the rear shock; dark oil streaks its body): “See this? It is weeping. The oil inside is what kills the bounce.”
Scene 2 · Dayo:“Kills the bounce? The spring is still strong.”
Scene 3 · Mallam Sule (pushing the new and the old damper side by side; the new one resists firmly and slowly, the old one slides freely): “A spring stores the energy and gives it back. The oil turns it to heat. Without oil, nothing takes the energy away.”
Scene 4 · Zainab:“So every time the bike bounces, the energy just keeps going round and round?”
Scene 5 · Mrs Okafor (in the physics lab, releasing a pendulum and a bouncing spring): “Let's watch where the energy is at every point — and what happens when we make it work through water.”
1Wonder · Your predictions
The Wonder Wall
Does a pendulum ever really stop by itself? Where would its energy go? Pin your idea before we measure it.
2Investigate · Activity 1
Kinetic, potential or both?
Drag or tap each of the nine moments into the right bin: KE greatest, PE greatest, or KE = PE.
KE greatest fastest
PE greatest at rest, at an end
KE = PE each half of E
2Investigate · 3D Practical · 25 min
Energy bars and a card vane in water
Load m = 0.20 kg on a spring of k = 24.5 N m⁻¹. Watch the energy swap, then make the load work through air, water and oil. Every number comes from the model.
01
Watch the energy bars
Release from 5.0 cm; watch KE, PE, total.
02
Fit the card vane
Damping: try air, then water, then oil.
03
Record each amplitude
Record the amplitude at five cycle ends.
04
Compare the periods
Compare 10-oscillation times in air and water.
3D bench · drag to rotate
The 3D bench couldn't start on this device. Use the expected results and the graph below.
Release, then watch KE and PE swap.
Your amplitudes
Cycle
Aair / cm
Awater / cm
Aoil / cm
Media recorded: 0 of 3 (air, water, oil)
Expected results illustrative model values, A = 5.0 cm
Cycle
0
1
2
3
4
5
Air (×0.97)
5.00
4.85
4.70
4.56
4.43
4.29
Water (×0.75)
5.00
3.75
2.81
2.11
1.58
1.19
Oil (×0.40)
5.00
2.00
0.80
0.32
0.13
0.05
Energy at the start: E = ½kA² = ½ × 24.5 × 0.050² = 0.0306 J. Ten oscillations: air 5.68 s, water about the same (≈ 5.7 s) — light damping barely changes T.
Live graph · plots as you record
Amplitude A against cycle number
0 of 3 media
Precautions and sources of error
Precautions (past tense): the vane was kept fully immersed at the same depth in each liquid; the amplitude was read with the eye level with the pointer to avoid parallax; the load was released from the same starting amplitude each time.
Sources of error: the vane touched the beaker wall (extra friction); the load swayed sideways instead of moving vertically; the moving pointer was misread at the end of a cycle.
Practical analysis questions
Answer these five questions, then compare with the model answers.
Where is KE greatest, and where is PE greatest?
Why did the energy (the total) shrink so fast in water?
Where did the lost energy go?
Did damping change the period noticeably?
Which medium would suit a shock absorber, and why?
Model answers unlocks after you try
KE is greatest as the load passes the equilibrium position (fastest); PE is greatest at the top and bottom of the bounce, where the load is momentarily at rest.
The water pushes back on the moving vane, so work is done against this resistance every cycle. The amplitude fell to 0.75 of its value each cycle, so the energy (∝ A²) fell to 0.75² ≈ 0.56 of its value each cycle.
It became internal energy (heat) in the water and the vane, plus a tiny amount of sound; it was not destroyed.
No: ten oscillations took 5.68 s in air and about 5.70 s in water. Light damping shrinks the amplitude but hardly changes the period.
A thick liquid such as oil: it removes a large share of the energy each cycle, so the bounce dies in one or two cycles, close to critical damping.
⏱Movement break · 45 min · 5 minutes
The energy swap squat
Stand up. Squat down slowly and say “PE stored!” Rise quickly and say “KE!” Freeze at the top for a moment: all PE again. Do it five times. Now “add damping”: repeat as if you were moving through thick oil, slower and smaller each time, until you stop. Shake out your legs and sit down.
3Make sense · 15 min
Energy sloshes; damping drains it
Now we connect the bars and the curves to the formulas WAEC, NECO and JAMB expect.
Energy interchange: in SHM without losses, the total energy stays fixed while it changes form — all potential at the ends, all kinetic at the centre. For a spring, E = ½kA²; at displacement x, PE = ½kx² and KE = ½k(A² − x²). Since k = mω², E = ½mω²A². The total energy is proportional to A²: double the amplitude, four times the energy.
Damping: a resistive force (air, a liquid, friction) takes energy out of the oscillation every cycle, so the amplitude dies away. Light damping: many shrinking oscillations; the period is almost unchanged. Heavy damping: a slow creep back to equilibrium with no oscillation. Critical damping: the quickest return to rest without overshooting — what a good shock absorber and a moving-coil meter's needle aim for.
The equations
E = ½kA² = ½mω²A²
PE = ½kx² · KE = ½k(A² − x²)
KE = PE at x = ±A/√2
E = mgh = ½mv_max²
Free and damped oscillations
Amplitude over time
Oscillates?
Example
Free (undamped)
stays constant
yes, for ever
ideal spring with no losses
Lightly damped
shrinks by the same ratio each cycle
yes, many cycles
pendulum in air; leaking shock
Heavily damped
creeps slowly back to zero
no
vane in thick oil
Critically damped
returns fastest, no overshoot
no
good shock absorber; meter needle
Position explorer
A 0.50 kg mass on a spring of k = 50 N m⁻¹ oscillates with A = 0.10 m (E = 0.25 J). Tap a position and read PE, KE and the speed.
Tap a position to see its energies.
3Make sense · 3D Zoom
Inside the shock absorber: oil turns bounce into heat
A cut-away coil-over: the spring outside, a piston with small holes moving through oil inside. Compare healthy and leaking, then give each ten pushes. The side panel draws the matching trace.
3D zoom · drag to rotate
The 3D zoom couldn't start on this device.
4Apply · Worked examples · 10 min
Worked examples, step by step
Try each on paper first, then check the working. Keep amplitudes in metres.
EXAMPLE 1
Energy at a point
A 0.50 kg mass on a spring of k = 50 N m⁻¹ oscillates with A = 0.10 m. Find the total energy, and PE, KE and speed at x = 0.060 m.
Total energy:E = ½kA² = ½ × 50 × 0.10² = 0.25 J
Potential energy at x:PE = ½kx² = ½ × 50 × 0.060² = 0.090 J
Kinetic energy:KE = E − PE = 0.25 − 0.090 = 0.16 J
Speed:v = √(2KE/m) = √(2 × 0.16 ÷ 0.50) = √0.64 = 0.80 m s⁻¹ (check: ω = 10 rad s⁻¹, v = 10√(0.010 − 0.0036) = 0.80)
E = 0.25 J · PE = 0.090 J · KE = 0.16 J · v = 0.80 m s⁻¹
EXAMPLE 2
Where is KE = PE?
For the same oscillator, where is the kinetic energy equal to the potential energy?
Each is half the total:½kx² = ½ × ½kA²
Simplify:x² = A²/2, so x = A/√2
Substitute:x = 0.10 ÷ 1.414 = 0.071 m
x = ±0.071 m (about 71 % of the way out, not halfway)
EXAMPLE 3
Pendulum energy
A 0.10 kg bob is released from 5.0 cm above its lowest point. Find its total energy and its greatest speed (g = 9.8 m s⁻²).
For the okada's rear (k = 3.16 × 10⁴ N m⁻¹), compare the energy in a gentle 2.0 cm bounce with that in a 6.2 cm lift-off bounce.
Gentle bounce:E₁ = ½ × 31600 × 0.020² = 6.3 J
Lift-off bounce:E₂ = ½ × 31600 × 0.062² = 61 J
Ratio:E₂/E₁ = (6.2/2.0)² = 9.6
about 6.3 J against 61 J — nearly ten times as much energy must build up for lift-off
Story numbers are a teaching model, not real measurements.
4Apply · The big idea
Mallam Sule's push test
In the zoom, feed the healthy and the leaking shock the same small push every half-second for ten pushes. Then answer the three steps.
Healthy or leaking: where does the energy go?
The big idea of Lesson 6: in SHM, E = ½kA² sloshes between KE and PE; damping drains it. A working shock absorber turns a large share of each cycle's energy into heat, so the road can never pile up the 61 J a 6.2 cm bounce needs.
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 / 14
5Reflect · 5 min
How did you do?
Rate your confidence from 1 to 4 on each goal of Lesson 6.
1 Not yet2 With help3 On my own4 I could teach it
I can describe the KE–PE interchange.
I can calculate E, KE and PE at a point.
I can find where KE = PE.
I can define and compare types of damping.
I can explain what a shock absorber does.
Exit ticket
The amplitude of an oscillator is doubled. By what factor does its energy change?
2 marks
Model answer unlocks after you try
E = ½kA², so E ∝ A² (1); doubling A multiplies the energy by 2² = 4(1).
5Assignment · 40 min at home
Assignment: Part A and Part B
Type your attempt first to unlock each model solution.
Part A · Home investigation: whose shock absorbers are tired?
Bounce test on two parked vehicles
You need: two parked vehicles (cars, a keke or motorcycles) with the owners' permission, a friend to watch, and a notebook. Each vehicle must be switched off, in gear or on its stand, with nobody in it or behind it.
Press down firmly on one corner (or the back of a motorcycle seat) and let go quickly.
Your friend counts how many times that corner rises and falls before it settles, and estimates the height of the first rebound.
Repeat three times for the same corner, then do the same on the second vehicle.
Ask each owner how old the shock absorbers are, if they know.
Sample results unlocks after you try
Vehicle
Bounces (3 trials)
Mean
First rebound
Neighbour's car (absorbers replaced last year)
1, 1, 2
1.3
Firm, comes straight back
Uncle's old car
4, 3, 4
3.7
Bouncy, overshoots several times
A good answer shows that the vehicle with healthy shock absorbers settles after about one rebound (close to critical damping), while the one with worn absorbers bounces several times with slowly shrinking amplitude (light damping), so less energy is turned into heat each cycle. A strong answer links this to Mallam Sule's leaking shock absorber: a lightly damped suspension lets energy from a bumpy road build up. It also gives a limitation, for example the push was not the same size each time, or counting depended on judging very small movements.
Part B · Written paper (30 marks)
1. Definitions
3 marks
Define: (a) damping; (b) critical damping; (c) a free oscillation.
Model solution unlocks after you try
(a) Damping is the gradual loss of energy from an oscillating system, caused by resistive forces such as friction or air or liquid resistance, so that the amplitude decreases. (1) (b) Critical damping is the amount of damping that returns a displaced system to its equilibrium position in the shortest possible time without oscillating. (1) (c) A free oscillation is one in which the system vibrates at its own natural frequency with no outside periodic force and no loss of energy, so its amplitude stays constant. (1)
Mark scheme note: In (a) both “energy loss” (or a resistive force) and “amplitude decreases” are expected. In (b) “without oscillating” and “shortest time” are both needed. In (c) “natural frequency” or “constant amplitude” must appear.
2. Energy Round a Swing
3 marks
A pendulum bob swings from end-point A through its lowest point O to the other end B. Describe the energy at A, at O, and halfway in energy between them, assuming no losses.
Model solution unlocks after you try
At A (and B) the bob is momentarily at rest: all its energy is potential, PE is greatest and KE is zero. (1) At O it moves fastest: all its energy is kinetic, KE is greatest and PE is least (zero, taking O as the reference level). (1) In between, the energy is shared, with KE + PE always equal to the same total; they are equal at x = A/√2. (1)
Mark scheme note: “Total energy is constant” must appear in some form for the third mark.
3. Deriving the Energy Expressions
4 marks
For a mass m on a spring of constant k oscillating with amplitude A: (a) using v = ω√(A² − x²) and ω² = k/m, show that KE = ½k(A² − x²); (b) show that KE + PE = ½kA², a constant; (c) show that KE = PE where x = A/√2.
Model solution unlocks after you try
(a) KE = ½mv² = ½mω²(A² − x²) (1); since mω² = k, KE = ½k(A² − x²). (1) (b) PE = ½kx², so KE + PE = ½kA² − ½kx² + ½kx² = ½kA², which does not depend on x. (1) (c) KE = PE gives ½k(A² − x²) = ½kx², so A² = 2x² and x = A/√2. (1)
Mark scheme note: The substitution mω² = k must be shown in (a). In (c) “x = A/2” scores zero.
4. Energy at a Point
5 marks
A 0.80 kg mass on a spring of spring constant 20 N m⁻¹ oscillates with amplitude 0.15 m. Calculate: (a) the angular frequency; (b) the total energy; (c) the maximum speed; (d) the potential and kinetic energies at x = 0.090 m; (e) the speed at x = 0.090 m.
Model solution unlocks after you try
(a) ω = √(k/m) = √(20 ÷ 0.80) = √25 = 5.0 rad s⁻¹ (1) (b) E = ½kA² = ½ × 20 × 0.0225 = 0.225 J (1) (c) v_max = ωA = 5.0 × 0.15 = 0.75 m s⁻¹ (check: √(2 × 0.225 ÷ 0.80) = 0.75) (1) (d) PE = ½ × 20 × 0.0081 = 0.081 J; KE = 0.225 − 0.081 = 0.144 J (1) (e) v = √(2 × 0.144 ÷ 0.80) = √0.36 = 0.60 m s⁻¹ (1)
Mark scheme note: Error carried forward allowed throughout. Both energies are needed for (d). Units required.
5. A Swinging Bob
4 marks
A 0.20 kg bob is pulled aside until it is 4.0 cm above its lowest point, then released (g = 10 m s⁻²). (a) Calculate its total energy. (b) Calculate its greatest speed. (c) Air resistance removes 20 % of its energy by the time it reaches the other side. How high does it rise there? (d) Where did the lost energy go?
Model solution unlocks after you try
(a) E = mgh = 0.20 × 10 × 0.040 = 0.080 J (1) (b) v = √(2gh) = √(2 × 10 × 0.040) = √0.80 = 0.89 m s⁻¹ (1) (c) Energy left = 0.80 × 0.080 = 0.064 J; h = E/mg = 0.064 ÷ 2.0 = 0.032 m = 3.2 cm. (1) (d) It was transferred to the surrounding air as heat (internal energy) and a little sound, by the work done against air resistance. (1)
Mark scheme note: In (c) giving 0.8 cm (the 20 % lost, not the height reached) or 4.8 cm loses the mark. In (d) “it was destroyed” scores zero.
6. Practical: Damping in Water
6 marks
A 0.30 kg load on a spring (k = 23.5 N m⁻¹) carried a card vane dipping into a beaker of water. Zainab recorded the amplitude at the end of each cycle: cycle 0 → 6.00 cm, 1 → 4.50, 2 → 3.38, 3 → 2.53, 4 → 1.90, 5 → 1.42 cm. Ten oscillations took 7.10 s with the vane in air and 7.14 s in water.
(a) Show that the amplitude falls by the same ratio each cycle, and find the ratio. (b) What fraction of the energy remains after each cycle? (c) Calculate the energy at cycle 0 and at cycle 5. (d) What do the two timings show about light damping and the period? (e) Describe the shape of a graph of amplitude against cycle number.
Model solution unlocks after you try
(a) 4.50 ÷ 6.00 = 0.75, 3.38 ÷ 4.50 = 0.75, 2.53 ÷ 3.38 = 0.75 (and so on): a constant ratio of 0.75. (1) (b) E ∝ A², so 0.75² = 0.56 (56 %) remains each cycle and about 44 % is lost. (1) (c) E₀ = ½ × 23.5 × 0.0600² = 0.042 J (1); E₅ = ½ × 23.5 × 0.0142² = 0.0024 J (1) (d) The periods are 0.710 s and 0.714 s: almost the same, so light damping hardly changes the period. (1) (e) A curve that falls steeply at first and then more gently, approaching zero but never quite reaching it (an exponential-type decay), not a straight line. (1)
Mark scheme note: In (a) at least three ratios must be shown. In (b) giving 0.75 (forgetting E ∝ A²) scores zero. In (c) amplitudes in centimetres lose the mark. In (e) “a straight line downwards” is not accepted.
7. The Shock Absorber
3 marks
(a) Explain how an oil-filled shock absorber removes energy from a vehicle's bounce. (b) Why are shock absorbers designed to be close to critical damping rather than heavily damped? (c) Give one other device designed to be critically damped.
Model solution unlocks after you try
(a) As the suspension moves, a piston is forced through oil; the oil is pushed through small holes, and the work done against this resistance turns the bounce energy into heat in the oil. (1) (b) Heavy damping would make the suspension return too slowly after each bump, so it would not be ready for the next one and the ride would be harsh; critical damping returns it quickly without bouncing. (1) (c) The pointer of a moving-coil meter (a door closer is also accepted). (1)
Mark scheme note: In (a) the conversion to heat must be stated. In (c) any correct example is accepted.
8. The Bridge: Mallam Sule's Verdict
2 marks
For the okada's rear spring (k = 3.16 × 10⁴ N m⁻¹), (a) calculate the energy in a 2.0 cm bounce and in a 6.2 cm bounce, and (b) explain why a working shock absorber stops the bounce reaching 6.2 cm.
Model solution unlocks after you try
(a) E₁ = ½ × 31600 × 0.020² = 6.3 J; E₂ = ½ × 31600 × 0.062² = 61 J (nearly ten times as much). (1) (b) The working damper turns a large share of each cycle's energy into heat in its oil, so the road can never pile up the 61 J needed; the leaking one removes very little, so energy can build up when the pushes arrive in step. (1)
Mark scheme note: Both energies are needed for (a). In (b) the energy-removal (heat) idea is essential.
TTeacher notes
For the teacher
Timing, damping note, competences, support tap to open
Timing: 0–5 Do Now · 5–15 Wonder · 15–45 Investigate · 45–50 movement break (energy-swap squat, then “add damping” through imagined thick oil) · 50–65 Make Sense · 65–75 Apply · 75–80 Reflect.
Damping note: textbooks describe shock absorbers as near-critical; real ones are set a little under critical (damping ratio about 0.2–0.4) for comfort, which the model's healthy value of 0.30 reflects. The bench ratios per cycle (0.97 / 0.75 / 0.40) and the okada numbers are an internally consistent teaching model.
Competences: L1 (where did the energy go?), L3 (worn shock absorbers are a road-safety hazard), L4 (energy and decay graphs), L6 (a mechanic's diagnosis).
Support: energy-bars worksheet. Stretch: show KE = ½mω²(A² − x²) from v = ω√(A² − x²).