What makes a motion simple harmonic, and why do two extra crates slow the okada's bounce?
Four quick cards on period, frequency, amplitude, ω and weight. Answer each in your head before you reveal it.
Hajia Ramatu wants more eggs delivered today. Dayo straps two extra crates onto the okada's carrier and Tunde is sure the heavier bike will ride steadier. Zainab is not so sure…
The 3D story couldn't start on this device. Read the dialogue below instead.
Scene 1 · Dayo: He straps two extra crates onto the carrier. “Hajia wants more eggs today. Heavier load, steadier ride — that's what my oga used to say.”
Scene 2 · Tunde: “Exactly. Heavy things don't bounce.”
Scene 3 · Zainab: She presses the loaded carrier down, lets go and times it. The bounce looks lazier. “It's still bouncing — just slower.”
Scene 4 · Mrs Okafor (lab): She hangs 50 g slotted masses on a spiral spring and the pointer drops step by step. “Every extra 50 grams pulls it down by the same amount. That even pull-back is the secret.”
Scene 5 · Mrs Okafor: She pulls the hanger down and releases it. “The further you pull, the harder the spring pulls back — always towards the middle. That is what makes the motion simple harmonic.”
Two extra crates are on the carrier. Will the okada bounce faster, slower or at the same rhythm? Why? Pin your idea before we measure.
Sort all 11 cards. Drag a card, or tap a card and then tap a bin. Ask: is there a force that grows with the displacement and points back to the middle?
Mrs Okafor's spring has k = 24.5 N m⁻¹ (a light spring; its own mass is ignored). The hanger itself is the first 50 g. Read the pointer for five loads, then time 20 oscillations for each.
Hang it and read the pointer.
Up to 250 g; record each extension x.
Time 20 oscillations for every load.
F against x, then T² against m.
The 3D bench couldn't start on this device. Use the expected results and worked examples below.
| m / kg | F = mg / N | x / cm | t (20) / s | T / s | T² / s² |
|---|
Loads read: 0 of 5 · timed: 0 of 5
| Okada load / kg | t (20) / s | T₀ / s | f₀ / Hz |
|---|
| m / kg | F / N | x / cm | t / s | T / s | T² / s² |
|---|---|---|---|---|---|
| 0.050 | 0.490 | 2.0 | 5.68 | 0.284 | 0.0806 |
| 0.100 | 0.980 | 4.0 | 8.03 | 0.401 | 0.161 |
| 0.150 | 1.47 | 6.0 | 9.83 | 0.492 | 0.242 |
| 0.200 | 1.96 | 8.0 | 11.35 | 0.568 | 0.322 |
| 0.250 | 2.45 | 10.0 | 12.69 | 0.635 | 0.403 |
Okada (model, k = 3.16 × 10⁴ N m⁻¹): 200 kg → 20 bounces in 10.0 s; 250 kg → 11.2 s.
T and T² are worked from the unrounded times, so squaring the displayed T can differ in the last digit: 0.2839 s gives T² = 0.0806 s², and 8.028 s gives T = 0.401 s. Either rounding is accepted.
Precautions: the pointer was read with the eye level with it to avoid parallax; the spring was not loaded beyond its elastic limit (it returned to its original reading when unloaded); the mass was pulled down only slightly so it moved vertically without swinging; timing started at the equilibrium position and was repeated.
Sources of error: the spring's own mass (makes T slightly longer than 2π√(m/k)); sideways swaying of the load; reaction time.
Five minutes, no screens. Stand with a partner. One of you holds both arms out straight in front. The other gently pulls those hands forward a little, then a little more. The further the pull, the harder the "spring" partner pulls back towards the middle. Let go and feel the arms spring back past the middle and return. Swap roles, shake out your arms, and sit down when your teacher calls you back.
Now we put names and equations on what the bench showed.
Hooke's law: provided the elastic limit is not exceeded, the extension of a spring is directly proportional to the stretching force: F = kx. The spring constant k (N m⁻¹) is the force needed for each metre of extension.
Restoring force and the SHM condition. Displace the mass from equilibrium and the spring pushes or pulls it back with F = −kx: proportional to the displacement and opposite to it. Newton's second law gives a = F/m = −(k/m)x. Any motion with a = −ω²x (ω² a constant) is simple harmonic: the acceleration is proportional to the displacement from a fixed point and always directed towards that point.
| Change | Effect on T | Effect on f |
|---|---|---|
| Heavier load (m ↑) | longer, T ∝ √m | lower |
| Stiffer spring (k ↑) | shorter, T ∝ 1/√k | higher |
| Bigger amplitude | no change | no change |
Tap a pin. The lab pins hang 50–250 g on the 24.5 N m⁻¹ spring; the okada pins load its rear spring (model k = 3.16 × 10⁴ N m⁻¹) with 200 kg or 250 kg.
The red arrow is the restoring force F = −kx. It is zero at the dashed equilibrium plane, longest at the ends, and always points back to the middle. Switch to the weighted test-tube: when it is pushed down, the extra upthrust is the restoring force.
The 3D zoom couldn't start on this device. The equations above carry the same idea.
Try each problem on paper first, then check the working. Take g = 9.8 m s⁻².
A 0.30 kg load stretches a spring by 6.0 cm. Find k and the period when the load bounces.
F = mg = 0.30 × 9.8 = 2.94 N, and x = 6.0 cm = 0.060 m.k = F ÷ x = 2.94 ÷ 0.060 = 49 N m⁻¹.T = 2π√(0.30 ÷ 49) = 2π√0.00612.T = 2 × 3.142 × 0.0782 = 0.49 s.For the spring in Example 1, find the acceleration of the load when it is 2.0 cm below equilibrium, and the angular frequency.
k ÷ m = 49 ÷ 0.30 = 163.a = −163 × 0.020 = −3.3 m s⁻²; the minus sign means it is directed upwards, towards equilibrium.ω = √163 = 12.8 rad s⁻¹.The okada's rear spring has k = 3.16 × 10⁴ N m⁻¹. Find the bounce period with 200 kg and with 250 kg on it.
T₁ = 2π√(200 ÷ 31600) = 2π√0.00633.T₁ = 2 × 3.142 × 0.0796 = 0.50 s, so f = 2.0 Hz.T₂ = 2π√(250 ÷ 31600) = 2π√0.00791.T₂ = 2 × 3.142 × 0.0889 = 0.56 s, so f = 1.79 Hz.The heavier load bounces more slowly, not less.
Story numbers are a teaching model, not real measurements.
A load of 0.20 kg on a spring has period 0.60 s. Find the period with 0.45 kg on the same spring.
T₂ ÷ T₁ = √(0.45 ÷ 0.20) = √2.25 = 1.5.T₂ = 1.5 × 0.60 = 0.90 s.Load the bench's okada with 200 kg, then 250 kg, and time 20 bounces each. Then answer three questions to settle Tunde's claim.
1 · With two extra crates, the okada…
2 · Its period becomes…
3 · So its bounce frequency is now about…
The big idea of Lesson 2: a restoring force proportional to displacement gives SHM; for a spring T = 2π√(m/k), so more mass means a longer period. Extra crates don't stop the okada's bounce — they slow its rhythm from 2.0 Hz to 1.8 Hz.
Choose an option for instant feedback and a full explanation. Two review questions from Lesson 1 come at the end.
Rate your confidence from 1 to 4 on each goal of Lesson 2.
I can state Hooke's law and find k.
I can state the SHM condition and explain the minus sign.
I can use T = 2π√(m/k).
I can find k from the T²–m slope.
I can solve T ∝ √m ratio problems.
F = mg = 0.25 × 9.8 = 2.45 N and x = 0.050 m (1); k = 2.45 ÷ 0.050 = 49 N m⁻¹ (1).
Type your attempt first to unlock each model solution.
You need: a spring from an old pen or a strong rubber band, a hook or nail, a light plastic bag or cup with a string handle, six identical keys (or coins), a 30 cm ruler and tape.
| Keys | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Reading / cm | 8.0 | 8.9 | 9.8 | 10.6 | 11.5 | 12.4 | 13.3 |
| Extension / cm | 0 | 0.9 | 1.8 | 2.6 | 3.5 | 4.4 | 5.3 |
For the spring, each extra key adds roughly the same extension (about 0.9 cm), so the graph is a straight line through the origin: extension ∝ load, which is Hooke's law. It returned to 8.0 cm, so the elastic limit was not passed. A rubber band usually gives a curve and may not return exactly, so it obeys Hooke's law poorly. This even pull-back is why a spring gives SHM, like the okada's rear spring. One error with its effect: reading the ruler from above made each reading slightly too small (parallax); a swinging cup made readings hard.
Define: (a) the spring constant of a spring; (b) the elastic limit; (c) restoring force.
(a) The force needed to produce unit extension (k = F/x); unit N m⁻¹. (1) (b) The greatest load a spring can carry and still return to its original length when the load is removed. (1) (c) A force on a displaced body that acts back towards its equilibrium position. (1)
Mark scheme note: "the stiffness of the spring" alone scores zero; "the point where it breaks" is the breaking point, not the elastic limit; in (c) the direction towards equilibrium is essential.
(a) State the condition for a body to move with simple harmonic motion. (b) Write it as an equation and explain the minus sign. (c) Give two examples, other than a loaded spring, of systems that move with SHM.
(a) The acceleration is directly proportional to the displacement from a fixed (equilibrium) point and always directed towards that point. (1) (b) a = −ω²x; the minus sign shows the acceleration is always opposite in direction to the displacement. (1) (c) Any two: a small-angle simple pendulum · a weighted test-tube bobbing in a liquid · the prong of a struck tuning fork · a plucked string. (1)
Mark scheme note: both "proportional" and "towards the fixed point" are needed in (a); "repeats its motion" scores zero; a bouncing ball is not accepted in (c).
A mass m hangs on a light spring of spring constant k and is pulled down a further distance x below equilibrium. (a) Write the restoring force. (b) Use Newton's second law to find the acceleration. (c) Hence show the motion is SHM and that T = 2π√(m/k).
(a) The extra stretch x gives an extra upward spring force kx, so F = −kx. (1) (b) a = F/m = −(k/m)x. (1) (c) k/m is a positive constant, so a ∝ x and opposite to it: this is a = −ω²x with ω² = k/m, so the motion is SHM. (1) Then ω = √(k/m) and T = 2π/ω = 2π√(m/k). (1)
Mark scheme note: the minus sign must appear in (a) or (b); writing the formula from memory without the comparison earns only the last mark.
A 0.20 kg mass stretches a spring by 4.0 cm. A 0.50 kg mass is then hung on the same spring and set bouncing (g = 10 m s⁻²). Calculate (a) k; (b) the period with 0.50 kg; (c) its frequency; (d) the size of its acceleration 2.0 cm from equilibrium.
(a) k = 2.0 ÷ 0.040 = 50 N m⁻¹ (1) (b) T = 2π√(0.50 ÷ 50) = 0.628 s (1) (c) f = 1 ÷ 0.628 = 1.59 Hz (1) (d) a = 100 × 0.020 = 2.0 m s⁻², towards equilibrium. (1)
Mark scheme note: error carried forward from (a); using 0.20 kg in (b) gives 0.40 s and loses the mark; x left in cm loses that mark; units required.
The okada's rear spring has k = 3.16 × 10⁴ N m⁻¹ (g = 9.8 m s⁻²). (a) Calculate the static sag with 200 kg. (b) Calculate the bounce period with 200 kg and with 250 kg. (c) Find the ratio of the periods and show it equals √(250/200). (d) How long do 20 bounces take with 250 kg?
(a) x = 1960 ÷ 31600 = 0.062 m (6.2 cm). (1) (b) T₁ = 2π × 0.0796 = 0.50 s (1); T₂ = 2π × 0.0889 = 0.56 s. (1) (c) T₂/T₁ = 0.559 ÷ 0.500 = 1.12 and √1.25 = 1.12, the same, because T ∝ √m. (1) (d) t = 20 × 0.559 = 11.2 s. (1)
Mark scheme note: using 200 kg as the force in (a) loses the mark; the √(m₂/m₁) statement must be shown in (c); accept 0.559 s or 0.56 s. Story numbers are a teaching model, not real measurements.
Zainab timed 20 oscillations for five loads: m / kg 0.100, 0.200, 0.300, 0.400, 0.500 and t / s 7.02, 9.94, 12.16, 14.06, 15.72. (a) Add rows for T (3 d.p.) and T² (3 d.p.). (b) State what goes on each axis and the shape of graph expected. (c) Using (0.100 kg, 0.123 s²) and (0.500 kg, 0.617 s²) from the best-fit line, find the slope. (d) Deduce k.
(a) T = t/20: 0.351, 0.497, 0.608, 0.703, 0.786 s (1); T²: 0.123, 0.247, 0.370, 0.494, 0.618 s². (1) (b) T² vertical, m horizontal; a straight line through the origin. (1) (c) slope = 0.494 ÷ 0.400 = 1.235 s² kg⁻¹ (2) (d) k = 39.48 ÷ 1.235 = 32 N m⁻¹. (1)
Mark scheme note: accept T² within ±0.002 s²; an inverted slope (Δm/ΔT²) loses both marks in (c); "k = slope" scores zero; accept 31.5–32.5 N m⁻¹ on error carried forward.
(a) State two precautions taken in the spiral-spring experiment. (b) The spring itself has some mass. Does this make the measured periods slightly too long or too short? Why?
(a) Any two (1 each): I read the pointer with my eye level with it to avoid parallax · I ensured the spring was not loaded beyond its elastic limit · I pulled the load down only slightly so it bounced vertically without swinging · I started timing as the load passed the equilibrium position and repeated each timing. (2) (b) Too long: part of the spring moves with the load, so the effective oscillating mass is a little more than m. (1)
Mark scheme note: past tense required in (a); in (b) both the direction and the reason are needed.
Dayo says that loading two extra crates makes the okada "steadier, so it won't bounce". Using T = 2π√(m/k), give two reasons why this is wrong.
(1) The spring still provides a restoring force proportional to displacement, so the loaded okada still oscillates in SHM; extra mass does not remove the bounce. (1) (2) Extra mass makes the period longer (0.50 s → 0.56 s) and the frequency lower (2.0 Hz → 1.79 Hz): the bounce becomes slower, not smaller. (1)
Mark scheme note: "heavy things still bounce" without a reason linked to the restoring force or to T ∝ √m scores zero.
Timetable (80 min): 0–5 Do Now · 5–15 Wonder · 15–45 Investigate · 45–50 movement break ("be the spring") · 50–65 Make Sense · 65–75 Apply · 75–80 Reflect.
Competences: L1 (test Tunde's claim), L4 (two graphs, two slopes), L5 (sliders and the 3D bench), L6 (a vehicle loader's real decision).
Support: pre-drawn axes for the F–x and T²–m graphs. Stretch: why does a spring's own mass make T slightly longer than 2π√(m/k)?
Model note: the okada values (k = 3.16 × 10⁴ N m⁻¹, 200 / 250 kg) are a constructed teaching model, not measurements.