Forced Vibration and Resonance

1. Every cone was pushed at the driver's frequency through the cord, but a pendulum's natural frequency depends on its length (T = 2π√(l/g)). Only the 0.60 m cone had a natural frequency equal to the driving frequency, so only it resonated.
2. The ridges are a fixed 2.5 m apart. The faster the okada goes, the less time it takes to reach the next ridge, so more ridges are met each second: f = v/d grows in step with v.
3. A was largest at f = 2.00 Hz (18 km/h), which equals the okada's natural frequency f₀ = 2.00 Hz: that is resonance.
4. It made the peak much lower (about 2.0 cm instead of 10.0 cm) and broader, at about the same frequency, so the bounce never reached the lift-off threshold.
5. More mass on the same spring lowers f₀ (f₀ ∝ 1/√m), from 2.00 Hz to 1.79 Hz. Resonance needs v = d f₀, so the dangerous speed falls to about 16 km/h.

v = d f₀ = 3.0 × 2.0 = 6.0 m s⁻¹ (1); 6.0 × 3.6 = 21.6 km/h (1).

A good answer gives the natural period (for example 2.5 s, f₀ ≈ 0.4 Hz) and explains that in-step pushes build the biggest swing because each push adds energy at the right moment (resonance: driving frequency = natural frequency), while pushes at other rhythms sometimes act against the motion and take energy out. A strong answer links this to the ridges pushing Dayo's okada once per bounce at 18 km/h, notes that friction and air resistance (damping) stopped the swing growing for ever, and gives a safety point about keeping the amplitude small.

(a) Vibration of a system at its own natural frequency, with no outside periodic force acting, after it is displaced and released (ideally with constant amplitude). (1) (b) Vibration whose amplitude decreases with time because energy is removed by resistive forces. (1) (c) Vibration of a system caused by an outside periodic force; the system vibrates at the frequency of that driving force. (1)

Mark scheme note: In (c) "at the driving frequency" is needed; "a vibration that is forced" scores zero.

(a) Resonance occurs when a system is driven by a periodic force whose frequency equals its natural frequency, so that it vibrates with the greatest amplitude. (1) (b) Useful (any one): tuning a radio · the hollow body of a guitar or talking drum amplifying the sound · pushing a swing in time (1). Harmful (any one): a bridge set swaying by marching feet or wind · a vehicle shaking violently at one speed (Dayo's okada) · machinery vibrating dangerously on its mounting (1).

Mark scheme note: In (a) both "frequency equals natural frequency" and "greatest amplitude" are needed. "A big vibration" alone scores zero.

(a) A heavy driver pendulum and several light paper-cone pendulums of different lengths hang from the same taut cord; one cone has the same length as the driver. (1) (b) All the cones begin to swing, but the one equal in length to the driver swings with by far the largest amplitude. (1) (c) The driver forces every cone at its own frequency through the cord; only the equal-length cone has a natural frequency equal to the driving frequency, so it resonates. (1) (d) Heavier cones are less affected by air resistance (relatively less damped), so the resonance becomes sharper: the matching cone swings even more widely compared with the others. (1)

Mark scheme note: In (c) the frequency-match idea is essential; "it is the same length" without linking length to natural frequency earns no mark.

(a) The time between ridges is the distance between them divided by the speed, d/v; the number met per second is the reciprocal, f = v/d. (1) (b) Resonance when f = f₀, so v = d f₀. (1) (c) v = 2.5 × 2.0 = 5.0 m s⁻¹ (1); 5.0 × 3.6 = 18 km/h. (1)

Mark scheme note: In (c) the conversion factor 3.6 must be applied correctly; 5.0 km/h loses the last mark.

(a) ω = √(k/m) = √(50 000 ÷ 400) = √125 = 11.2 rad s⁻¹; f₀ = 11.18 ÷ 6.283 = 1.78 Hz. (1) (b) v = d f₀ = 3.0 × 1.78 = 5.34 m s⁻¹ = 5.34 × 3.6 km h⁻¹ = 19.2 km h⁻¹. (1) (c) f₀ = (1/2π)√(50 000 ÷ 500) = 10 ÷ 6.283 = 1.59 Hz (1); v = 3.0 × 1.59 = 4.77 m s⁻¹ = 17.2 km/h. (1) (d) Extra load lowers the natural frequency, so the resonant speed moves lower; a speed chosen to avoid resonance when empty may now be close to the new dangerous speed. (1)

Mark scheme note: Error carried forward allowed. Forgetting to change m s⁻¹ to km/h loses one mark only once. In (d) the link "more mass → lower f₀ → lower dangerous speed" is required.

(a) v in m s⁻¹ is 2.78, 3.89, 4.44, 5.00, 5.56, 8.33; f = v/2.5 = 1.11, 1.56, 1.78, 2.00, 2.22, 3.33 Hz. (1) (b) A rises slowly, then steeply to a sharp peak, and falls away again: the peak is at f = 2.00 Hz. (1) (c) The peak frequency equals the natural frequency f₀ = 2.0 Hz: resonance. (1) (d) Only at 18 km/h (10.0 cm > 6.2 cm). (1) (e) The peak is much lower (about 2.0 cm, never reaching 6.2 cm) and broader, at about the same frequency. (1) (f) At 30 km/h the forcing frequency (3.33 Hz) is far from f₀, so the amplitude is only 0.6 cm. (1)

Mark scheme note: All six frequencies are needed in (a). In (c) the word resonance is required. In (e) both "lower" and "broader" (or "never reaches 6.2 cm") are needed.

(a) T = 10.0 ÷ 20 = 0.50 s; k = 4π²m/T² = 39.48 × 200 ÷ 0.25 = 3.16 × 10⁴ N m⁻¹. (1) (b) sag = mg/k = 200 × 9.8 ÷ 31600 = 0.062 m = 6.2 cm. (1) (c) Lift-off needs ω²A ≥ g, so A ≥ g/ω² = g × m/k = mg/k: exactly the static sag, 6.2 cm. (1)

Mark scheme note: In (a) accept 3.2 × 10⁴ N m⁻¹. In (c) the step ω² = k/m must be used; the numerical coincidence alone is not credited.

(1) Replace the leaking rear shock absorber: a healthy damper turns bounce energy into heat and keeps the resonant amplitude near 2 cm, well below the 6.2 cm lift-off threshold. (1) (2) Until it is replaced, cross the ridge stretch slowly (about 10 km/h or below), well away from the resonant speeds of 18 km/h (usual load) and 16 km/h (extra crates), so the ridges do not push in step with the bounce. (1)

Mark scheme note: Advice without a reason scores zero. "Ride faster" alone is not credited: it means passing through the resonant speeds on the way and gives no margin.