Oscillations, SHM, damping and resonance: worked solutions

Answer: a = −(2g/ℓ)x, so ω² = 32.7 s⁻²; T = 1.10 s. At x = +0.0150 m, a = −0.491 m/s².

1. a = F/m = −2ρSgx/(ρSℓ) = −(2g/ℓ)x.
2. g and ℓ are constant, so acceleration magnitude is proportional to displacement magnitude. Its sign is opposite to x: it acts back towards equilibrium. These together establish SHM for the supplied ideal model.
3. ω² = 2g/ℓ = 32.7 s⁻²;
   T = 2π√[ℓ/(2g)] = 1.10 s.
4. a = −32.7(0.0150) = −0.491 m s⁻². The left surface accelerates downwards.

Check: Density and area cancel because they multiply both the restoring force and moving mass. ω² has unit s⁻². Both liquid surfaces move, which accounts for the factor of 2 in the supplied force.

Mark checklist · 5 marks

• 1 mark — Uses F = ma and the stated moving mass to obtain a = −(2g/ℓ)x.
• 1 mark — Identifies the constant coefficient and proportional dependence on x.
• 1 mark — Explains acceleration/restoring resultant opposes displacement, hence SHM.
• 1 mark — Identifies ω² = 2g/ℓ and obtains T ≈ 1.10 s.
• 1 mark — Converts mm to m and obtains signed a ≈ −0.491 m/s².

Common mistake: Repeating “a is proportional to −x” without deriving it from these forces does not show that this system qualifies. Use the given mass and explain which parameters are constant.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/1, June 2023 — Q06.1; report p. 6

Answer: A = 0.040 m; T = 0.80 s; ω = 7.85 rad/s; vmax = 0.314 m/s; amax = 2.47 m/s². Maximum positive velocity is at 0.60 s. x and a are in antiphase.

1. A = 0.040 m; T = 0.80 s;
   ω = 2π/0.80 = 7.85 rad s⁻¹.
2. vmax = ωA = 0.314 m s⁻¹;
   amax = ω²A = 2.47 m s⁻².
3. At 0.20 s the displacement decreases fastest, so v is at its negative minimum. At 0.60 s it increases fastest, so v is at its positive maximum.
4. x = 0.040 cos(7.85t) m;
   v = −0.314 sin(7.85t) m s⁻¹;
   a = −2.47 cos(7.85t) m s⁻². Use radians and unrounded ω when calculating.
5. v starts at zero and initially becomes negative. a starts at −amax. x and a differ by π rad (half a cycle); x and v have a quarter-cycle separation in their variations. The signed equations or graph slopes establish the direction, rather than an ambiguous “leads/lags” statement.

Check: At t = 0, v = 0 but acceleration is most negative. At both equilibrium crossings a = 0, while speed is greatest. This independently checks the signs.

Mark checklist · 7 marks

• 1 mark — Reads A = 0.040 m and T = 0.80 s, not the peak-to-peak displacement or half-period.
• 1 mark — Obtains ω ≈ 7.85 rad/s.
• 1 mark — Uses ωA to obtain vmax ≈ 0.314 m/s.
• 1 mark — Uses ω²A to obtain amax ≈ 2.47 m/s².
• 1 mark — Identifies 0.60 s from the maximum positive slope.
• 1 mark — Correct signed v–t curve with zero endpoints and labelled ±0.314 m/s extrema.
• 1 mark — Correct a–t curve with labelled ±2.47 m/s² extrema, antiphase to x; describes quarter-cycle x/v separation.

Common mistake: Graph height is not velocity. Use its gradient: x = 0 can accompany either positive or negative maximum velocity. A turning point has zero velocity and a nonzero restoring acceleration.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/1, June 2023 — Q06.2; report p. 6

Answer: First crossing t = 0.200 s, v = −0.272 m/s and a = −0.822 m/s². Next crossing t = 1.00 s, with positive velocity +0.272 m/s.

1. ω = 2π/1.20 = 5.24 rad s⁻¹;
   cosωt = 0.0300/0.0600 = 0.500.
   First phase = π/3, so t = (π/3)/ω = 0.200 s.
2. |v| = ω√(A² − x²) = 0.272 m s⁻¹.
3. At the first crossing the displacement is decreasing: v = −0.272 m/s. Positive x means the restoring acceleration is negative.
4. a = −ω²x = −0.822 m s⁻².
5. The later phase is 5π/3: t = (5π/3)/ω = 1.00 s.
   It is moving out towards +A, so v = +0.272 m/s.

Check: The two times sum to T = 1.20 s, as expected for a cosine starting at maximum positive displacement. Equal x gives equal acceleration and speed magnitude, but opposite velocity.

Mark checklist · 7 marks

• 1 mark — Sets cosωt = x/A = 0.500 with ω = 2π/T.
• 1 mark — Chooses first phase π/3 and obtains 0.200 s.
• 1 mark — Obtains speed ≈ 0.272 m/s from the SHM velocity relation.
• 1 mark — Chooses negative velocity for the first crossing.
• 1 mark — Finds a ≈ −0.822 m/s² with the restoring sign.
• 1 mark — Chooses later phase 5π/3 and time 1.00 s.
• 1 mark — Chooses positive velocity at the later crossing.

Common mistake: An inverse cosine returns one phase. It does not identify every crossing; use the cycle and direction. If ωt is in radians, set the calculator accordingly.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Answer: Gradient = −64.0 s⁻²; ω = 8.00 rad/s; T = 0.785 s; vmax = 0.400 m/s. Half-KE points: (+0.0354 m, −2.26 m/s²) and (−0.0354 m, +2.26 m/s²).

1. Gradient = (−3.20 − 3.20)/(0.0500 − (−0.0500)) = −64.0 s⁻².
   a = −ω²x, so ω = √64.0 = 8.00 rad s⁻¹.
2. Draw a single straight line through the origin from (−0.0500, +3.20) to (+0.0500, −3.20). It occupies the second and fourth quadrants, with both axes showing signed values and units.
3. T = 2π/ω = 0.785 s;
   vmax = ωA = 0.400 m s⁻¹.
4. Ek/Ek,max = (A² − x²)/A² = 0.500.
   x² = 0.500A², so x = ±A/√2 = ±0.0354 m.
5. a = −64.0x gives ∓2.26 m s⁻². Mark both points on the straight line, rather than halfway between origin and endpoint.

Check: At x = A/2, kinetic energy would be 75%, not 50%, of maximum. Half kinetic energy means speed vmax/√2, not vmax/2.

Mark checklist · 7 marks

• 1 mark — Finds signed gradient −64.0 s⁻² from the correctly scaled straight line/table.
• 1 mark — Uses gradient = −ω², giving 8.00 rad/s.
• 1 mark — Finds T ≈ 0.785 s.
• 1 mark — Finds vmax = 0.400 m/s.
• 1 mark — Sets Ek/Ek,max = 1 − x²/A² = 0.500.
• 1 mark — Obtains both x = ±0.0354 m.
• 1 mark — Marks the corresponding opposite-sign a values ±2.26 m/s² on the negative-gradient line.

Common mistake: A sine curve is appropriate for a–t, not a–x. Halfway along this line does not mean half energy because the energy contains x².

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/1, June 2025 — Q03.1; report p. 4 · AQA 7408/1, June 2025 — Q03.2; report p. 4

Answer: E = 0.0800 J; T = 0.497 s. At +0.0300 m: Ep = 0.0288 J, Ek = 0.0512 J, v = ±0.506 m/s. Energy repeats every T/2 = 0.248 s.

1. E = ½(64.0)(0.0500)² = 0.0800 J;
   T = 2π√(0.400/64.0) = 0.497 s.
2. Ep = ½(64.0)(0.0300)² = 0.0288 J;
   Ek = 0.0800 − 0.0288 = 0.0512 J.
3. v = ±√(2 × 0.0512/0.400) = ±0.506 m s⁻¹. Choose the sign only if the direction is known.
4. Against x, Ep is an upward parabola and Ek a downward parabola over −A to +A. Total energy is a horizontal line. The same energy occurs at equal positive and negative |x|.
5. For release at +A, Ep = E cos²ωt and Ek = E sin²ωt. Their maxima alternate; each repeats twice per displacement cycle. Total remains constant and their repetition period is T/2 = 0.248 s.

Check: Ep + Ek = 0.0800 J and |v| is below vmax = √(k/m)A = 0.632 m/s. Energies never become negative in this stated range.

Mark checklist · 7 marks

• 1 mark — Finds E = 0.0800 J.
• 1 mark — Finds T ≈ 0.497 s.
• 1 mark — Finds Ep = 0.0288 J.
• 1 mark — Finds Ek = 0.0512 J by conservation.
• 1 mark — Finds both velocities ±0.506 m/s.
• 1 mark — Describes correct parabolic x dependence and constant total.
• 1 mark — Describes alternating twice-per-cycle time energies and T/2 ≈ 0.248 s.

Common mistake: Energy is proportional to x², not x. Mechanical energy stays constant while kinetic energy changes; neither kinetic nor potential energy alone is conserved.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Answer: e₀ = 0.123 m; F = −ky; T = 0.702 s. At release the spring pulls up with 3.90 N and a = −3.20 m/s². Oscillation potential is combined elastic plus gravitational energy relative to equilibrium.

1. At equilibrium ke₀ = mg, so e₀ = (0.300 × 9.81)/24.0 = 0.123 m.
2. Fdown = mg − k(e₀ + y) = −ky, giving a = −(k/m)y. Gravity balances the equilibrium part of the spring force, not the extra restoring part.
3. T = 2π√(m/k) = 0.702 s.
   At release: spring force k(e₀ + 0.0400) = 3.90 N upwards;
   Fdown = −24.0(0.0400) = −0.960 N;
   a = F/m = −3.20 m s⁻².
4. ΔUelastic = ½k[(e₀ + y)² − e₀²] = ke₀y + ½ky².
   ΔUgravity = −mgy. Since ke₀ = mg, their linear terms cancel:
   ΔUtotal = ½ky².
5. This is the combined potential energy relative to equilibrium. The actual spring’s elastic energy there is ½ke₀², not zero.

Check: The minimum extension over this oscillation is e₀ − A = 0.0826 m, so the spring remains stretched. Oscillation energy is ½kA² = 0.0192 J.

Mark checklist · 7 marks

• 1 mark — Obtains e₀ ≈ 0.123 m from ke₀ = mg.
• 1 mark — Writes signed resultant mg − k(e₀ + y) = −ky.
• 1 mark — Finds T ≈ 0.702 s.
• 1 mark — Finds spring-force magnitude ≈ 3.90 N upwards.
• 1 mark — Finds a = −3.20 m/s² (or 3.20 m/s² upwards).
• 1 mark — Includes both elastic change and gravitational change in the potential accounting.
• 1 mark — Shows cancellation using ke₀ = mg and identifies equilibrium reference; not actual elastic energy alone.

Common mistake: Using total extension e₀ + y in a = −(k/m)x predicts nonzero acceleration at equilibrium. Use displacement from loaded equilibrium. Do not identify ½ky² as elastic energy alone for this vertical system.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Answer: k = 24.7 N/m. Intercept = 1.60 × 0.0500 = 0.0800 s². Mean T = 0.492 s, ΔT = 0.010 s; approximate percentage uncertainty in T² = 4.07%.

1. Clamp the spring securely, use a small vertical displacement and release without a push. Count 20 complete cycles between same-direction crossings of a marker, repeat and average. Keep the same spring and a small amplitude; avoid sideways swing and exceeding the proportional range.
2. T² = (4π²/k)(mₐ + mh) = (4π²/k)mₐ + (4π²/k)mh.
   Plot T² vertically against added mass horizontally. Gradient = 1.60 s² kg⁻¹;
   k = 4π²/1.60 = 24.7 N m⁻¹.
3. Intercept = (1.60)(0.0500) = 0.0800 s². It is the hanger contribution; plotting against total moving mass would put this ideal-model line through the origin.
4. Mean 20-cycle time = (9.80 + 9.88 + 9.84)/3 = 9.84 s;
   T = 9.84/20 = 0.492 s; ΔT = 0.20/20 = 0.010 s.
5. % uncertainty in T = 100(0.010/0.492) = 2.03%;
   % uncertainty in T² ≈ 2 × 2.03% = 4.07%.
6. This is the approximate uncertainty in the squared period at one mass. It is not automatically the uncertainty in k from a multi-point fit; that requires allowed-gradient analysis and any relevant mass uncertainty.

Check: The repeated mean gives T² = 0.242 s², close to the 0.240 s² central value and within the stated uncertainty. A stiffness calculated from one total mass is also close to the fitted value.

Mark checklist · 9 marks

• 1 mark — Times many complete same-direction cycles, repeats and averages, then divides by cycle count.
• 1 mark — Gives a contextual control/reliability point: same spring, small amplitude, no push/sideways motion or proportional range.
• 1 mark — Chooses T² versus mₐ and relates gradient to 4π²/k.
• 1 mark — Finds k ≈ 24.7 N/m.
• 1 mark — Explains hanger mass contributes to total moving mass.
• 1 mark — Checks intercept 1.60 × 0.0500 = 0.0800 s².
• 1 mark — Averages timings then obtains T = 0.492 s.
• 1 mark — Obtains ΔT = 0.010 s by dividing the supplied whole-run uncertainty by 20.
• 1 mark — Doubles percentage T uncertainty to obtain approximately 4.07% for T².

Common mistake: Using mₐ alone as the total moving mass or forcing the fit through the origin biases the result. Do not leave ΔT at 0.20 s or assume repeats remove the supplied timing allowance.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/3A, November 2020 — Q01.5–01.6; report pp. 3–4

Answer: g = 9.80 m/s²; allowed g interval 9.61–9.99 m/s². A +0.0200 m length offset leaves gradient 4.03 s²/m but shifts intercept to −0.0796 s².

1. Measure from the pivot to the bob centre, allowing for bob radius and reading without parallax. Use a small initial angle and release without a push; keep motion in one vertical plane.
2. Place a fixed fiducial mark at equilibrium, where the bob moves fastest, so the crossing instant is more sharply defined. Time many complete same-direction cycles, repeat and average; change length while retaining a small angle.
3. T² = (4π²/g)ℓ. Plot T² vertically against ℓ horizontally, using uncertainty bars and a best-fit line rather than a single point or forcing the origin.
   g = 4π²/4.03 = 9.80 m s⁻².
4. gmin = 4π²/4.11 = 9.61 m s⁻²;
   gmax = 4π²/3.95 = 9.99 m s⁻².
5. Let ℓrecorded = ℓtrue + 0.0200 m.
   T² = 4.03(ℓrecorded − 0.0200) + 0.001
   = 4.03ℓrecorded − 0.0796 s².
6. This systematic offset shifts the line horizontally and changes its intercept, not its gradient. An intercept alone does not uniquely diagnose an error; check the measurement model.

Check: The interval includes 9.81 m/s². Pendulum bob mass does not appear in the small-angle period. Large angles invalidate the simple formula’s exact amplitude independence.

Mark checklist · 9 marks

• 1 mark — Measures pivot to bob centre with bob-radius/parallax allowance.
• 1 mark — Uses a small angle, releases without pushing and keeps one-plane motion.
• 1 mark — Uses an equilibrium marker because speed is greatest there, with many same-direction cycles and repeat averaging.
• 1 mark — Links T²–ℓ gradient to 4π²/g.
• 1 mark — Finds g ≈ 9.80 m/s².
• 1 mark — Uses inverse-gradient extremes in the correct order.
• 1 mark — Obtains approximately 9.61–9.99 m/s².
• 1 mark — Substitutes ℓtrue = ℓrecorded − 0.0200 so gradient is unchanged.
• 1 mark — Obtains new intercept −0.0796 s².

Common mistake: A marker at a turning point gives a poorly defined instant because the bob moves slowly there. A positive length error produces a negative T² intercept in this chosen-axis convention, not a larger gradient.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/3A, November 2020 — Q01.1; report p. 3 · AQA 7408/3A, November 2020 — Q01.5–01.6; report pp. 3–4

Answer: Gradient ln q = −0.1116 per cycle; q = 0.894. By n = 6, 26.2% of initial energy remains and 73.8% is lost.

1. ln(Aₙ/mm) = ln(A₀/mm) + n ln q.
   This is a straight line: gradient ln q, intercept ln(50.0). The logarithm argument is dimensionless.
2. q² = 40.0/50.0 = 0.800, so q = √0.800 = 0.894.
   Gradient = ln q = −0.1116 per cycle.
3. E6/E0 = (25.6/50.0)² = 0.262144.
   Energy retained = 26.2%; energy lost = 73.8%.
4. Damping transfers mechanical energy to the surroundings, for example as thermal energy. Its force depends on motion as well as displacement; the net acceleration is not determined by x alone as −ω²x. Lightly damped oscillations can resemble SHM, but their decaying amplitude is not ideal constant-amplitude SHM.

Check: Each pair of cycles multiplies amplitude by 0.800 and energy by 0.640. The logarithmic slope is negative as amplitude decays.

Mark checklist · 7 marks

• 1 mark — Takes logs of the given law with a dimensionless amplitude ratio.
• 1 mark — Identifies gradient = ln q and intercept ln(A₀/mm).
• 1 mark — Uses q² = 0.800 or an equivalent multi-cycle route.
• 1 mark — Finds q ≈ 0.894 and gradient ≈ −0.1116.
• 1 mark — Uses squared amplitude ratio to find retained energy fraction 0.262.
• 1 mark — Subtracts from initial energy to obtain 73.8% lost.
• 1 mark — Explains damping adds a motion-dependent force so acceleration is not solely proportional to −x.

Common mistake: A 48.8% amplitude loss by n = 6 is not the energy loss. Square the retained amplitude fraction first. Do not take ln of a dimensional length.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/3A, November 2020 — Q01.7; report p. 4

Answer: Lightly damped free motion is close to 2.00 Hz; settled forced motion is at 1.20 Hz. Resonance gives maximum response near natural frequency. Greater damping lowers and broadens the peak.

1. After a single displacement and release without continuing drive, L oscillates freely at a frequency close to its 2.00 Hz undamped natural frequency while its amplitude decays. With a continuing 1.20 Hz drive, the settled forced motion is at 1.20 Hz; initial transients need not be.
2. Resonance is the maximum steady amplitude response when the driving frequency is close to the natural frequency for light damping, with effective energy transfer from the driver. On L the peak is close to 2.00 Hz. Exact peak equality is the negligible-damping approximation.
3. Greater damping dissipates more mechanical energy, lowering the peak. The curve becomes broader, so resonance is less sharp. With appreciable damping the displacement-response peak can shift lower; very strong damping may give no nonzero-frequency peak.
4. This comparison holds mass, stiffness and driving-force amplitude fixed. Changing stiffness as well as damping also changes natural frequency; it would be a different comparison. No quantitative driven-oscillator equation or phase-lag formula is required here.

Check: The two curves share one amplitude reference and frequency scale. A lower peak is not proof that the natural frequency has increased.

Mark checklist · 6 marks

• 1 mark — Distinguishes free vibration at approximately the natural frequency after release.
• 1 mark — Gives settled forced frequency 1.20 Hz, with transients distinguished.
• 1 mark — Identifies near-natural driving frequency at resonance.
• 1 mark — Links resonance to maximum steady amplitude/effective energy transfer.
• 1 mark — Explains lower peak through greater energy dissipation.
• 1 mark — Describes broader/less sharp response with the other variables fixed.

Common mistake: “More damping stops resonance” is too broad. Explain lower amplitude and broader response. Do not attribute a higher-frequency peak to damping alone when stiffness has also been changed.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/1, June 2023 — Q06.3; report p. 6 · AQA 7408/1, June 2017 — Q07.3; report p. 5

Answer: Driving frequencies are 1.40, 1.60 and 2.00 Hz; 4.00 m/s gives the largest response. Stated large-response range: 3.75–4.25 m/s.

1. fdrive = v/d:
   3.50/2.50 = 1.40 Hz;
   4.00/2.50 = 1.60 Hz;
   5.00/2.50 = 2.00 Hz.
2. The 4.00 m/s case matches the natural frequency and lies in the stated high-response band. It has the greatest response of these three within this model; the fastest case does not.
3. vlow = 2.50(1.50) = 3.75 m/s;
   vhigh = 2.50(1.70) = 4.25 m/s.
4. One approach is a damper acting between the platform and its support. Relative motion dissipates mechanical energy, for example into heat, reducing the resonance amplitude. Alternatively change stiffness/mass so the natural frequency is farther from the entire drive range; verify the resulting response, rather than assume any change is helpful.

Check: v/d has unit s⁻¹. Doubling strip spacing at fixed speed halves the disturbance frequency. Greater damping reduces the peak but broadens response; compare amplitudes rather than assuming the old threshold band is unchanged.

Mark checklist · 6 marks

• 1 mark — Calculates all three frequencies: 1.40, 1.60 and 2.00 Hz.
• 1 mark — Identifies 4.00 m/s as the near-natural largest-response case with a frequency-based reason.
• 1 mark — Uses v = dfdrive to map the interval.
• 1 mark — Finds inclusive 3.75–4.25 m/s bounds.
• 1 mark — Proposes a specific damper or justified stiffness/mass retuning.
• 1 mark — Explains energy dissipation/lower peak or a checked shift away from the relevant driving frequencies.

Common mistake: Calling the fastest speed “most dangerous” ignores the driving-frequency match. Use the supplied repeated-disturbance model and support the comparison with numbers.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.

Related examiner evidence: AQA 7408/1, June 2023 — Q06.4; report p. 6

Answer: c = 150 m/s; f1 = 100 Hz. A 200 Hz drive selects the second harmonic with nodes at 0, 0.375 and 0.750 m; adjacent loops are in antiphase. With four times the tension it selects the first harmonic.

1. c = √[18.0/(8.00 × 10⁻⁴)] = 150 m s⁻¹;
   f1 = c/(2ℓ) = 150/(2 × 0.750) = 100 Hz.
2. 200 Hz = 2f1, so the drive excites the second harmonic. Reflected waves travelling in opposite directions superpose to form a stationary pattern at an allowed resonant frequency.
3. Nodes lie at both ends and the centre: 0, 0.375 and 0.750 m. The two antinodes are at 0.1875 and 0.5625 m. Points in adjacent loops oscillate in antiphase, π rad; a node has zero amplitude, so do not assign it an oscillation phase.
4. New tension = 72.0 N, so c doubles to 300 m/s and f1 doubles to 200 Hz.
5. The unchanged 200 Hz drive now excites the first harmonic: one loop with only the end nodes, rather than the original second-harmonic pattern.

Check: For the second harmonic λ = ℓ = 0.750 m and c = fλ = 150 m/s. Particle vibration frequency is 200 Hz; this is not the wave speed.

Mark checklist · 6 marks

• 1 mark — Finds c = 150 m/s with μ in SI units.
• 1 mark — Finds f1 = 100 Hz.
• 1 mark — Identifies second harmonic and nodes at 0, 0.375 and 0.750 m.
• 1 mark — Gives adjacent-loop antiphase/π rad, with nodes not assigned a phase.
• 1 mark — Explains fourfold tension doubles wave speed and harmonic frequencies.
• 1 mark — Finds new f1 = 200 Hz and identifies first harmonic at the unchanged drive.

Common mistake: Doubling wave speed does not keep the harmonic number fixed at a fixed drive. Recalculate allowed frequencies. Stationary-wave phase is shared within each loop and reverses across a node.

Exam technique: Define displacement from equilibrium and keep directions, units and model assumptions explicit. These are original teaching questions with PhysicsUK’s own mark guidance.