Generated ERQ

✓ passed C.4 Standing waves and resonance × D.4 Induction (HL) 14 marks HL 3 passes 122.25s $0.7829
``` ## ERQ · 14 marks · Topics: C.4 Standing waves and resonance + D.4 Induction (HL) · Archetype: theory_application **Integration:** primary=C.4 Standing waves and resonance, secondary=D.4 Induction (HL) (strength: supporting) **Stem.** A student investigates standing waves in an air column using a vertical glass tube, open at the top and connected at the bottom to a water reservoir whose height can be lowered to extend the air column. A tuning fork of frequency f = 512 Hz is held just above the open end. To detect the instant at which the water surface passes a given height, a small cylindrical permanent magnet of mass 4.20 g floats on the water surface, and a flat circular coil of N = 240 turns and mean radius r = 1.50 cm is fixed around the tube at a known height. As the reservoir is lowered, the magnet descends with the water and the coil registers an induced EMF on a data-logger. The student records the two successive air-column lengths at which loud acoustic resonance is heard: L₁ = 0.162 m and L₂ = 0.494 m. The speed of sound in air at the laboratory temperature is to be determined. Assume the magnet behaves as a point dipole of moment p = 0.082 A m². ### Part (a) Define [2 marks] · AO1 · Topic: C.4 State the boundary conditions, in terms of displacement nodes and antinodes, that apply at (i) the open end of the tube and (ii) the water surface, for a standing sound wave inside the air column. ### Part (b)(i) Calculate [3 marks] · AO2 · Topic: C.4 Using L₁ and L₂, calculate the wavelength of the sound in the air column and hence determine the speed of sound v. ### Part (b)(ii) Determine [3 marks] · AO2 · Topic: C.4 The position of the open end of the tube is not exactly the position of the displacement antinode; there is an "end correction" e such that the effective length of the air column is L + e. Determine the value of e from the student's data. ### Part (c) Show that [4 marks] · AO2/AO3 · Topic: C.4 + D.4 As the floating magnet descends through the coil at near-constant speed u, the axial magnetic flux linkage through the coil is approximately Φ(z) = (μ₀ N p)/(2(r² + z²)^{3/2}) · r², where z is the distance from the magnet to the plane of the coil. Show that the induced EMF reaches its maximum magnitude when the magnet is at z = r/2, and not when the magnet is in the plane of the coil (z = 0). ### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: C.4 + D.4 Using the result of part (c), suggest how the student's value of the speed of sound from (b)(i) is affected if the data-logger is used to mark each resonance position from the EMF peak rather than from the audible resonance, and justify whether the student's calculated v is altered. --- ## Mark Scheme ### Part (a) [2 marks] — Define - M1: At the open end (top) — displacement antinode (pressure node) [no ECF] - M2: At the water surface (closed end) — displacement node (pressure antinode) [no ECF] ### Part (b)(i) [3 marks] — Calculate - M1: Recognises L₂ − L₁ = λ/2, so λ = 2(0.494 − 0.162) = 0.664 m [no ECF] - M2: Applies v = fλ with f = 512 Hz [substitution] [ECF from M1] - M3: v = 512 × 0.664 = 340 m s⁻¹ (accept 339–340 m s⁻¹) [ECF from M1] ### Part (b)(ii) [3 marks] — Determine - M1: Recognises L₁ + e = λ/4 (fundamental antinode-to-node distance) [no ECF] - M2: e = λ/4 − L₁ = 0.166 − 0.162 [substitution, ECF from (b)(i) λ] - M3: e = 0.004 m = 4 × 10⁻³ m (accept 3–5 mm) [ECF from (b)(i)] ### Part (c) [4 marks] — Show that (derivation, ~1 step per mark) - M1: EMF ε = −dΦ/dt = −(dΦ/dz)(dz/dt) = −u · dΦ/dz, so |ε| is maximised when |dΦ/dz| is maximised [identifies Faraday's law applied to moving magnet] - M2: Differentiates: dΦ/dz = (μ₀ N p r²/2) · d/dz[(r² + z²)^{−3/2}] = −(3μ₀ N p r² z)/[2(r² + z²)^{5/2}] - M3: Sets d²Φ/dz² = 0 (extremum of dΦ/dz); obtains (r² + z²)^{5/2} − 5z²(r² + z²)^{3/2} = 0, i.e. r² + z² = 5z², giving z² = r²/4 - M4: Therefore z = r/2; since this is non-zero, the EMF peak occurs before the magnet reaches the coil plane (z = 0 gives dΦ/dz = 0, i.e. zero EMF, not maximum) [conclusion] ### Part (d) [2 marks] — Suggest (proposal + physics-based warrant per §4.4.1) - M1 [2 marks combined]: Proposes that each EMF-marked position is offset from the true water-surface (acoustic-node) position by a fixed amount Δ = r/2 = 0.75 cm above the coil plane, **with** the physics warrant that, by part (c), the induction signal peaks when the magnet is at z = r/2 above the coil — this same geometric offset is added to both L₁ and L₂ readings, so the difference L₂ − L₁ (and hence λ and v) is unchanged; only the end correction e in (b)(ii) absorbs the shift. Award full 2 marks for a proposal that explicitly states (i) a constant offset and (ii) cites the z = r/2 induction result to justify why v is unaffected. - Partial credit (1 mark): correct conclusion that v is unchanged but without explicit reference to the z = r/2 offset from part (c); OR identifies the r/2 offset but does not conclude that v survives. ### Marker notes - Alternative method accepted for (c): students may set dε/dz = 0 directly (equivalent to d²Φ/dz² = 0) and reach z² = r²/4 in one step — award M2+M3 together; M1 and M4 still required. - Part (b)(ii) ECF: accept e in range 3–5 mm using student's λ from (b)(i). - Part (d): accept proposals that note the offset Δ may differ between L₁ and L₂ only if the descent speed u changes — credit M1 fully if student notes that for constant u the offset is identical at both resonances. Do NOT credit answers that invoke buoyancy or response-time without linking to the z = r/2 induction result. - Show-that target in (c): z = r/2 given exactly; any student-derived expression algebraically equivalent to z² = r²/4 acceptable. ```