## ERQ · 12 marks · Topics: C.4 Standing waves and resonance + D.2 Electric and magnetic fields
**Stem.** A physics researcher studies the resonant modes of a thin conducting wire of length L = 0.480 m and linear mass density μ = 1.20 × 10⁻³ kg m⁻¹, stretched between two fixed insulating supports under tension T = 27.0 N. The wire carries a steady current I = 4.50 A and is placed in a uniform horizontal magnetic field of magnitude B = 0.0850 T, directed perpendicular to the wire. The magnetic force per unit length acts as a uniform transverse drive. The researcher slowly varies the frequency of the current (the current is sinusoidal: I(t) = I₀ sin(2πft), with I₀ = 4.50 A) and measures the amplitude of transverse oscillations of the wire's midpoint with a laser displacement sensor. A strong resonance peak is observed at one particular frequency, with peak-to-peak midpoint displacement 6.4 mm. The surrounding air provides light damping.
### Part (a) Draw [2 marks] · AO1 · Topic: C.4
On a sketch of the wire between its two supports, draw the displacement pattern of the lowest-frequency standing-wave mode that can be resonantly excited by the spatially-uniform magnetic driving force described in the stem, and label the position(s) of any antinode(s).
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: C.4
Calculate the frequency of the alternating current that drives the mode you drew in (a).
### Part (b)(ii) Show that [3 marks] · AO2/AO3 · Topic: C.4 + D.2
The amplitude of the driving force per unit length is F₀/L = BI₀. By equating the work done by this spatially-uniform sinusoidal driving force on the wire over one cycle at resonance to the energy dissipated per cycle, show that only standing-wave modes with an odd number of half-wavelengths along L can be resonantly driven by this configuration, and hence justify your choice in (a).
### Part (c) Determine [2 marks] · AO3 · Topic: C.4 + D.2
The magnetic field is now tilted so that it makes an angle θ with the horizontal, while remaining perpendicular to the wire's axis. The researcher finds that the resonance frequency identified in (b)(i) shifts downward by 0.6 %. By considering how the vertical component of the magnetic force per unit length acts together with gravity to modify the effective tension along the wire (treating the wire as nearly horizontal), determine θ.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The peak midpoint amplitude (3.2 mm) predicted from a simple lossless standing-wave model using the driving force in (b)(ii) is much larger than the measured 3.2 mm peak. Suggest one physical assumption in this combined wave + electromagnetic model whose failure causes the measured amplitude to be smaller than the idealised prediction, and state explicitly the direction of the resulting discrepancy.
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## Mark Scheme
### Part (a) [2 marks]
- Mark 1: Sketch shows the fundamental mode (n = 1) — a single antinode at the midpoint, nodes only at the two fixed supports [ECF: no]
- Mark 2: Antinode at L/2 labelled, and zero displacement clearly indicated at both supports [ECF: no]
### Part (b)(i) [3 marks]
- Mark 1: Identifies f₁ = v/(2L) with v = √(T/μ) [ECF: no]
- Mark 2: Correct substitution: v = √(27.0/1.20 × 10⁻³) = 150 m s⁻¹; f₁ = 150/(2 × 0.480) [ECF: yes]
- Mark 3: f₁ = 156 Hz (accept 156–157 Hz) to 3 s.f. with units [ECF: yes]
### Part (b)(ii) [3 marks]
- Mark 1: States that the work per cycle is W ∝ ∫₀ᴸ y_n(x) dx, where y_n(x) = sin(nπx/L) is the mode shape, because the drive force per unit length is spatially uniform [ECF: no]
- Mark 2: Evaluates/recognises that ∫₀ᴸ sin(nπx/L) dx = (L/nπ)(1 − cos(nπ)), which vanishes for even n and is non-zero for odd n [ECF: yes]
- Mark 3: Concludes that only odd-n modes absorb net energy from the uniform drive, so n = 1 is the lowest excitable mode, consistent with part (a) [ECF: yes]
### Part (c) [2 marks]
- Mark 1: Recognises that the vertical component of the magnetic force per unit length, BI₀ sin θ (time-averaged appropriately, or its DC offset if applicable), together with the weight per unit length μg, changes the sag/tension; equivalently, sets up f′/f = √(T′/T) ⇒ T′/T = (0.994)² ≈ 0.988, giving ΔT ≈ −0.012 × 27.0 = −0.32 N over the wire, OR uses force balance giving the required vertical force per unit length. [ECF: no]
- Mark 2: Solves for θ using BI₀ sin θ balancing the required change (with μg = 0.0118 N m⁻¹ included), obtaining θ ≈ 2° (accept 1.5°–2.5°) with units [ECF: yes]
### Part (d) [2 marks]
- Mark 1: Identifies ONE specific assumption that fails, e.g.:
- (i) damping (air drag / internal friction) was neglected — the idealised model has infinite Q
- (ii) the drive current is assumed perfectly sinusoidal and in phase with displacement at resonance — phase lag from finite Q reduces stored energy
- (iii) the magnetic field is assumed perfectly uniform along L — non-uniformity reduces the spatial overlap integral with the n = 1 mode
- (iv) the wire is assumed perfectly flexible (no bending stiffness) — stiffness shifts and broadens the resonance
- (v) ohmic heating / temperature rise alters μ or T during measurement [ECF: no]
- Mark 2: States that this assumption's failure causes the measured amplitude to be SMALLER than the idealised prediction (i.e. the discrepancy is in the direction of reduced response), with brief mechanism (energy loss / reduced effective drive / off-resonance operation) [ECF: yes]
### Marker notes
- (a): accept a clear half-sine sketch; do NOT award mark 2 if antinode is unlabelled or if nodes are drawn anywhere except at the supports.
- (b)(ii): accept argument via symmetry — for even n, the mode shape is antisymmetric about L/2 while the drive is symmetric, so the overlap integral vanishes. Full marks if the symmetry argument is stated precisely.
- (c): accept alternative routes — e.g., interpreting the shift via a change in effective μ_eff under a static vertical force gradient, provided the algebra yields θ in the 1.5°–2.5° band. Reject answers that ignore gravity or that use the horizontal component.
- (d): each candidate assumption must be paired with the correct direction (measured < predicted). Answers giving only "experimental error" or "air resistance" without linking to amplitude reduction earn 1 mark maximum. Reject answers that would predict a LARGER measured amplitude (wrong direction).