NBPhO 2026

Part (i) — 1.0 / 1.0 pts

Criterion	Points	Result
Correct sketch	1.0	✓ ASCII sketch shows the circle of radius $R = g/(4\omega^2)$ centred at $(0, R)$, the pivot $O$ on the circle at its lowest point, the labelled $x$- and $y$-axes, and the downward gravity arrow — matches the official figure

Part (ii) — 1.0 / 1.0 pts

Scored against the Solution 4 (rotating-frame ODE) grading column, which is the route Claude takes.

Criterion	Points	Result
Correct equation of motion in the rotating frame	0.2	✓ $\ddot{x} - \omega^{2}\,x = -g\sin(\omega t)$, derived by projecting $\ddot{\vec R}$ onto $\hat r$ (Coriolis lies along $\hat\theta$ and is absorbed by the constraint, as the official notes)
States implicitly or explicitly that the general solution to the differential equation is the sum of the homogeneous and particular solutions	0.2	✓ Overview: “The homogeneous part is exponentially unstable… but a perfectly tuned particular solution exists at frequency $\omega$, and it generates the circular trajectory of parts (i)–(ii).”
Correct homogeneous solution	0.1	✓ identified qualitatively in the Overview as exponentially unstable with rates set by $-\omega^{2}$; the explicit exponential form $\xi \propto e^{\pm\sqrt{-\omega_{0}^{2}}\,t}$ is written out in the part-(iii) regime table, which reduces to $A e^{\omega t} + B e^{-\omega t}$ at $k=0$
Correct particular solution	0.2	✓ $x_{p}(t) = \dfrac{g}{2\omega^{2}}\sin(\omega t)$, found by the ansatz $A\sin(\omega t)$
States that homogeneous solution corresponding to periodic motion is 0	0.1	✓ educational remark: “every other initial condition develops an exponentially growing component along the unstable mode of the homogeneous equation” — i.e. only the $A=B=0$ choice gives the bounded circle
Correct answer	0.2	✓ $R = g/(4\omega^{2})$

Part (iii) — 2.0 / 2.0 pts

Scored against the Solution 1 stability column plus the Solution 2 radius column (the ODE / particular-amplitude path Claude takes); both alternatives sum to the part’s 2.0 pts.

Criterion	Points	Result
Equation of motion including the spring	0.2	✓ $\ddot{x} + \omega_{0}^{2}\,x = -g\sin(\omega t)$ with $\omega_{0}^{2} \equiv k/m - \omega^{2}$
Idea of looking at (small) displacements to analyse stability	0.3	✓ “A small perturbation $\xi(t) \equiv x(t) - x_{\text{circle}}(t)$ obeys the homogeneous equation”
Identifying the homogeneous equation governing perturbations	0.5	✓ $\ddot{\xi} + \omega_{0}^{2}\,\xi = 0$, with the three-regime table making the mapping from sign of $\omega_{0}^{2}$ to stability character explicit
Correct stability criterion $k > m\omega^{2}$	0.3	✓ boxed; bead can be Lyapunov stable iff “the centrifugal anti-spring is overpowered”
Solving for particular-solution amplitude $A'$	0.5	✓ $A = mg/(2m\omega^{2} - k)$, with the $k = 2m\omega^{2}$ resonance singled out as the additional restriction $k \neq 2m\omega^{2}$
Correct radius formula $R = mg/[2\lvert 2m\omega^{2} - k\rvert]$	0.2	✓ boxed, with explicit discussion of $k \lessgtr 2m\omega^{2}$ flipping the circle from above to below the pivot

Part (iv) — 1.4 / 4.0 pts

Claude offers a partial write-up that derives the rod-frame equation of motion, the linear decomposition $f = f_1 + f_2$, and both particular-solution amplitudes — picking up most of the “setup” points — but follows it with a rose-curve / rhodonea analysis that is non-unique and lands on the wrong frequency ratio: $\Omega = 5\omega$ instead of the official $\Omega = 9\omega/5$. The numerical extraction of $g$ and $a_E$ is recorded as “methodology, not extracted” and forfeits the figure-reading points.

Criterion	Points	Result
Equation of motion including the electric field	0.1	✓ $\ddot{x} - \omega^{2}x = -g\sin(\omega t) + a_E\cos((\Omega-\omega)t + \phi_0)$, derived by projecting $\vec E\cdot\hat r$ onto the rod
Writing $f(t) = f_{1}(t) + f_{2}(t)$	0.1	✓ “Linearity lets us superpose particular solutions”; gravity contribution $x_g$ and field contribution $x_E$ split additively
Deducing $f_{1} = \dfrac{g}{2\omega^{2}}\sin(\omega t)$	0.2	✓ recycled verbatim from part (ii)
Guessing $f_{2} = A\sin(\Omega' t + \phi)$	0.2	✓ Claude takes the cosine form $B\cos((\Omega-\omega)t + \phi_0)$, equivalent up to phase to the official’s sinusoidal ansatz at frequency $\Omega' = \Omega - \omega$
Finding $A = a_{E}/(\Omega'^{2} + \omega^{2})$	0.2	✓ $\lvert B\rvert = a_E/[(\Omega-\omega)^2 + \omega^2]$, matching the official up to sign convention
Realising $f_{2}$ dominates with proper reasoning	0.5	~ 0.2 / 0.5 — the rose-curve framing implicitly treats $f_2$ as primary (“phasor $A$ … is what rounds the teardrops out into the closed circles seen in the figure”), but the official’s symmetry argument (the figure being almost reflection-symmetric across the $x$-axis, which a pure-$f_1$ circle is not, hence $\lvert f_2\rvert \gg \lvert f_1\rvert$) is absent — the dominance is used, not argued
Noting periodicity	0.2	✓ “all integer multiples of $\omega$, so the trajectory closes after one rod period $T = 2\pi/\omega$” — the closure condition is identified, just with the wrong period (a downstream consequence of the wrong $\Omega$)
$T_{f} = n T_{f_{1}} = p T_{f_{2}}$	0.2	✗ this commensurability relation is not written; Claude bypasses it for the rhodonea $r(\theta) = C\cos((p/q)\theta + \phi_0)$ petal-count formula, which is mathematically equivalent but not in the form the criterion asks for
Writing a restrictive enough inequality for $n$ (or $p$) from the figure	0.8	~ 0.2 / 0.8 — the rhodonea petal-count formula is a legitimate alternative procedure, but Claude derives $\Omega/\omega = 1 + 4/q$ for $q$ odd and then picks $q = 1$ “as the simplest” without further restriction, leaving $q = 3, 5, 7, \ldots$ as unjustified-but-also-not-ruled-out alternatives. The procedure is therefore not restrictive enough to find $n$ uniquely; falls under the scheme’s “0.2 pt for a correct idea (in case it is executed wrong)” rule
$n = 5$	0.1	✗ Claude’s framework gives $n = 1$ (the trajectory closes in one rod period under his $\Omega = 5\omega$); the official’s $n = 5$ corresponds instead to the $q = 5$ branch he discarded
$\Omega = \tfrac{9}{5}\omega$	0.2	✗ Claude got $\Omega = 5\omega$ (the $q = 1$ branch) — the correct value is on the $q = 5$ branch he flagged as “not ruled out” but did not investigate
Deducing electric field points along $y$-axis when bead reaches extremal $y$	0.5	✗ the figure’s left-right reflection symmetry, used by the official to pin the field’s direction at the extremal $y$-points, is not invoked
Writing $\lvert f\rvert_{b}$ and $\lvert f\rvert_{t}$ as the difference/sum of $f_{1}$ and $f_{2}$ amplitudes	0.3	✗ Claude writes $r_{\max} \approx C + R_g$ as a lab-frame radial extent from “pattern centre” — the wrong geometric quantity (the official’s $\lvert f\rvert_{t/b}$ are rod-frame distances from the rotation axis), and only the sum branch, not the difference $\lvert f\rvert_b = \lvert f_2\rvert - \lvert f_1\rvert$. He notes that both extrema would pin $C$ and $R_g$ separately but does not write the formula
Solving for $g$ and $a_{E}$	0.1	✗ explicitly deferred (“the algebra here is what I did not get right inside the time cap”); no numerical values are extracted
$g \approx d\omega^{2}/2$ and $a_{E} \approx 5\,d\omega^{2}$	0.3	✗ not produced

Discrepancies

The part-(i)–(iii) numerical answers match the official key exactly: $R = g/(4\omega^{2})$ in part (ii) and $R = mg/[2\lvert 2m\omega^{2} - k\rvert]$ with $k > m\omega^{2}$ (and $k \neq 2m\omega^{2}$) in part (iii).

In part (iv), $\Omega = 5\omega$ (Claude) vs $\Omega = 9\omega/5$ (official): a factor of $25/9 \approx 2.78$ disagreement, well outside any rounding tolerance. The error traces to the unjustified choice $q = 1$ in the rhodonea-petal formula $\Omega/\omega = 1 + 4/q$ ($q$ odd) — the correct branch is $q = 5$. With the wrong $\Omega$, the prefactor $(\Omega-\omega)^2 + \omega^2$ that converts $C$ to $a_E$ is also wrong (Claude’s $17\omega^2$ vs the correct $41\omega^2/25$), so even if Claude had read the figure, the numerical $a_E$ would have come out wrong too. No numerical $g, a_E$ are quoted, so there is no second discrepancy to record beyond the $\Omega$ value itself.

Overall score: 5.4 / 8.0 pts

Full marks on parts (i)–(iii) (4.0 / 4.0); 1.4 / 4.0 on part (iv). The 2.6-point loss in part (iv) splits roughly into: (a) ~0.6 pts on the procedure-for-finding-$n$ tier, where the rhodonea idea is right but the execution does not pin $\Omega$ uniquely and lands on the wrong branch; (b) ~0.5 pts on the field-direction / amplitude-sum-and-difference tier, which Claude does not develop; and (c) ~1.0 pts on the final figure-reading and numerical extraction, which Claude explicitly defers as “methodology only, not extracted numerically”.

Numerical answers: $R = g/(4\omega^{2})$ (parts ii) and $R = mg/[2\lvert 2m\omega^{2} - k\rvert]$ (part iii) match the official key exactly; $\Omega = 5\omega$ (part iv) disagrees with the official $\Omega = 9\omega/5$.

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads the structural fact that controls everything in parts (i)–(iii) — the equation along the rod is a negative-stiffness harmonic oscillator forced at exactly the frequency of its (would-be) positive-stiffness sibling, so the homogeneous mode is exponentially unstable while a perfectly tuned particular solution at frequency $\omega$ generates a bounded circle. Part (i) augments the explicit ODE-and-trajectory derivation with an inscribed-angle-theorem reading of the $2\omega$ angular rate, and an “educational remark” unpacking the at-first-sight paradox that gravity pulls down yet the steady orbit lies entirely above the axis — the explanation hinges on $x_{p}$ and $\hat r$ flipping sign together. Part (ii) carries four consistency checks (dimensional, $\omega \to \infty$ pinning the bead at the axis, $\omega \to 0$ blowing the orbit up, and an energetic remark on how the rotor delivers exactly the energy needed to keep the orbit circular). Part (iii) goes well past what the scheme rewards: the three-regime table for the homogeneous mode separating $k < m\omega^{2}$, $k = m\omega^{2}$ and $k > m\omega^{2}$ as exponential, linear-in-$t$, and oscillatory respectively; the explicit observation that $A < 0$ above the resonance flips the steady circle through the pivot to the below-axis region; the second condition $k \neq 2m\omega^{2}$ derived as a forcing-resonance and named “parametric resonance with gravity”; and the limits $k = 0 \Rightarrow R = g/(4\omega^{2})$ and $k \to \infty \Rightarrow R \to 0$. Part (iv) brings two genuinely additional ideas not in the official: a lab-frame phasor decomposition $\vec R(t) = i R_g - i R_g e^{2i\omega t} - (C/2)e^{i\phi_0}e^{i\Omega t} - (C/2)e^{-i\phi_0}e^{i(2\omega-\Omega)t}$ that splits the trajectory into four rotating phasors at frequencies $\{0, 2\omega, \Omega, 2\omega-\Omega\}$ and exposes the symmetry that the two field-driven phasors carry equal amplitude $C/2$; and a rhodonea reading $r(\theta) = C\cos(k\theta + \phi_0)$ with $k = (\Omega-\omega)/\omega$ that recasts the rosette as a rose curve. Both observations are correct and structurally illuminating, even though the second one is what then misfires.

Where the official solution is sharper. Four places, in increasing order of weight. (1) Eero’s “Solution 1” to part (i) is a one-paragraph zero argument: at the lowest point of the trajectory the rod’s force on the bead must be vertical, so the bead must coincide with the pivot at that instant; since the rod sweeps a vertical plane the lowest point is reached when the rod is horizontal, which forces the trajectory to be a circle through the origin tangent to the horizontal there. Claude’s route via $X(t), Y(t)$ trigonometric parametrisation is correct but several times longer for the same conclusion. (2) Eppu’s “Solution 1” to part (ii) extracts the radius in a single arithmetic line: at the topmost point the rod constraint forces $v = 2R\omega$, the only force is gravity so $a = g$, and $a = v^{2}/R$ gives $R = g/(4\omega^{2})$. The same idea generalises in one stroke to part (iii) — at the top, $a = g + 2Rk$ — yielding $R = mg/[2(2m\omega^{2} - k)]$ with the absolute-value branch handled by inspection. Claude derives the radius via the particular-solution amplitude in both parts, fully credited but the longer route. (3) The crux of part (iv) is the uniqueness of the frequency ratio, and the official’s procedure pins it down where Claude’s does not. The official argues from the near-symmetry of the figure across the $x$-axis that $f_2$ dominates $f_1$, identifies the eight visible loops as sign changes of $f_2$ (so $4 T_{f_2} = T_f$), and then bounds $T_{f_2}$ from above and below by reading off the figure how far the bead travels in roughly $3 T_{f_2}/4$ versus $T_{f_2}$ along a fixed ray from the origin — concluding $\tfrac{4}{3} > n/4 > 1$, which has only $n = 5$ as an integer solution. That is the 0.8-pt argument. Claude derives the rose-curve constraint $\Omega/\omega = 1 + 4/q$ for $q \in \{1, 3, 5, \ldots\}$, then picks $q = 1$ “as the simplest” with no figure-based further restriction; he flags $q = 3$ as unresolved but never reaches the $q = 5$ branch, which is where the correct answer $\Omega = 9\omega/5$ lives. The official’s procedure constrains $\Omega$ uniquely from the figure; Claude’s leaves a one-parameter family of options and resolves it by aesthetics, which gives the wrong answer. (4) The final extraction of $g$ and $a_E$. The official uses the figure’s reflection symmetry across the $y$-axis to argue that the field points along the rod when the bead is at its extremal $y$-positions, which forces $\dot f_1 = 0$ and aligns the two contributions; reading $\lvert f\rvert_b \approx 2.8\,d$ and $\lvert f\rvert_t \approx 3.3\,d$ from the figure and writing them as $\lvert f\rvert_t = a_E/(\omega^2+\Omega^2) + g/(2\omega^2)$ and $\lvert f\rvert_b = a_E/(\omega^2+\Omega^2) - g/(2\omega^2)$ yields $g \approx d\omega^2/2$ and $a_E \approx 5\,d\omega^2$ in two lines of arithmetic. Claude writes $r_{\max} \approx C + R_g$ in the lab frame (a different geometric quantity), notes that both extrema would in principle separate $C$ and $R_g$, then explicitly defers the algebra and the figure reading. This is the 1.0-pt block at the end of the grading scheme that Claude leaves on the table.