NBPhO 2026

Part (i) — 2.0 / 2.0 pts

Criterion	Points	Result
Silence before $t_0$, explained as the sound not reached the receiver yet	0.2	✓ “While the observer is outside this cone she hears nothing. The cone surface reaches her at observer time $t_{\mathrm{cone}}$” — silence identified as the pre-Mach-cone interval
Sonic boom at $t_0$, explained as the shock-wave front producing a short broadband pressure peak	0.5	✓ Jacobian argument: ”$dt/d\tau$ vanishes, so a finite range of source-times collapses onto a single observer-time. The intensity received per unit time is proportional to $\|d\tau/dt\|$, which diverges at $t=t_{\mathrm{cone}}$… a vertical bright streak spanning the frequency axis — the audible boom”
Associating the changes in frequency with the Doppler effect	0.2	✓ supersonic Doppler factor written as $f_o = f_e/\|1-M\cos\theta\|$ and called out by name (with the absolute-value pitfall of the $\tau<\tau_*$ branch made explicit)
Explanation of two simultaneous tones for $t > t_0$, with two emission events reaching the observer at the same time	0.6	✓ “the same emitted tone $f_e$ is heard at two simultaneously-Doppler-shifted frequencies… two source-times mapping to it — call them branch 1 ($\tau<\tau_*$) and branch 2 ($\tau>\tau_*$)“
Explanation of ridges diverging at the boom and tending asymptotically to finite values as $t \to \infty$	0.5	✓ “The two curves diverge from the same singular point (the boom at $t=t_{\mathrm{cone}}$, $f_o=\infty$) and relax monotonically toward the two horizontal asymptotes $f_e/(M-1)$ above and $f_e/(M+1)$ below”

(The official’s heading “Part i) (1.5 points)” appears to be a typo: the listed grading items add to 2.0 pts and the problem’s frontmatter awards 2 pts to part (i). The 2.0-pt total is used here.)

Part (ii) — 2.8 / 3.0 pts

Criterion	Points	Result
Recognising that the radial velocity tends to $\pm v$ in the asymptotic limits	0.5	✓ “the line of sight is along the velocity. These are the two ‘axial’ rays, head-on (branch 1, $\tau\to-\infty$) and tail-on (branch 2, $\tau\to+\infty$)… the geometry forgets $d$, and the asymptotic frequencies depend only on $M$“
Correct Doppler formula $f = f_0/\|1 \pm v_r/c\|$	0.5	✓ $f_o = f_e/\|1-M\cos\theta\|$ with the absolute value, and explicit branch-by-branch sign accounting
Asymptotic upper-ridge value $f_2 = f_0/(M-1)$	0.2	✓ $f_{+} = f_e/(M-1)$
Asymptotic lower-ridge value $f_1 = f_0/(M+1)$	0.2	✓ $f_{-} = f_e/(M+1)$
Reading $f_1 \in [310, 350]\,\text{Hz}$ and $f_2 \in [1500, 1700]\,\text{Hz}$ from the spectrogram (0.2p each)	0.4	~ 0.2 / 0.4 — $f_{+} \approx 1500\,\text{Hz}$ sits at the lower edge of the upper band ✓; $f_{-} \approx 300\,\text{Hz}$ falls 10 Hz below the keyed band $[310, 350]\,\text{Hz}$ ✗
Correct symbolic form $M = (f_2 + f_1)/(f_2 - f_1)$	0.8	✓ $M = (f_{+}+f_{-})/(f_{+}-f_{-})$, derived from the asymptote ratio $f_{+}/f_{-} = (M+1)/(M-1)$
Numerical answer $M \in [1.35, 1.65]$	0.4	✓ $M = 1800/1200 = 1.5$ — the 5:1 asymptote ratio Claude reads happens to land on exactly the same $M$ as the official’s 5:1 ratio (the lucky cancellation that masks the $f_{-}$ underread)

Part (iii) — 2.5 / 3.0 pts

Criterion	Points	Result
Computing $f_0 = 2f_1f_2/(f_2 - f_1)$ with $f_0 \in [720, 910]\,\text{Hz}$ using the asymptotes from part (ii)	0.4	✓ $f_e = 2f_{+}f_{-}/(f_{+}+f_{-}) = 2\cdot 1500\cdot 300/1800 = 750\,\text{Hz}$, in band; identified as the harmonic mean of the two asymptotes
Using Mach cone and the corresponding condition $\sin\alpha = 1/M$	0.3	✓ ”$\sin\alpha = 1/M$, $\tan\alpha = 1/\sqrt{M^{2}-1}$… the Mach cone — a half-angle-$\alpha$ cone… opening backwards along $-\hat v$“
Recognising that at the point of closest approach, $f = f_0$ (no Doppler shift)	0.3	✗ Claude does not use the no-Doppler-shift insight to identify the overhead-pass moment from the spectrogram; he calibrates the time origin via an assumed visual sighting at $t = 0$ instead (see “Discrepancies”)
Expressing the time $t_0$ between the emission (event $A$) and reception of the sonic boom in terms of $d$	0.5	✓ $R(\tau_*)/c = dM/(c\sqrt{M^{2}-1})$ — the propagation time of the boom-emitting wavefront, derived as the cross-check on $t_{\mathrm{cone}}$
Expressing the time $t_1$ between event $A$ and reception of the wave emitted at $B$ in terms of $d$	1.0	✓ Equivalent geometric work: Claude derives $t_{\mathrm{cone}} = d\sqrt{M^{2}-1}/(cM)$ from the Mach-cone-meets-altitude-line construction. The specific quantity is the visual-sighting-to-boom interval rather than the official’s $\|AB\|/v + d/c$, but the Mach-cone geometry, the $\sqrt{M^{2}-1}$ factor, and the $/v$-then-$/c$ decomposition are all present and correct
Reading $t_0$, $t_1$ from the spectrogram and computing $\tau \in [0.85, 1.15]\,\text{s}$	0.2	✗ Claude reads only the boom timestamp ($t_{\mathrm{cone}} \approx 3\,\text{s}$) and uses it directly; he does not identify a second timestamp ($t_1$) or compute the audio interval $\tau$
Numerical answer if approach correct $d \in [1.0, 1.7]\,\text{km}$	0.3	✓ $d \approx 1.37\,\text{km}$, in band

Discrepancies

• $f_{-}$: Claude $300\,\text{Hz}$ vs official band $[310,\,350]\,\text{Hz}$. Below band by 10 Hz at the lower edge — visually $\sim 1.3\%$ of the y-axis on the log scale, but the band is explicit. The downstream $M = 1.5$ is unaffected because Claude’s $f_{+}/f_{-} = 5$ ratio matches the official’s $1600/320 = 5$ ratio exactly.
• $f_e$: Claude $750\,\text{Hz}$ vs official $800\,\text{Hz}$. Both in the keyed $[720, 910]\,\text{Hz}$ band. The $50\,\text{Hz}$ gap is the direct consequence of the underread $f_{-}$, since $f_e = f_{-}(M+1)$ scales linearly with $f_{-}$.
• $d$: Claude $1.37\,\text{km}$ vs official $1.5\,\text{km}$. Both in the keyed $[1.0,\,1.7]\,\text{km}$ band. The $\sim 9\%$ gap is partly the alternative time interval ($t_{\mathrm{cone}} \approx 3\,\text{s}$ visual-sighting-to-boom vs $\tau \approx 1.1\,\text{s}$ boom-to-overhead-arrival; the two are linked geometrically by $\tau + t_{\mathrm{cone}} = d/c$) and partly the spectrogram-reading uncertainty on the boom location.
• Time-origin assumption: Claude calibrates the spectrogram’s $t = 0$ as “visual sighting of jet overhead”, on the grounds that this is “the only operationally accessible” zero. The official treats $t = 0$ as the recording start and uses two purely-audio events ($t_0$ at the boom, $t_1$ at the lower-ridge-crosses-$f_0$ moment) whose interval $\tau$ is calibration-free. Claude’s interpretation requires the observer to have started recording exactly at the overhead-pass — an assumption the problem statement does not pin down.

Overall score: 7.3 / 8.0 pts

Full marks on part (i). In part (ii) a 0.2-pt dock for the $f_{-} = 300\,\text{Hz}$ readout sitting just below the keyed $[310, 350]\,\text{Hz}$ band, although the resulting $M = 1.5$ matches the official key exactly. In part (iii) a 0.5-pt dock distributed across two criteria: 0.3 pts for not invoking the ”$f = f_0$ at closest approach” insight as the operational hook for identifying the overhead-pass on the spectrogram, and 0.2 pts for reading only one timestamp instead of computing the audio interval $\tau$ — both consequences of an alternative time-origin convention that Claude assumes rather than derives.

Numerical answers: $M = 1.5$ (exact match to official), $f_e = 750\,\text{Hz}$ (vs official $800\,\text{Hz}$, in band), $d \approx 1.37\,\text{km}$ (vs official $1.5\,\text{km}$, in band).

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads the structural picture — kinematic non-monotonicity of $t(\tau)$ as the root cause of two-arrival, with the Mach number identified as the single dimensionless control parameter and the timescale hierarchy $t_{\mathrm{cone}} \sim d/v \sim d/c$ stated up-front. Part (i) earns its boom-explanation points via an unusually careful Jacobian argument: $dt/d\tau = 1 - M\cos\theta$ is differentiated, its zero is located at the Mach angle $\cos\theta_* = 1/M$, and the intensity divergence $|d\tau/dt| \to \infty$ is identified as the physical content of the broadband pile-up — a derivation that anchors the boom in the integrable singularity rather than just naming it. The “education remark” on why both ridges come from a single emitted tone (not two distinct emission lines, but two propagation paths) pre-empts the most common student misreading and connects to the broadband-noise generalisation of a real flyby. Part (ii) flags the harmonic-mean identity $f_e = 2f_{+}f_{-}/(f_{+}+f_{-})$ as a “pleasant little algebraic fact”, checks both $M \to 1^{+}$ (upper branch diverges, two-arrival regime collapses to the standard subsonic Doppler swoosh) and $M \to \infty$ (both asymptotes redshift to zero) limits, and verifies $f_e$ via the cross-product $f_{+}(M-1) = f_{-}(M+1)$. Part (iii) carries dimensional, $M \to 1^{+}$ (cone flattens, $d \to \infty$), and $M \to \infty$ (cone collapses onto velocity axis, $d \to ct_{\mathrm{cone}}$, $v$ drops out) consistency checks, plus a plausibility statement (km-scale standoff is in the right ballpark for fighter-jet airshow). The closing geometric remark — “the jet is always seen before it is heard, and the gap grows with lateral distance” — frames the whole problem as a memorable physical phenomenon rather than a solved exercise.

Where the official solution is sharper. Three places. (1) Operational grounding of the time axis. The official identifies a purely-audio time interval $\tau = t_1 - t_0$ between two events visible on the spectrogram alone (boom onset at $t_0$, lower ridge crosses $f_0$ at $t_1$) and inverts $\tau = (d/c)(1 - \sqrt{1 - 1/M^{-2}})$ to get $d$. This is calibration-free and self-contained. Claude’s alternative — assume the spectrogram’s $t = 0$ is the visual-sighting moment, read $t_{\mathrm{cone}} \approx 3\,\text{s}$ directly — relies on an unstated assumption (the observer started recording exactly when she spotted the jet overhead) that the problem does not justify; the geometry is correct given the assumption, but the assumption is the weak link. The two approaches are linked by $\tau + t_{\mathrm{cone}} = d/c$ and yield the same $d$ within tolerance, but the official’s path is the operationally cleaner of the two. (2) The ”$f = f_0$ at closest approach” insight. The official explicitly recognises that at the overhead-pass instant the radial velocity is zero, so the un-Doppler-shifted tone $f_0$ is heard once and only once on the lower ridge. This is the bridge that ties part (ii)‘s $f_0$ to part (iii)‘s time-axis calibration. Claude derives $f_e = 750\,\text{Hz}$ but never uses it as a measurable feature on the spectrogram — a missed connection that costs the 0.3-pt criterion. (3) Geometric economy. The official’s similar-triangles argument ($\triangle JOA \sim \triangle OBA$ via the right angles at $O$ and $B$ and the shared angle at $A$) reads off $|BA| = ct/M$ and $d = ct\sqrt{M^{2}-1}/M$ in one step. Claude’s Cartesian-coordinate derivation — set up $\vec r_s(\tau)$, find $\tau_*$, substitute $R(\tau_*)$ into $t = \tau + R/c$ — arrives at the same result over several lines.