NBPhO 2026

Overview

A supersonic source flying past a stationary observer produces a spectrogram that looks completely unlike the textbook subsonic Doppler “swoosh.” The reason is purely kinematic: when $v>c$, the map from source-emission time $\tau$ to observer-arrival time $t$ is non-monotonic. The same instant of observer time picks up two different emission events — one from while the jet was still approaching, and one from after it had already passed. The two arrivals are differently Doppler-shifted, so a single emitted tone $f_e$ shows up as a pair of curves on the spectrogram. Their high-frequency end is the sonic boom (where the two branches merge into a divergent pile-up of arrivals); their tails settle into the head-on and tail-on Doppler factors

$$f_{+} \;=\; \frac{f_e}{M-1}, \qquad f_{-} \;=\; \frac{f_e}{M+1}.$$

Once these are read off the asymptotes, $M$ is fixed by their ratio, $f_e$ falls out as a by-product, and the time elapsed between “jet directly overhead” and “boom heard” sets the geometric scale $d$.

The key dimensionless quantity is the Mach number $M=v/c$; everything is parameterised by it. Two physical scales separate:

Symbol	Meaning	Value
$\alpha$	Mach-cone half-angle, $\sin\alpha = 1/M$	depends on $M$
$t_{\mathrm{cone}}$	delay between jet-overhead and boom	$\sim d/c$
$d/v$	timescale on which Doppler curves relax to asymptotes	$\sim d/c$

The delay-to-asymptote ratio is fixed by $M$ alone, so the time-axis is calibrated by a single length, $d$.

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Part (i) — Qualitative features of the spectrogram

Geometry and the two-arrival phenomenon

Let the observer sit at the origin of a Cartesian frame, with the jet flying along the line $y=d$, $z=0$ at constant velocity $v\hat x$. Set source-time $\tau=0$ at the closest-approach instant, so the jet is at $\vec r_s(\tau)=(v\tau,d)$ and its distance to the observer is

$$R(\tau) \;=\; \sqrt{v^{2}\tau^{2}+d^{2}}.$$

A wavefront emitted at source-time $\tau$ travels at $c$ in still air and reaches the observer at observer-time

$$t(\tau) \;=\; \tau + \frac{R(\tau)}{c}.$$

The angle $\theta$ between the jet’s velocity and the line of sight from source to observer satisfies $\cos\theta = -v\tau/R$ (positive when the observer is ahead of the jet, $\tau<0$). Differentiating,

$$\frac{dt}{d\tau} \;=\; 1 - M\cos\theta.$$

For a subsonic source ($M<1$) this is always positive: arrivals at the observer occur in the same order as emissions. For a supersonic source it changes sign at the Mach angle $\cos\theta_*=1/M$, i.e. at

$$\tau_{*} \;=\; -\frac{d}{v\sqrt{M^{2}-1}}\quad(<0).$$

So $t(\tau)$ is not monotonic: it decreases on $\tau<\tau_*$, reaches a minimum at $\tau_*$, and increases on $\tau>\tau_*$. For any observer time $t>t_{\mathrm{cone}}\equiv t(\tau_*)$ there are two source-times mapping to it — call them branch $1$ ($\tau<\tau_*$) and branch $2$ ($\tau>\tau_*$). For $t<t_{\mathrm{cone}}$ there is no source-time at all: the observer is in silence.

The four features visible in the plot

(a) Initial silence. The leading edge of all sound emitted by a supersonic jet is the Mach cone — a half-angle-$\alpha$ cone of half-opening angle $\sin\alpha=1/M$, attached to the jet, opening backwards along $-\hat v$. While the observer is outside this cone she hears nothing. The cone surface reaches her at observer time $t_{\mathrm{cone}}$, when the jet has already moved a horizontal distance $d/\tan\alpha = d\sqrt{M^{2}-1}$ past the closest-approach point.

(b) The bright vertical band — the sonic boom. Near $\tau=\tau_*$ the Jacobian $dt/d\tau$ 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 $t=t_{\mathrm{cone}}$. Physically this is signal pile-up: a continuous burst of emissions (over a range of $\tau$) arrives all at once, with all spectral content. On a spectrogram this is a vertical bright streak spanning the frequency axis — the audible “boom.”

(c) Two descending curves. Past the boom, the observer is in the two-arrival regime. The same emitted tone $f_e$ is heard at two simultaneously-Doppler-shifted frequencies,

$$f_o \;=\; \frac{f_e}{|1-M\cos\theta|}.$$

Branch 1 ($\tau<\tau_*$) corresponds to sound emitted while the jet was still well ahead of the Mach moment, hence at small angle $\theta$ to the line of sight, hence with $\cos\theta>1/M$ — so $1-M\cos\theta<0$, and $f_o = f_e/(M\cos\theta-1)$. As observer time $t$ runs forward, this branch traces back to ever earlier emissions (i.e. $\tau$ runs backward from $\tau_*$ toward $-\infty$). The observer is essentially hearing the jet’s history played in reverse, on a Doppler-blueshifted carrier. As $\cos\theta \to 1$ (jet originally far ahead, head-on), $f_o\to f_e/(M-1)$ — the upper asymptote.

Branch 2 ($\tau>\tau_*$) corresponds to emissions after the Mach moment, when the line of sight has rotated past the cone angle; here $\cos\theta<1/M$, so $1-M\cos\theta>0$ and $f_o=f_e/(1-M\cos\theta)$. As $\tau\to+\infty$ the jet recedes tail-on ($\cos\theta\to-1$) and $f_o\to f_e/(M+1)$ — the lower asymptote.

(d) Both curves descend monotonically. On both branches, $|1-M\cos\theta|$ grows as $|\tau|$ increases beyond $\tau_*$. So $f_o$ decreases on both branches as observer time $t$ marches forward. 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 fact that exactly two curves are seen, rather than four, means the spectrogram is dominated by a single emitted tone — the engine’s blade-pass frequency (or whatever produces the discrete tone in the jet noise).

Educational remark — why both curves come from one emitted frequency

A common confusion is to treat the upper and lower curves as separate emission lines. They are not: they are two sample paths of the same physical frequency $f_e$, picked up via two different propagation routes (one going through space ahead of the Mach surface, one behind it). For a generic spectrogram of a real flyby with broadband noise plus several discrete tones, one therefore sees a family of curve pairs, one per emitted spectral line, all with the same Mach-number-controlled shape. Fitting any pair gives the same $M$.

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Part (ii) — Mach number from the asymptote ratio

Algebra

The two asymptotes are

$$f_{+} \;=\; \frac{f_e}{M-1}, \qquad f_{-} \;=\; \frac{f_e}{M+1}.$$

Their ratio depends only on $M$:

$$\frac{f_{+}}{f_{-}} \;=\; \frac{M+1}{M-1}.$$

Solving,

$$\boxed{\; M \;=\; \frac{f_{+}+f_{-}}{f_{+}-f_{-}} \;}$$

The corresponding emitted frequency is then

$$f_{e} \;=\; f_{+}(M-1) \;=\; f_{-}(M+1) \;=\; \frac{2\,f_{+}f_{-}}{f_{+}+f_{-}}$$

— the harmonic mean of the two observed asymptotes (a pleasant little identity).

Numerical readout

Reading the asymptotic plateaus from the spectrogram (logarithmic frequency axis):

$$f_{+} \;\approx\; 1500\,\text{Hz}, \qquad f_{-} \;\approx\; 300\,\text{Hz}.$$

Therefore

$$M \;=\; \frac{1500 + 300}{1500 - 300} \;=\; \frac{1800}{1200} \;=\; \boxed{\;1.5\;}$$

and as a by-product

$$f_{e} \;=\; f_{+}(M-1) \;=\; 1500 \times 0.5 \;=\; 750\,\text{Hz}.$$

(Cross-check: $f_{-}(M+1) = 300\times 2.5 = 750\,\text{Hz}$. ✓)

Consistency checks

• Subsonic limit. As $M\to 1^{+}$, $f_{+}\to\infty$: the upper branch exists only for genuinely supersonic motion, since head-on the source catches up with its own previous wavefronts. Sensible — the two-arrival phenomenon disappears in the subsonic limit, and the spectrogram would show a single descending curve, the standard subsonic flyby Doppler.
• High-Mach limit. As $M\to\infty$, both $f_{\pm}\to 0$: a hypersonic source leaves all its sound far behind; what little reaches the observer is enormously redshifted on both branches.
• Ordering. Since $M>1$, $1/(M-1)>1/(M+1)$, so $f_{+}>f_{-}$ as labelled. ✓

Educational remark — why the ratio is so clean

The supersonic Doppler factor $|1-M\cos\theta|$ reaches its two extreme values when $\cos\theta=\pm 1$ — i.e. when 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, $\tau\to+\infty$). Both correspond to the source being infinitely far away, so the geometry “forgets” $d$, and the asymptotic frequencies depend only on $M$. That is precisely why $M$ can be read off the asymptotes without any other information — neither $d$ nor $f_e$ enter the ratio.

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Part (iii) — Closest-approach distance from the boom delay

Mach-cone arrival time

The Mach cone is the geometric envelope of all wavefronts. It is a half-angle-$\alpha$ cone with apex at the present jet position, opening backwards along $-\hat v$, where

$$\sin\alpha \;=\; \frac{1}{M}, \qquad \tan\alpha \;=\; \frac{1}{\sqrt{M^{2}-1}}.$$

The cone surface intersects the observer’s altitude line $y=d$ at horizontal distance $d/\tan\alpha = d\sqrt{M^{2}-1}$ behind the jet’s apex. The cone passes through the observer the instant the jet has flown that far past closest approach, i.e. at source-time

$$\tau_{\mathrm{cone}} \;=\; \frac{d\sqrt{M^{2}-1}}{v} \;=\; \frac{d\sqrt{M^{2}-1}}{cM}.$$

Because the cone surface is the leading wavefront, its passage is the boom — the source-time and the observer-time of the boom coincide:

$$t_{\mathrm{cone}} \;=\; \tau_{\mathrm{cone}} \;=\; \frac{d\sqrt{M^{2}-1}}{cM}.$$

(One can verify this directly: substitute $\tau=\tau_{*}=-d/(v\sqrt{M^{2}-1})$ into $t=\tau+R/c$. Using $R(\tau_{*})=dM/\sqrt{M^{2}-1}$ one recovers exactly the formula above.)

Why the time origin on the spectrogram is “jet overhead”

The spectrogram’s $t$-axis must have a physically calibrated zero for the boom delay to be meaningful. The natural — and only operationally accessible — zero is the moment the observer sees the jet pass directly overhead: light from the jet reaches her essentially instantaneously, so visual sighting at closest approach defines $t=0$. The first thing she hears, the boom, then arrives at $t=t_{\mathrm{cone}}>0$ given by the formula above. (For $0<t<t_{\mathrm{cone}}$ she sees the jet but hears nothing — the dramatic feature of supersonic flyby.)

Reading the boom delay

From the spectrogram the bright vertical band sits at

$$t_{\mathrm{cone}} \;\approx\; 3\,\text{s}.$$

Solving for $d$

Inverting,

$$\boxed{\; d \;=\; \frac{cM\,t_{\mathrm{cone}}}{\sqrt{M^{2}-1}} \;}$$

Numerically, with $c=340\,\text{m s}^{-1}$, $M=1.5$, $t_{\mathrm{cone}}=3\,\text{s}$:

$$\sqrt{M^{2}-1} \;=\; \sqrt{1.25} \;=\; \frac{\sqrt{5}}{2} \;\approx\; 1.118,$$ $$d \;\approx\; \frac{340 \times 1.5 \times 3}{1.118} \;=\; \frac{1530}{1.118} \;\approx\; 1.37\,\text{km}.$$

Consistency checks

• Dimensions. $[cMt/\sqrt{M^{2}-1}] = (\text{m s}^{-1})\cdot 1\cdot \text{s}/1 = \text{m}$. ✓
• Limits.
  • $M\to 1^{+}$: $\sqrt{M^{2}-1}\to 0$, so $d\to\infty$ for fixed $t_{\mathrm{cone}}$. Makes sense — at marginal supersonic speed the cone is nearly flat ($\alpha\to 90^{\circ}$), so a finite delay implies an enormous lateral distance.
  • $M\to\infty$: $\sqrt{M^{2}-1}\to M$, so $d\to ct_{\mathrm{cone}}$. The cone collapses onto the velocity axis ($\alpha\to 0$), and the boom delay is just the lateral travel time of sound at speed $c$ — the jet’s own velocity drops out. Sensible.
• Plausibility. A few-kilometre standoff for a low-altitude supersonic flyby is in the right ballpark for fighter-jet airshows.

Educational remark — the geometry of the boom delay

Equivalently, the boom delay is the time taken by the jet to fly from “directly overhead” to the apex position from which the cone touches the observer:

$$t_{\mathrm{cone}} \;=\; \frac{d}{v\,\tan\alpha} \;=\; \frac{d}{c\tan\alpha\,\cdot M} \;=\; \frac{d\sqrt{M^{2}-1}}{cM}.$$

So the delay measures how far past overhead the jet had already flown by the time its first sound reached the ground — a striking illustration that for a supersonic source, the jet is always seen before it is heard, and the gap between the two grows with lateral distance.

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Summary of results

Part	Quantity	Formula	Numerical value
(i)	features	silence + boom + two Doppler branches with asymptotes $f_e/(M\mp 1)$	—
(ii)	Mach number	$M = (f_{+}+f_{-})/(f_{+}-f_{-})$	$M\approx 1.5$
(ii)	emitted tone	$f_e = 2f_{+}f_{-}/(f_{+}+f_{-})$	$f_e \approx 750\,\text{Hz}$
(iii)	distance	$d = cMt_{\mathrm{cone}}/\sqrt{M^{2}-1}$	$d\approx 1.4\,\text{km}$