University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.11
Shock Waves & Mach Number
Push a source through air faster than the air can carry the news and the wavefronts stop spreading and start stacking. Their envelope is a cone, one number fixes its angle, and the boom trails the aircraft the whole way.
Build the model
Connect the measurement to the mechanism.
Every wavefront, once emitted, expands at the medium's speed v about the point it left, whatever the source does next. Below that speed the fronts crowd ahead of the source without ever crossing, which is the Doppler shift. At exactly that speed they share one tangent plane at the nose.
Above it the source is outside all of them, neighbouring fronts intersect, and the envelope of the whole family is a cone trailing the source with half-angle μ set by sin μ = v/vₛ = 1/M. The Mach number M = vₛ/v carries the entire geometry: it fixes the cone angle, the strip of ground a boom sweeps, and the lag between an aircraft passing overhead and the bang arriving. What geometry cannot give is amplitude.
Overlapping fronts add to a finite-amplitude disturbance, and finite amplitude is exactly what the linear wave model behind this extension assumes away.
- Simple definition
- The Mach number M = vₛ/v is a source's speed divided by the wave speed in the medium it moves through. For M > 1 the wavefronts overlap into a cone trailing the source, with half-angle μ satisfying sin μ = 1/M.
- Example
- A rifle bullet at 850 m s⁻¹ through air at 343 m s⁻¹ has M = 2.48, so its cone half-angle is arcsin(1/2.48) = 23.8°. The sharp crack a bystander hears is that cone sweeping past, not the muzzle blast.
Source speed in units of the medium's wave speed: M < 1 subsonic, M > 1 supersonic.
Dimensionless; v is the local sound speed, not a fixed 343 m s⁻¹
Faster source, tighter cone: 90° at M = 1, 41.8° at M = 1.5, 30.0° at M = 2.
μ measured from the flight path; defined only for M ≥ 1
The same airspeed counts as a higher Mach number in colder air.
γ = 1.40, Mₘ = 0.0290 kg mol⁻¹; 343 m s⁻¹ at 20 °C, 295 at −56 °C
The divergence is the linear model announcing its boundary, not an infinite pitch.
Returns a negative frequency for vₛ > v, so it does not apply there
At M = 2.00 and h = 10.0 km with v = 300 m s⁻¹, the bang arrives 28.9 s later.
Level flight at height h; uniform v, refraction ignored
Impulsive, not sinusoidal — scored as a steady tone it would read near 130 dB.
Large aircraft at cruise; pₐₜₘ = 1.01 × 10⁵ Pa
A wavefront stops caring about its source the instant it leaves
Each front emitted by a source becomes an independent sphere, expanding at radius v t about the point of emission at the speed the medium sets. Moving the source moves the centres of successive spheres; it does not change how fast any of them grows. Below the sound speed each new front starts inside the previous one and never catches it, so the fronts bunch ahead and stretch behind while the source stays inside every sphere it has made — that bunching is the Doppler shift of the previous topic. At exactly vₛ = v the source keeps pace with its own leading edge: all the fronts touch at one point at the nose, and their common tangent is a plane across the direction of motion. Push past that and the picture inverts. The source now sits outside all of its earlier fronts, each front lies behind the one emitted after it, and neighbouring spheres cut through one another. Where they cut, their disturbances add.
The cone falls out of similar triangles
Take t = 0 as the moment a supersonic source passes A, and let it reach B a time t later, so AB = vₛ t. The front emitted at A has by then grown to radius v t. Draw from B the tangent line to that sphere, touching at P: angle APB is a right angle, so sin(angle ABP) = v t/(vₛ t) = 1/M. That ratio does not contain t, so the front emitted at any intermediate point — centre a distance vₛ t₁ from A, radius v(t − t₁) — is tangent to the same line. Every front touches it, so the envelope of the family is a cone with its apex at the source, its axis along the path, and half-angle μ given by sin μ = 1/M. None of this is dynamics: it is circles with a moving centre, so it holds for any source that outruns the waves it makes. The cone tightens with speed — 90° at M = 1, 41.8° at M = 1.5, 30.0° at M = 2, 19.5° at M = 3. Read backwards it is a measurement: photograph a bullet's shock, measure μ = 23.8°, and M = 1/sin μ = 2.48.
M is a ratio, and both of its halves move
The Mach number is not a speed, and neither ingredient is fixed. The denominator is the local sound speed, v = √(γRT/Mₘ), which in air depends on temperature alone: 343 m s⁻¹ at 20 °C, but 295 m s⁻¹ at the −56 °C of the tropopause. An aircraft holding 250 m s⁻¹ through the air is at M = 0.73 near the ground and M = 0.85 at 11 km, having changed nothing about its own motion. The numerator is speed through the medium, not over the ground: a 50 m s⁻¹ tailwind adds 50 m s⁻¹ to ground speed and leaves M untouched, because the air the wavefronts travel in is carried along with it. This is also why M is the useful variable rather than vₛ. Whether fronts merely crowd, touch, or overlap depends on how close the source is to outrunning its own signals, and that comparison — not any absolute speed — is what M reports.
What a ground observer hears, and when
Outside the cone, no sound from the source has arrived at all; the cone's edge is the first signal to reach any point, and ahead of it there is real silence. For level flight at height h, that edge reaches an observer once the aircraft has travelled a horizontal distance x = h/tan μ = h√(M² − 1) past overhead, so the lag is Δt = h√(M² − 1)/(M v). At M = 1.5 and h = 12.0 km with v = 295 m s⁻¹, x = 13.4 km and Δt = 30.3 s: the bang lands half a minute after the aircraft was overhead, by which time it is far down the sky. The same delay is why a supersonic aircraft is always heard well behind where it is seen. Notice what never enters Δt: any instant of crossing M = 1. The cone exists for as long as M > 1, so its intersection with the ground drags a strip along the whole supersonic track, and every observer under that strip gets one boom as the edge passes.
Overlap fixes the geometry; amplitude needs new physics
The construction says where the disturbance is and nothing about how big it is. Linear acoustics — every result in this extension so far — assumes the pressure perturbation is small next to ambient pressure and the fluid's particle speed small next to v, so the wave speed belongs to the undisturbed medium. Where fronts pile onto each other, neither assumption survives. A compression is warmer than its surroundings and its fluid is already moving forwards, so it travels faster than the trough behind and overtakes it. The profile steepens until viscosity and heat conduction stop it, leaving a front a few mean free paths thick across which pressure, density and temperature jump and entropy rises. That is a shock, not a sound wave. Far from the aircraft the airframe's details wash out and the ground signature is an N: sharp rise, linear fall through ambient into rarefaction, sharp recompression — two bangs, roughly 0.3 s apart for an airliner-sized body, at an overpressure of order 100 Pa.
Where this topic stops, and it is meant to stop
This extension is institution-dependent, and inside it this topic is enrichment: take it if the local syllabus wants shock geometry, leave it out and the sequence still closes cleanly on beats and Doppler. The boundary is easy to draw. The kinematics is fair game — the wavefront construction, sin μ = 1/M, recovering M from a photographed cone, boom timing and carpet geometry — because it needs only circles, a moving centre, and the wave speed you already have. Everything about a shock's strength belongs to compressible gas dynamics: jump conditions across the front, entropy production, wave drag. Two cautions ride along with the geometry. It assumes a non-dispersive medium, so it does not carry over to deep-water waves, whose Kelvin wake holds a half-angle near 19.5° at every boat speed. And the same triangle reappears in electromagnetism, where a particle faster than light in a medium radiates a Cherenkov cone with cos θ = 1/(nβ).
Change one variable at a time
Make the relationship visible.
Raise the source speed, or drop the sound speed to mimic cold air at altitude — either one raises M and tightens the cone. Outside that cone no sound has arrived at all.
MACH NUMBER M1.98
SIN μ = 1/M0.504
CONE HALF-ANGLE μ30.3 °
BOOM LAG, h = 10.0 km25.2 s
Live interpretationMACH NUMBER M: 1.98. SIN μ = 1/M: 0.504. CONE HALF-ANGLE μ: 30.3 °. BOOM LAG, h = 10.0 km: 25.2 s
Catch the common trap
Explain before calculating.
An aircraft flies level at 10.0 km altitude at Mach 2.00, through air in which the sound speed is a uniform 300 m s⁻¹. How long after it passes directly overhead does an observer on the ground first hear the sonic boom?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA jet flies at 528 m s⁻¹ through air in which the sound speed is 330 m s⁻¹. Find its Mach number and the half-angle of its Mach cone.
- M = v(s)/v = 528/330 = 1.60 — greater than 1, so a cone exists at all.
- sin μ = 1/M = 1/1.60 = 0.625.
- μ = arcsin 0.625 = 38.7°, measured from the flight path, opening backwards.
AnswerM = 1.60; cone half-angle μ = 38.7°
MediumA shadowgraph of a bullet in air at 15 °C, where v = 340 m s⁻¹, shows a shock cone whose half-angle measures 21.0°. Find the bullet's Mach number and its speed.
- The construction runs backwards as easily as forwards: sin μ = 1/M gives M = 1/sin μ.
- sin 21.0° = 0.3584, so M = 1/0.3584 = 2.79.
- v(s) = M v = 2.79 × 340 m s⁻¹ = 949 m s⁻¹.
- A tighter cone would mean a faster bullet — μ shrinks as M grows, never the other way.
AnswerM = 2.79; v(s) = 949 m s⁻¹
HardAn aircraft cruises level at 11.0 km, where the air is −56 °C and the sound speed is 295 m s⁻¹, holding a true airspeed of 590 m s⁻¹. Find M, the cone half-angle, and how long after it passes overhead the boom reaches the ground.
- M = v(s)/v = 590/295 = 2.00 — the airspeed alone would not tell you this, since v is only 295 m s⁻¹ at that temperature.
- sin μ = 1/2.00 = 0.500, so μ = 30.0°.
- The cone edge reaches the observer once the aircraft has flown x = h/tan μ = h√(M² − 1) = 11.0 km × √3 = 19.05 km past overhead.
- Δt = x/v(s) = 19 050 m ÷ 590 m s⁻¹ = 32.3 s, which is the formula Δt = h√(M² − 1)/(Mv) written out.
- By then the aircraft is that same 19.05 km down-track — the bang always lands well behind where the aircraft is seen.
AnswerM = 2.00, μ = 30.0°, and the boom arrives 32.3 s after the aircraft is overhead