Mach number M = V - a — subsonic ( - 1), transonic (~1), supersonic ( - 1), hypersonic ( - 5): Worked examples

This page is the "no surprises" drill for Mach number. We march through every kind of question the topic can throw at you: each flow regime, the degenerate cases ( $M \to 0$ , $M = 1$ , $M \to \infty$ ), the Mach-cone geometry, a real-world word problem, and an exam-style twist. Every symbol used here was built in the parent note; where a new idea sneaks in, we build it from scratch first.

The scenario matrix

Each row is a distinct case class. The worked examples that follow are tagged with the cell they cover.

#	Case class	What is tricky	Covered by
A	Subsonic, find $M$ from $V, T$	order of operations ( $a$ first)	Ex 1
B	Supersonic, find $M$	recognising regime boundary	Ex 2
C	Same $V$ , different $T$ → different $M$	$M$ tracks the gas state	Ex 3
D	Inverse problem: given $M, T$ , find $V$	rearranging the formula	Ex 4
E	Limiting case $M \to 0$ (incompressible)	when density change is ignorable	Ex 5
F	Degenerate case $M = 1$ (sonic)	Mach angle collapses to $9 0^{\circ}$	Ex 6
G	Mach-cone geometry ( $sin μ = 1/ M$ ) + limit $M \to \infty$	angle → 0, needle cone	Ex 7
H	Real-world word problem: sonic-boom delay	mixing geometry + kinematics	Ex 8
I	Exam twist: local vs free-stream Mach	one number ≠ whole field	Ex 9

Ex 1 — Subsonic: find $M$ (Cell A)

Worked example A light aircraft flies at

$V = 110 m/s$ where $T = 280 K$ . Air: $γ = 1.4$ , $R = 287 J/(kg\cdotK)$ . Find $M$ and name the regime.

Forecast: Guess — is $M$ closer to $0.3$ or $0.9$ ? (Everyday small planes are slow.)

Step 1 — Compute the local sound speed first. $a = γ R T = 1.4 \cdot 287 \cdot 280 = 112, 504 \approx 335.4 m/s$ Why this step? $M$ is a ratio; you cannot form it until you know the messenger speed $a$ . And $a$ depends on temperature alone (parent note), so $T$ is all we need.

Step 2 — Form the ratio. $M = \frac{V}{a} = \frac{110}{335.4} \approx 0.328$ Why this step? $M$ answers "how many multiples of the local sound speed am I moving?"

Conclusion: $M \approx 0.33$ → subsonic, and just above the $M \approx 0.3$ line where compressibility starts to matter (see Compressibility & Bernoulli's Limits).

Verify: Units: $(J/(kg\cdotK)) \cdot K = J/kg = m^{2} / s^{2} = m/s$ ✓. And $M < 1$ matches "slow plane" ✓.

Ex 2 — Supersonic: find $M$ (Cell B)

Ex 3 — Same $V$ , different $T$ → different $M$ (Cell C)

Worked example A cargo jet flies at true airspeed

$V = 240 m/s$ . Compare its Mach number (a) near the ground, $T = 300 K$ , and (b) at cruise, $T = 220 K$ .

Forecast: The speedometer reads the same both times. Will the Mach number be the same? Guess before computing.

Step 1 — Ground sound speed. $a_{gnd} = 1.4 \cdot 287 \cdot 300 = 120, 540 \approx 347.2 m/s$ Why this step? Warmer air → faster molecular collisions → faster sound (parent note's $T$ scaling).

Step 2 — Cruise sound speed. $a_{cr} = 1.4 \cdot 287 \cdot 220 = 88, 396 \approx 297.3 m/s$

Step 3 — Two Mach numbers. $M_{gnd} = \frac{240}{347.2} \approx 0.691, M_{cr} = \frac{240}{297.3} \approx 0.807$ Why this step? Same $V$ , but $a$ shrank, so the ratio grew.

Conclusion: Identical airspeed, yet the plane climbs from $M \approx 0.69$ (comfortably subsonic) to $M \approx 0.81$ (edge of transonic) — purely because the air got colder. Mach is about the gas state, not just your speed.

Verify: $a_{gnd} > a_{cr}$ because $300 > 220$ , so $M_{gnd} < M_{cr}$ ✓ (bigger denominator → smaller ratio). See Speed of Sound in Gases.

Ex 4 — Inverse problem: given $M$ and $T$ , find $V$ (Cell D)

Ex 5 — Limiting case $M \to 0$ : when is flow "incompressible"? (Cell E)

Worked example Air moves at

$V = 90 m/s$ at $T = 288 K$ . Estimate the fractional density change and decide whether Bernoulli's incompressible formula is safe.

Forecast: Below or above the famous $M \approx 0.3$ threshold?

Step 1 — Sound speed and Mach number. $a = 1.4 \cdot 287 \cdot 288 = 115, 718 \approx 340.2 m/s, M = \frac{90}{340.2} \approx 0.265$ Why this step? The density-change rule is written in terms of $M$ , so we need $M$ first.

Step 2 — Apply the density-change estimate. The parent note gives $\frac{d ρ}{ρ} \sim \frac{1}{2} M^{2} .$ Why this tool and not Bernoulli directly? This one number tells us how much the gas squishes; it is the yardstick for whether the constant- $ρ$ assumption behind incompressible Bernoulli is acceptable. $\frac{d ρ}{ρ} \sim \frac{1}{2} (0.265)^{2} \approx 0.0351 \Rightarrow \approx 3.5%$

Conclusion: About a $3.5%$ density change — below the $\sim 5%$ line at $M = 0.3$ . Incompressible Bernoulli is a fair approximation here. This is the == $M \to 0$ == corner of the matrix: the compressible terms vanish.

Verify: At the threshold $M = 0.3$ : $\frac{1}{2} (0.3)^{2} = 0.045 = 4.5%$ ✓ (matches the parent's "~5%"). Our $3.5% < 4.5%$ , so being under $M = 0.3$ is consistent ✓. See Compressibility & Bernoulli's Limits.

Ex 6 — Degenerate case $M = 1$ : the sonic wall (Cell F)

Worked example At exactly

$M = 1$ , what is the Mach-cone half-angle $μ$ , and what does the wavefront picture look like?

Forecast: As the body speeds up toward sound, does the cone get sharper or flatter? Predict the angle at $M = 1$ .

Step 1 — Recall the Mach-angle relation. From the parent note, $sin μ = \frac{1}{M} .$ Why this tool? $sin$ turns the wavefront geometry (a right triangle of sides $a t$ and $V t$ ) into an angle — it is the ratio "opposite over hypotenuse" for the cone (look at the triangle in the figure).

Step 2 — Set $M = 1$ . $sin μ = \frac{1}{1} = 1 \Rightarrow μ = arcsin (1) = 9 0^{\circ}$ Why this step? $arcsin$ asks "which angle has this sine?" — the inverse of $sin$ — and the answer is a right angle.

Conclusion: At $M = 1$ the "cone" opens all the way to $9 0^{\circ}$ : the wavefronts pile into a single flat wall of sound perpendicular to the motion (left panel of the figure). This is the transonic knife-edge.

Verify: $sin (9 0^{\circ}) = 1 = 1/ M$ with $M = 1$ ✓. For $M$ slightly above 1, say $M = 1.05$ , $sin μ = 0.952$ , $μ \approx 72. 2^{\circ} < 9 0^{\circ}$ — the wall just begins to tilt into a cone ✓.

Ex 7 — Mach cone geometry and the $M \to \infty$ limit (Cell G)

Worked example Find the Mach half-angle for (a)

$M = 2$ , (b) $M = 5$ , (c) the limit $M \to \infty$ .

Forecast: Does the cone get narrower or wider as speed climbs? Guess the shape at extreme hypersonic speed.

Step 1 — $M = 2$ . $sin μ = \frac{1}{2} = 0.5 \Rightarrow μ = arcsin (0.5) = 3 0^{\circ}$ Why this step? Straight application of the wavefront-envelope geometry (figure).

Step 2 — $M = 5$ (start of hypersonic). $sin μ = \frac{1}{5} = 0.2 \Rightarrow μ = arcsin (0.2) \approx 11.5 4^{\circ}$ Why this step? Same tool, larger $M$ ⇒ smaller $sin μ$ ⇒ smaller angle.

Step 3 — The limit. As $M \to \infty$ , $sin μ = 1/ M \to 0$ , so $μ \to 0^{\circ}$ . Why this step? We take a limit — "what does $μ$ approach as $M$ grows without bound?" — to describe hypersonic behaviour we can never reach exactly but can approach.

Conclusion: Faster ⇒ narrower cone. From the flat wall at $M = 1$ ( $9 0^{\circ}$ ) → $3 0^{\circ}$ at $M = 2$ → $\approx 11. 5^{\circ}$ at $M = 5$ → a needle-thin cone hugging the flight path as $M \to \infty$ . See Oblique Shocks & Mach Cone.

Verify: Angles decrease monotonically as $M$ rises: $9 0^{\circ} > 3 0^{\circ} > 11. 5^{\circ} > 0^{\circ}$ ✓. And $sin (3 0^{\circ}) = 0.5 = 1/2$ ✓.

Ex 8 — Real-world word problem: sonic-boom delay (Cell H)

Worked example A jet flies

level at altitude $h = 1500 m$ and $M = 1.8$ (local $a = 320 m/s$ ). It passes directly overhead an observer at time $t = 0$ . How long after that does the observer hear the sonic boom?

Forecast: Do you hear the boom before, at, or after the plane is overhead? (The plane outran its own sound...)

Step 1 — Find the Mach angle. $sin μ = \frac{1}{M} = \frac{1}{1.8} \approx 0.5556 \Rightarrow μ = arcsin (0.5556) \approx 33.7 5^{\circ}$ Why this step? The boom arrives when the trailing cone surface sweeps over the observer; the cone's half-angle sets its geometry (figure).

Step 2 — Flight speed. $V = M a = 1.8 \times 320 = 576 m/s$ Why this step? We need how fast the plane travels so we can turn a horizontal distance into a time.

Step 3 — Horizontal distance the plane has flown when the cone reaches the ground point. The cone trails behind at angle $μ$ from the flight path; when the observer (a vertical distance $h$ below the flight line) is just touched by the cone, the plane is a horizontal distance $d$ ahead, where $tan μ = \frac{h}{d} \Rightarrow d = \frac{h}{t a n μ} .$ Why the tangent here (not sine)? We have the vertical leg $h$ (opposite the angle) and want the horizontal leg $d$ (adjacent) — that ratio "opposite over adjacent" is exactly what $tan$ encodes on the right triangle in the figure. $d = \frac{1500}{t a n ( 33.7 5 ^{\circ} )} = \frac{1500}{0.6682} \approx 2245 m$

Step 4 — Convert distance to time. $t = \frac{d}{V} = \frac{2245}{576} \approx 3.90 s$ Why this step? The plane covers $d$ at speed $V$ ; time = distance / speed.

Conclusion: The observer hears the boom about ==3.9 s after== the jet is overhead — the boom always lags, because the aircraft outran its own pressure signals.

Verify: Geometry check — $tan μ = h / d = 1500/2245 \approx 0.668$ , and $tan (33.7 5^{\circ}) \approx 0.668$ ✓. Units: $m / (m/s) = s$ ✓. "After overhead" matches the supersonic-lag intuition ✓.

Ex 9 — Exam twist: local vs free-stream Mach (Cell I)

Worked example A wing flies at free-stream

$M_{\infty} = 0.85$ in air where $a = 300 m/s$ . Over the thickest part of the wing the flow accelerates to $V_{local} = 345 m/s$ (and the local sound speed there has dropped to $a_{local} = 298 m/s$ ). Is any part of the flow supersonic?

Forecast: The whole plane is subsonic ( $0.85 < 1$ ). Trick question — can the air still go supersonic somewhere?

Step 1 — Free-stream Mach (given). $M_{\infty} = 0.85 < 1$ → subsonic far ahead of the wing. Why this step? Establish the "one number" everyone quotes — and show it isn't the whole story.

Step 2 — Local Mach over the wing crest. $M_{local} = \frac{V _{local}}{a _{local}} = \frac{345}{298} \approx 1.158$ Why this step? Air speeds up over the curved upper surface (Bernoulli), and the local sound speed can fall as the gas cools in that fast region — both push the local ratio above $1$ .

Conclusion: Even though $M_{\infty} = 0.85$ , the local flow reaches $M \approx 1.16$ → a supersonic pocket ends in a small shock. This is precisely the transonic difficulty: the free-stream Mach is one number; the field has many. This is why designers use the Prandtl–Glauert Correction near $M \approx 1$ .

Verify: $M_{local} = 345/298 \approx 1.158 > 1$ while $M_{\infty} = 0.85 < 1$ ✓ — subsonic overall, supersonic locally, exactly the transonic paradox.

Active recall

Recall Predict, then reveal

Which comes first when finding $M$ , computing $a$ or dividing? ::: Compute $a = γ R T$ first, then $M = V / a$ . Same true airspeed, colder air — does $M$ rise or fall? ::: Rises, because $a = γ R T$ falls, shrinking the denominator. At $M = 1$ the Mach half-angle is? ::: $9 0^{\circ}$ (a flat wall of sound; $sin μ = 1$ ). As $M \to \infty$ the Mach cone half-angle approaches? ::: $0^{\circ}$ (needle-thin cone; $sin μ = 1/ M \to 0$ ). Given $M$ and $a$ , how do you get $V$ ? ::: $V = M a$ . Does the sonic boom reach an observer before or after the plane is overhead? ::: After — the aircraft outran its own sound; the trailing cone arrives later. Can a subsonic free-stream have supersonic local flow? ::: Yes — air accelerates over the wing; local $M$ can exceed 1 (the transonic paradox).