Outrunning your own sound
Tonight's idea
A sound spreads out from where it was made as a growing ball, about 340 m every second in air near the ground. A slow plane stays inside the balls of sound it has made, so you hear it coming.
A plane faster than sound escapes from each ball as soon as it makes it. No sound gets ahead of it. Draw the balls from earlier moments as circles: their edges line up along two straight lines that meet at the plane. In space, that is a cone with the plane at its tip. Inside the cone you can hear the plane. Outside, nothing yet. The boom is the surface of the cone sweeping over you.
A plane with speed \(v\) in air where sound has speed \(c\) has Mach number \(M = \dfrac{v}{c}\). If \(M \gt 1\), the angle \(a\) between the flight line and the edge of the cone has \(\sin a = \dfrac{c}{v} = \dfrac{1}{M}\).
This is a model: a very small plane, flying straight at a steady speed, through air where sound has the same speed everywhere.
Watch one: Mach 2
A plane flies at Mach 2, so \(v = 2 \times 340 = 680\) m each second. Measure in units of 340 m, the distance sound goes in 1 s. Then the plane moves 2 units each second.
The sound it made 4 s ago started \(4 \times 2 = 8\) units behind where the plane is now. Since then it has spread into a circle of radius \(4 \times 1 = 4\) units. The sound from 3 s ago started 6 units back and has radius 3, and so on.
Draw a line from the plane that just touches the biggest circle. A line that touches a circle meets the radius at a right angle. So there is a right-angled triangle with the angle \(a\) at the plane, the side 8 along the flight line as hypotenuse, and the radius 4 opposite \(a\). So \(\sin a = \frac{4}{8} = \frac{1}{2}\) and \(a = 30^\circ\). Every other circle gives the same ratio (\(\frac{3}{6}\), \(\frac{2}{4}\), \(\frac{1}{2}\)), so the same line touches them all. The whole cone is \(60^\circ\) wide.
Your turn
- A plane flies at Mach 1.5. Use a scale where sound goes 1 cm each second. (a) How many cm does the plane go each second? (b) On spare paper, draw a flight line and mark the plane \(P\). Mark where it was 1, 2, 3 and 4 s ago, and draw the circle of sound from each of those moments. (c) Draw the two lines from \(P\) that just touch the circles. Measure the angle between one of them and the flight line. (d) Check with \(\sin a = \frac{1}{M}\).
- (a) Find the angle \(a\) at Mach 3. (b) A cone has \(a = 45^\circ\). What is the Mach number, and how fast is that if \(c = 340\) m each second? (c) A plane speeds up from Mach 2 to Mach 4. Does the angle \(a\) halve?
- A plane at Mach 2 flies level, 6 km up, and passes straight over your head. In this model \(c = 340\) m each second at every height. (a) When the boom reaches you, the edge of the cone runs from the plane down to you. Sketch it from the side and mark \(a\) and the 6 km. (b) How far past the point above your head is the plane at that moment? (c) How far is the plane from you in a straight line? (d) How long ago was it overhead?
- (a) What does the rule give at exactly Mach 1? What shape has the cone become? (b) Try the rule at Mach 0.8. What goes wrong, and what do the circles look like now? (c) True or false: "The boom happens once, at the moment the plane first goes faster than sound."
Stretch: the boat that is not a jet
A boat on deep water trails a V-shaped wake. Each arm makes an angle of about \(19.5^\circ\) with the boat's track. (a) If this were a Mach cone, what would the boat's Mach number be? (b) A real boat doubles its speed, and the angle of the V stays at about \(19.5^\circ\). What would a Mach cone have done? (c) Sounds of every pitch travel at the same speed. On deep water, long waves travel faster than short ones. Why does that stop the rule \(\sin a = \frac{1}{M}\) from working for the boat?
Answers
- (a) 1.5 cm. (b) The marks are 1.5, 3, 4.5 and 6 cm behind \(P\), and the circles have radius 1, 2, 3 and 4 cm. (c) You should measure about \(42^\circ\). (d) \(\sin a = \frac{1}{1.5} = \frac{2}{3}\), so \(a \approx 41.8^\circ\). In your drawing that is the triangle with hypotenuse 6 and opposite side 4. A slower plane has a wider cone than the \(30^\circ\) of Mach 2.
- (a) \(\sin a = \frac{1}{3}\), so \(a \approx 19.5^\circ\). (b) \(M = \dfrac{1}{\sin 45^\circ} = \sqrt 2 \approx 1.41\), which is about \(1.41 \times 340 \approx 480\) m each second. (c) No. At Mach 2, \(a = 30^\circ\). At Mach 4, \(\sin a = \frac{1}{4}\) and \(a \approx 14.5^\circ\), a little less than half. Doubling the speed halves \(\sin a\), which is not the same as halving \(a\).
- (a) A right-angled triangle: you at one corner, the point 6 km above you at the right angle, the plane at the third corner, with \(a = 30^\circ\) at the plane. (b) The 6 km is opposite \(a\), and the distance \(d\) along the flight line is next to it: \(\tan 30^\circ = \frac{6}{d}\), so \(d = \dfrac{6}{\tan 30^\circ} = 6\sqrt 3 \approx 10.4\) km. (c) The hypotenuse: \(\dfrac{6}{\sin 30^\circ} = 12\) km. (d) The plane covers 0.68 km each second, so \(10.4 \div 0.68 \approx 15\) s. The plane goes over in silence and the boom comes about 15 s later. In real air sound is slower high up, where it is colder, so the true numbers differ a little.
- (a) \(\sin a = 1\), so \(a = 90^\circ\). The cone has opened out flat: a wall of sound straight across the flight line at the plane's nose. Every circle passes through the plane, because the sound and the plane have travelled the same distance. (b) \(\sin a = \frac{1}{0.8} = 1.25\), and no angle has a sine bigger than 1. There is no cone. Each circle is now bigger than the distance the plane has moved, so the plane is inside all of them. The circles sit one inside another without touching, bunched up in front and spread out behind. (c) False. The cone follows the plane for as long as it flies faster than sound. Each person under the flight path hears a boom when the cone reaches them, so a plane at Mach 2 for 10 minutes drags its boom along about \(0.68 \times 600 \approx 400\) km of ground.
Stretch. (a) \(M = \dfrac{1}{\sin 19.5^\circ} \approx 3\). (b) Twice the speed would mean Mach 6 and \(a \approx 9.6^\circ\): the V would close up to about half as wide. The wake does not do that. (c) The rule needs one wave speed \(c\) to compare with \(v\). Sound has one. Deep water does not: a wave 1 m long travels at about 1.25 m each second and a wave 100 m long at about 12.5 m each second. So a boat has no single Mach number. Whatever its speed, it makes waves of many lengths, and the mix always spreads into the same V. Go faster and the whole pattern gets bigger, but its angle stays the same. This pattern is called a Kelvin wake. The full working gives \(\sin a = \frac{1}{3}\) exactly, the same number as the Mach 3 cone, for a different reason. That too is a model: deep water and a steady boat on a straight course.