I'm reading this physics book, and in it contains this problem. Quoted:

An airplane is flying at Mach 1.75 at an altitude of 8000 m, where the speed of sound is 320 m/s. How long after the plane passes directly overhead will you hear the sonic boom?

Obviously the first steps are to find the speed of the plane, and the cone angle, which are: $$v_p = 1.75*320 m/s = 560 m/s$$ and $$\alpha = \arcsin(\frac{1}{1.75}) = 34.8\deg$$

Considering the shock wave is moving at the same speed as the plane, we can simply use some trigonometry to deduce that

$$\tan\alpha=\frac{8000 m}{560 m/s * t}$$ so $t$ can be calculated to be 20.5 seconds. This is the solution given by the book. My question is, doesn't the shock wave have to take time to appear? Like, from the moment the shock is created it must take time until it reaches the ground? My hypothesis stems from the idea of taking the limit of the Mach number to 1. Then, the plane velocity approaches 320 m/s, and the angle approaches 90 degrees. Plugging in these limits to the final equation yield:

$$\lim_{v_p→320, \alpha → 90 \deg}t=\frac{8000m}{340m/s*\tan(90)}=0$$ which makes no sense. How can the shock wave created by the plane reach the ground in no time at all? Shouldn't the lower bound be at least $\frac{8000}{320}$? Thank you for any help given.


The time it takes is certainly not zero, that would violate the speed of sound, among other things.

To solve your problem, the easiest thing to do is assume that the speed of sound is approximately constant in all your air medium. So that you can use the Mach cone. Considering this, and assuming that the airplane has been flying at constant velocity for some time before reaching over your head, the initial condition looks like this:

Initial condition

The shaded red region is the Mach cone of the airplane, and you are located at the origin $(0, 0)$. What the question is asking can be rephrased as "how much time does it take for the Mach cone to reach you?". If the velocity and altitude of the airplane remains constant, this would look like this:

Final condition

The airplane is on the tip of the cone, so you know the distance it traveled, and you also know the speed.

With this you should be able to calculate the time fairly easily.

  • $\begingroup$ Oh, I get it now. Thank you. The shock travels with the plane, because it was already created a long time ago. I would assume that if the plane accelerated up to Mach 1.75 in a very short amount of time, the time it takes for the shock wave to reach a person would be delayed as opposed to the example given, right? $\endgroup$ – Gabe Aug 26 '19 at 21:14
  • $\begingroup$ @Gabe I think you can also imagine the edge of the Mach cone as a single wave front that propagates perpendicularly to the cone. If it starts accelerating the surface would no longer be a cone because the faster you go the smaller the angle of the "envelope" is (for a section of constant acceleration I think it would be like fitting two lines with different slopes with a parabola section in the middle). If you start slower and accelerate, the sound actually gets there earlier because you give more time to the first waves to approach the location of the observer before the plane passes overhead $\endgroup$ – S V Aug 26 '19 at 23:59

My question is, doesn't the shock wave have to take time to appear? Like, from the moment the shock is created it must take time until it reaches the ground?

Of course. The cone is the geometric realization of this. The shock is created by the surfaces of the airplane disturbing the air. The sound of this event then moves outward at the speed of sound in the air.

The surface of the cone represents the motion of this shock wave through the air. The portions near the vertex of the cone were created most recently. The portions farther away are older and have traveled away from where the plane was. But assuming the speeds remain constant, the shape of the surface also is constant and we can imagine it moving forward at the same speed as the plane.

But this cone is not a physical object. It's not really moving forward.

How can the shock wave created by the plane reach the ground in no time at all?

It can't. Even in the first case that doesn't happen. It takes $(8000\text{m}/(320\text{m/s}))$ or $25\text{s}$ for the generated sound to reach the ground. But you can hear sounds from the plane that were created before it flew overhead. Above mach 1, the plane will always pass you before any previously-generated sound can reach you. Below mach 1, some earlier sounds can "beat" the plane to reach you first.

  • $\begingroup$ I thought it would take 25 seconds to reach the ground, but if you use the formulas given in the book, it would take zero seconds to reach the ground. Since arctan of 90 approaches infinity and the reciprocal approaches zero. $\endgroup$ – Gabe Aug 26 '19 at 21:15
  • 3
    $\begingroup$ It takes the sound 25 seconds to reach the ground. But you can hear the plane as it flies overhead because you are hearing sounds created earlier. Do not confuse the time difference between the position of the plane and hearing it with the speed the sound took to reach that point. $\endgroup$ – BowlOfRed Aug 26 '19 at 21:17
  • $\begingroup$ Thank you. I presume then, that when you hear the sound, although the plane may have created it directly overhead, it is now hundreds of meters away. $\endgroup$ – Gabe Aug 26 '19 at 21:25
  • $\begingroup$ The first sound you hear (the boom) was created before the plane flew overhead. You then continue to hear sounds that were created both before and after that point (as later generated sounds move back to you and as earlier generated sounds move forward to you). Sounds created when the plane flies directly overhead will be heard 25 seconds later. $\endgroup$ – BowlOfRed Aug 26 '19 at 21:29
  • $\begingroup$ This answer is actually incorrect, the actual time taken is closer to $20.52 s$. As BowlOfRed pointed out, the sound you hear first is one that was created earlier. If you look at my answer, there are two ways to calculate that time. One is the method proposed in the answer, and the other is to consider a line perpendicular to the lower envelope that passes through the origin, the length between the intersections of such line (which is $6566 m$, rather than $8000 m$) is the one with which you calculate the time. $\endgroup$ – S V Aug 27 '19 at 0:17

Consider a situation where the aircraft is traveling at a subsonic speed, below the speed of sound. In that scenario, the sound waves from the aircraft will reach you before it flies overhead (as they are traveling faster than the aircraft). In this case, the $t$ in your equation will be negative.

So, given that t is positive for an aircraft traveling at $M > 1$ and negative for $M < 1$, it stands to reason that there must be a speed of the aircraft at which the sound waves reach you at the same time as it flies overhead, in which case $t = 0$.

It is true that, in theory, when the aircraft is traveling at $M = 1$, the shock will be a flat plane coincident with the nosecone of the aircraft; however, bear in mind that the sound waves that are reaching the ground in that situation will have left the aircraft some time before that. The shock front is moving with the aircraft, but that doesn't mean the sounds waves have traveled instantaneously. Remember, sound waves travel perpendicular to the shock, not along it.


You are ignoring the fact that the shock wave dissipates as it spreads further from the plane, and its initial strength close to the plane is greater as the Mach number increases.

So in real life, the very weak shock from a plane flying at say Mach 1.001 would not propagate very far, and if the plane was at 8000m it would not be heard at all on the ground.

Close to the plane, the shock wave at Mach 1.001 would indeed be almost vertical.

The question assumes that the shock at Mach 1.75 is strong enough to reach the ground and be heard. In fact, the shocks from military fast jets flying at high altitude are unlikely to be heard at all on the ground, unless you are in a quiet situation and know what they sound like so you recognise the characteristic "double shock" when you hear it. Of course the shocks from bigger planes, like Concorde, were more powerful and caused a ban on supersonic flight over land.


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