In this video, we’re going to learn about shock waves. We’ll learn what shock waves are. And we’ll also study motion that’s supersonic, that is, the motion of an object that’s moving faster than the speed of sound.
To get started with this idea of shock waves, imagine that you’re standing still and you’re whistling a particular note. That is, you’re creating a steady sound. We know that sound travels in waves. And one way to represent these sound waves that you’re creating is with a series of concentric circles that start at the sound source. Each one of these rings represents a peak in the sound wave that’s traveling radially outward from the source of the sound.
If we were to look at this situation more closely, if we were to zoom way in to the molecular level, we would see that the sound is actually transmitted by air molecules that run into one another and thereby transmit the sound energy. We can imagine that the faster these air molecules are moving, the more space they’ll cover in a given amount of time and the faster sound will transmit through that air mass. Similarly, if the molecules are moving very slowly, the speed of sound will be slowed in that medium. So sound in air is transmitted by air molecules that come in contact with one another. And that overall speed of sound has to do with how fast those air molecules are moving.
Returning now to our whistler, let’s imagine that, instead of standing still, they now start to move at a steady speed to the right, continuing to whistle the same note. We can imagine that our walking whistler emits one wavelength of this sound wave while stationary. But then it goes in motion so that when the next wavelength is emitted, they’re in a different location.
In the time that passes between the whistler being in these two locations, the first sound wavefront emitted expands. That is, it moves radially out at the speed of sound. So by the time our second wavefront is emitted, the first wavefront may’ve expanded to look something like this. Notice something interesting about this expanded sound wave. The center of the wave is at its point of origin. It hasn’t moved along with our moving walker to be centered on the second sound wave. This may be counterintuitive when we consider concepts in physics like relative motion.
We know that if we had a runner, say, running along with a speed 𝑣 while holding an object in their hand, then that object would also be moving forward with the same speed 𝑣 as the runner. So that if the runner dropped it, then the object would travel along with the runner and land at his or her feet. But sound waves, it turns out, aren’t like that. They don’t move with the object that emits them. Rather, they have an origin point and they move outward from that point.
If our whistler keeps walking ahead at a steady pace, this pattern continues. A wavefront is emitted and the wavefronts behind that emitted wavefront continue to expand. By this point, the original wavefront is greatly expanded. And the second one emitted continues to grow.
Looking at these three wavefronts, we’re just starting to see the beginning of a trend. If we consider the distance from the center of the wavefront being emitted in the direction of the walker’s motion to the original wavefront. And if we compare that distance with the distance between the same two wavefronts, but in the opposite direction. We can see that, in the direction of the walker’s motion, that distance is significantly less than opposite that direction. This means that, in the direction of the walker’s motion, the wavefronts are stacked more closely to one another. If we wanted to stack them even more closely, the way we would do that is to increase the speed of our walker’s motion.
We know that each one of our sound wavefronts moves outward at the speed of sound, which in air is about 340 meters per second, depending on temperature and air density. Imagine for a moment that our walker was capable of moving forward at the speed of sound. What would the sound waves emitted by the walking whistler look like then?
Here’s something that’s pretty amazing. If the speed of our walker was equal to the speed of the sound waves the whistler is giving off, if we looked at the leading edge of the wavefronts given off, we would see that they all overlap perfectly one on top of the other. Just imagine what the air is like right around this point where the waves overlap. If we were to plot air pressure versus position, we would see as we approach these stacked sound waves that the pressure in the air was at some low-level atmosphere pressure.
But then as we arrive at these multiply stacked waves with pressure maxima, the pressure would discontinuously jump up. These wavefronts stacked one on top of another on top of another create a significant disturbance as they move forward. We’ve talked about how when our ears hear sound, what we’re hearing is variations in air pressure. And we’ve seen that our ears are very sensitive at detecting slight pressure differences as sound.
This means that if a pressure change as big as that created by all these stacked sound waves reached our ears, we would hear it as a very loud noise, a booming sound. And actually, that’s the name given to this phenomenon of sound waves stacked exactly one on top of another on top of another. When we hear sound like that, we say that we’re hearing a sonic boom. And this leading wave of stacked sound waves is known as a shock wave.
Shock waves get their name from the fact that they’re so disruptive when they pass by. In this example we’ve imagined, of course it’s not realistic because a person isn’t able to move at the speed of sound. But other objects can. Jet aircraft, for example, are able to move at and above the speed of sound. And even something as simple and old fashioned as a whip is actually able to break the sound barrier. The cracking sound we hear from a whip is a sonic boom created by a shock wave reaching our ears.
If we switch out our walking whistler with a jet aircraft moving at the speed of sound, if our jet aircraft is moving at the speed of sound, then, as before, the wavefronts pile up one on top of another at the leading edge of the wave in the direction of the jet’s motion. When objects begin to move at speeds comparable to the speed of sound, there is a special notation for those speeds. With object speeds around the speed of sound, we often refer to their speeds by a number called a Mach number.
The Mach number is a ratio. It’s the ratio of the object’s speed to the speed of sound, which we sometimes abbreviate 𝑣 sub 𝑠. If an object is moving at the speed of sound, like our aircraft here, then it has a Mach number of one. For an object moving with a Mach number of one, it’s traveling at what is called a transonic speed, or it’s traveling with transonic motion. If an object’s Mach number is less than one, that means its speed is subsonic. And greater than one means it’s moving at a supersonic speed.
If we consider the shape or the profile of the shock wave created by our jet aircraft moving at Mach one, we see that it’s nearly a horizontal line. That is, the shock wave is almost flat. The shape of our shock wave though will change as we move into supersonic speeds.
Let’s imagine for a moment that our jet is moving at a supersonic speed, greater than the speed of sound. That means that the fronts of the sound waves that it produces don’t catch up with one another because they’re being produced by a source that’s moving faster than the wave itself. If we were to draw in the shock wave the place where all these different sound waves overlap, we’d see there still is a shock wave. But it’s not in a vertical line as before when we were moving at a transonic speed.
Now thanks to our supersonic motion, the shock wave has a backward tilt to it. We can characterize the angle of the shock wave by an angle from the horizontal. We’ll call it 𝜃. Given some information about the speed of our jet, the speed of sound, and the time elapsed, we’re able to solve for that angle 𝜃, the angle of the shock wave created. Let’s do that in a short example.
A plane is flying at Mach 1.2. And an observer on the ground hears the sonic boom 15 seconds after the plane is directly overhead. What is the altitude of the plane? Use a value of 340 meters per second for the speed of sound.
In this scenario, we have a plane flying along at a steady altitude with a speed of 1.2 times the speed of sound, which we’re calling 𝑣 sub 𝑠. As the plane flies along, its engines produce sound. And the sound waves from these engines expand outward as time elapses. Since the plane is moving at a speed greater than the speed of sound, that means that it creates a shock wave, a line at which the sound waves produced by the plane stack on top of one another. This shock wave slowly moves out. And we’re told that an observer on the ground hears the shock wave 15 seconds after the plane is directly overhead.
Based on this and knowing that the speed of sound is 340 meters per second, we want to solve for the altitude of the plane, capital 𝐴. To solve for the attitude 𝐴, we’re first gonna solve for the angle created by the shock wave and a horizontal line through the center of the plane. We’ll call that angle 𝜃. And to solve for it, let’s consider the triangle that 𝜃 is a part of.
If we start at the center of the first sound wave we’ve drawn in and move from that center to the shock wave so that the two lines meet perpendicularly, then we can say that the length of this line we’ve drawn in is equal to the speed of sound, 𝑣 sub 𝑠, times the time elapsed, what we’ll call 𝑡. Likewise, we can draw a line starting at that same center moving in a horizontal direction to the current position of the plane and say that the length of that line segment is equal to 𝑣, the speed of the jet engine, times 𝑡, the time elapsed.
As we look at this right triangle then, we see that 𝜃 is related to 𝑣 sub 𝑠 times 𝑡 and 𝑣 times 𝑡. In particular, we can write that the sin of the angle 𝜃 is equal to 𝑣 sub 𝑠 times 𝑡 over 𝑣 times 𝑡. We see that the time values cancel out from this fraction. And since 𝑣 is equal to 1.2 𝑣 sub 𝑠, that means our fraction reduces to one divided by 1.2. If we take the inverse sine of both sides of this equation, we find that 𝜃 is approximately equal to 56.44 degrees.
Now that we know 𝜃, let’s consider how this can help us solve for the altitude 𝐴. With our diagram as shown, the observer on the ground won’t hear the shock wave until the shock wave has advanced a bit further until it’s reaching the ground. When that takes place, we’ll have a right triangle with an angle 𝜃. And the side length opposite 𝜃 is the altitude 𝐴. In addition, the entire adjacent side of this triangle is equal to the speed of our jet engine, 𝑣, times the time elapsed, 𝑡. This means we can once again use the trigonometry of this right triangle to solve for what we want to find.
This time, we can say that it’s the tangent of 𝜃 which is equal to, in this case, 𝐴 over 𝑣 times 𝑡. Rearranging, we can write that the altitude of the jet is equal to its speed times the time that’s elapsed since it was overhead the observer times the tangent of the angle 𝜃, the angle its shock wave makes with the horizontal. The speed, 𝑣, is equal to the speed of sound, 340 meters per second, times 1.2. The time, 𝑡, is equal to 15 seconds. And 𝜃 as we’ve seen is 56.44 degrees. Plugging all these values in to our calculator, we find that, to two significant figures, 𝐴 is 9.2 kilometers. That’s the elevation of this plane above ground level.
Let’s review what we’ve learned so far about shock waves. We’ve seen that a shock wave is a sonic disturbance that’s caused by an object moving at or faster than the speed of sound. We’ve also seen that Mach number, sometimes abbreviated with a capital 𝑀, characterizes object speed relative to the speed of sound. As an equation, 𝑀 is equal to object speed divided by the speed of sound. And finally, we saw that objects in motion can move at subsonic, 𝑀 is less than one; transonic, 𝑀 equals one; or supersonic, 𝑀 is greater than one, speeds.