Video Transcript
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.
