Video Transcript
In this video, we’re going to learn
about modeling one-dimensional transverse waves. We’ll see examples of these
transverse waves. And we’ll also see how to
mathematically represent them.
To start out, imagine that as a
surfer on the lookout for the biggest possible waves to find and then ride, you make
a point of carefully studying the ways that individual waves in the ocean combine
and mix with one another. To be able to know when these
individual waves will combine to create the ultimate rideable wave, it’s helpful to
know something about modeling one-dimensional transverse waves.
A transverse wave is a wave that
displaces matter perpendicular to its direction of motion. So if we have a wave that is
transverse, then even though it’s moving in this example to the right, it’s
displacing the matter of the medium it’s travelling through in the direction up and
down, that is, perpendicular to that direction of propagation. This is different from the kind of
wave called a longitudinal wave, where the wave motion is in the same direction as
the direction of the wave travel. Transverse waves are very common
waves. Waves in the ocean are transverse,
and so our waves along the string of an instrument, like a guitar. In fact, light itself is a
transverse wave where the electric and magnetic fields that make up the light wave
move transversely to the direction of the wave’s travel.
Even though there are many
different kinds of transverse waves with many different shapes to them, there is a
one- dimensional wave equation we can use to mathematically encapsulate or describe
this transverse wave motion. Say that we picked a transverse
wave like a water wave and we took a snapshot of the wave’s profile at one instant
in time. Then, we take that wave profile and
we plot it on a pair of axes, where vertically, we’re looking at wave height in
units of meters and horizontally, we’re seeing the wave position also in units of
meters. If we wanted to model the wave
profile of this transverse wave mathematically, we might be interested in something
like this: given a horizontal position of the wave in meters, what’s the wave height
at that point?
If we call the horizontal position
of the wave 𝑥, then we’re saying that given 𝑥, what is ℎ? That is, ℎ as a function of 𝑥. Assuming that our transverse wave
is sinusoidal, like it would be for a harmonically oscillating system, we can recall
that in modeling a wave like this, we start with the amplitude of the wave—that is,
its maximum displacement from equilibrium—and let that lead off our function that
describes the height of the wave given a position 𝑥. Looking at our wave function so
far, we know there’s more to it because the wave amplitude is modulated by a
sinusoidal function. That function makes it dip down,
cross over to zero, and achieve a value as negative as it was positive.
The next question we wanna answer
is what’s inside the argument of the sine function. We know there’s at least gonna be
an 𝑥 there because we want to be able to input our horizontal position and then get
our height of the wave as an output. But will there be anything else
along with that 𝑥? To answer that question, let’s
think about the sine function by itself for a second. If we plot the sine function all by
itself, just versus 𝑥, its argument, we know that it starts at zero where 𝑥 is
equal to zero, then goes up to its maximum of one, crosses back through zero at 𝑥
is equal to 𝜋. And then when 𝑥 is equal to two
𝜋, it goes back to its original value and slope.
This means that if we go back to
our wave equation, if we left the argument inside the sine function simply as 𝑥,
that means we’d be requiring our transverse wave to cycle back on itself after 𝑥
had advanced two 𝜋. But we can see even by looking at
this curve that that constraint doesn’t work. After all, two 𝜋 is just a little
bit more than six. And we see that this wave doesn’t
meet that condition. It loops back on itself to have the
same value and slope to the line after eight meters, not two 𝜋 meters. This means we will indeed need to
add something else inside the sine argument in order to make it fit our actual
wave.
When we look at our curve, our
transverse water wave that we’re modeling, we see that its wavelength is equal to
eight meters. That wavelength, represented by the
Greek letter 𝜆, is the horizontal distance it takes for our wave to loop back on
itself. Knowing that, we might guess that
𝜆 is something we’ll want to include in the argument of our sine function. In fact, let’s consider this
factor: two 𝜋 divided by 𝜆. Here’s what this accomplishes: the
two 𝜋 factor fits with the cycle of our sine function. It cycles back on itself every two
𝜋 radians.
We also know that we want the
argument of this function to be unitless so that overall the unit of our height will
be meters, which is given to it by the units of our amplitude. That suggests putting 𝜆 in the
denominator so the units of 𝜆 meters can cancel out with the units of position,
also meters. This representation of our
transverse wave at one snapshot in time is looking promising. So let’s test out a few values from
our curve and see if it gives us back the wave height that it should.
Let’s say we test out this point
that’s already highlighted, where 𝑥 is equal to eight meters. We can see from our graph that the
amplitude is three meters. So we say three meters times the
sin of two 𝜋 over our wavelength, which we’ve said is eight meters, multiplied by
the 𝑥 position of eight meters. The eight meters cancels out and
the sin of two 𝜋 is equal to zero, just like the sin of zero. So given an input horizontal
distance of eight meters, the output height that this function gives us is zero
meters. And that agrees with what we see on
our curve.
Now let’s try a different value on
our curve. Let’s try where 𝑥 is equal to 10
meters. We expect our output for height
then to be three meters, the amplitude of this wave. When we insert this value for 𝑥
into our height function, we see in the argument that the units of meters cancel
out. The remaining numbers inside this
argument simplify to five 𝜋 divided by two. And since the sine function loops
back onto itself every two 𝜋 radians, that means that five 𝜋 over two is the same
as 𝜋 over two. The sin of 𝜋 over two or 90
degrees is equal to one. So this means when we evaluate our
height of our wave for 𝑥 equals 10 meters, it gives us three meters. That’s the value we expect when
we’re looking at the representation of this curve on our plot.
So our one-dimensional transverse
wave model is looking good for representing the wave at a particular instant in
time. But as we know, most waves that we
experience, like water waves, don’t stand still but actually translate or move in
time. Say that we added a third dimension
to our axes. This dimension is the time
dimension in units of seconds. Now imagine instead of just
plotting one snapshot of this wave, we plotted a snapshot at every one-second
interval. It gets kinda of confusing looking
if we plot every single second. But the general idea is that we see
the wave is starting to move left to right. That fits with our general
understanding that the wave is in motion. So as time advances, the position
of the wave advances along the 𝑥-axis in the horizontal direction.
The question is can we modify our
function which gives us the accurate height of the wave given its position 𝑥 at the
time instant 𝑡 equals zero to also give us an accurate wave height, not just as a
function of 𝑥, but as a function of 𝑡 as well. In other words, can we model both
the spatial, that is position, and temporal, that is time, evolution of this
wave? The answer is yes. And we’re gonna to do it by
changing once more what’s inside the sine function in a similar way that we did it
for the position 𝑥. Something interesting to notice as
we look at this curve is that in order to include our time evolution, all the wave
is doing is shifting a steady amount every second to the right.
Essentially, what we do then is we
add a phase shift that corresponds to the speed of the wave. And by doing that, we’ll account
for its behaviour in time and the fact that it’s in motion. Now here’s an interesting fact. Let’s say as our first guess for
the way to include temporal information in this function we simply add 𝑡 to it. This is actually a little bit like
when we just had 𝑥 as the argument of our sine function. It was a good starting point. But then we modified it. We can do the same thing here.
For one thing, our sign, a positive
sign, would mean that our wave is moving to the left. But we want it to move to the right
in the positive 𝑥-direction. So we’ll change that to a
negative. Then as we consider this 𝑡-value,
time in seconds all by itself, we see we’ll want to modify it a little bit like we
modified the position 𝑥 with this factor of two 𝜋 over a physical parameter of the
wave. We chose 𝜆 when it came to
position 𝑥 because those are corresponding values and they both have the same
units. What’s a corresponding value for 𝑡
that has the same units as it? The period, capital 𝑇, of our wave
has units of seconds. And this represents the time it
takes for one complete wave cycle.
Just like we did with the position
𝑥, we’ll insert two 𝜋 in the numerator to satisfy the fact that the sine function
comes back on itself every two 𝜋 radians. And in the denominator, we’ll put
the wave period, capital 𝑇, which has the same units as time 𝑡 so that those units
will cancel out and our argument in our parentheses will ultimately be unitless. We’ve now developed a mathematical
model of this one-dimensional transverse wave. What this model does is it tells us
that for a given position and at a given time what the wave height will be. This model relies on the two
fractions that we’ve inserted inside the sine function.
It turns out that these two
fractions each have special names. Two 𝜋 over the wavelength 𝜆 is
known as the wavenumber of a wave. That’s a helpful parameter to
recall. And two 𝜋 over the period, capital
𝑇, is known as the angular frequency of a wave. Knowing these names for these
fractions helps us identify wave properties simply from looking at the 1D wave
equation that represents the wave. It’s common for wavenumber to be
represented with the letter lowercase 𝑘 and angular frequency to be represented by
𝜔. It can be helpful to recall that 𝑘
is equal to two 𝜋 divided by 𝜆 and that 𝜔, the angular frequency, is equal to two
𝜋 over the wave period or two 𝜋 times its frequency 𝑓. There is also a helpful
relationship between wavelength 𝜆 and frequency 𝑓 for moving waves. And that is that the wave speed 𝑣
is equal to their product. Let’s practice these ideas a bit
with an example of a one-dimensional transverse wave.
A wave is modeled by the
wavefunction 𝑦 as a function of 𝑥 and 𝑡 is equal to 0.30 meters times the sin of
two 𝜋 over 4.50 times the quantity 𝑥 minus 18.00𝑡, where 𝑥 is measured in meters
and 𝑡 is measured in seconds. Find the amplitude of the wave. Find the wavelength of the
wave. Find the speed of propagation of
the wave. Find the frequency of the wave. Find the period of the wave.
We can label our wave amplitude
capital 𝐴, the wavelength 𝜆, the wave speed 𝑣, the wave frequency 𝑓, and the
period capital 𝑇. We can start solving for these
various wave properties by recalling that a one-dimensional transverse wave is
modeled by the function 𝑦 is a function of 𝑥 and 𝑡 is equal to wave amplitude
times the sine of wavenumber 𝑘 times position minus angular frequency of the wave
times time.
Looking back at our given wave
function, we see right away that the amplitude of our wave is given as 0.30
meters. To solve next for the wavelength of
our wave 𝜆, we can recall that wavelength and wavenumber are related. Wavenumber 𝑘 is equal to two 𝜋
over wavelength. And as we look at our given
function, we see that this is two 𝜋 over 4.50. This indicates that our wavelength
𝜆 is 4.50 meters.
Next, we want to solve for wave
speed 𝑣. And to do this, we’ll recall two
pieces of information. First, we recall that angular
frequency is equal to two 𝜋 times frequency 𝑓. Working off of our given wave
function, we see that two 𝜋 times 18.00 divided by 4.50 is equal to 𝜔, which is
two 𝜋 times 𝑓. The factors of two 𝜋 cancel
out. And 𝑓 is equal to 4.00 hertz. We can fill that in for our
frequency 𝑓 in advance. But now we wanna use the
relationship that says 𝑣 is equal to 𝜆 times 𝑓 to solve for 𝑣. 𝑣 is equal to 𝜆𝑓 or 4.50 meters
times 4.00 hertz, which is equal to 18.00 meters per second. That’s the speed of propagation of
this wave.
Finally, we want to solve for the
wave period 𝑇. Recalling that wave period is equal
to the inverse of wave frequency, the period of this wave is equal to one over 4.00
hertz or 0.250 seconds. That’s the wave period determined
using the wave function for a wave as well as relationships among wave
parameters.
Let’s summarize what we’ve learnt
so far about modeling one-dimensional transverse waves.
We’ve seen that one-dimensional
transverse waves oscillate perpendicularly to their direction of motion. Examples of transverse waves
include water waves and light waves. We’ve also seen that the
mathematical function 𝑦 as a function of position and time is equal to wave
amplitude 𝐴 times the sine of wavenumber 𝑘 times its position minus angular
frequency times time. And that this function describes
wave height in position as well as time. And finally, we learned several
mathematical relations among and between wave parameters. These included that wavenumber 𝑘
is equal to two 𝜋 over wavelength, wave angular frequency 𝜔 is equal to two 𝜋
over wave period, and that wave speed 𝑣 is equal to wavelength times frequency.
