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.