Hey, guys. So up until now, we've seen some pretty basic equations for the wave speed of waves. But in some problems, you're going to have to use something called the wave function to solve problems. So I'm going to introduce you to the wave function in this video and show you exactly what it is. Now if you've never seen it before, it can be kind of scary. So what I'm going to show you is that there's actually only 3 variables you really need to know. Let's check this out here. So the idea here is that the wave function is really just a sinusoidal equation. Remember sinusoidal just means sine or cosine, and it describes the shape of an oscillating wave. When you whip a string up and down, you're producing these sort of sinusoidal graphs that go up and down like this, so they can be described by sine or cosine graphs. Alright? So this is the first time you're going to see something like this in physics, so I want to go a little bit carefully. The equation for this wave function is going to be y of x and t. So you're going to have 2 things inside the parentheses. What it's doing here is it's actually giving you an output. It's giving you the displacement value when you plug into inputs, the position and the time. So you plug in x of t in your equation, and it's going to give you the y value of a particle on that string at a certain position and time. That's what it's telling you. Alright? So depending on the textbook that you're using, you might gonna see this, you might see this written in 2 different ways. So you might see this written as a⋅sinkx±ωt, and in some textbooks you'll see this written as a⋅coskx±ωt. Your professor might also have a preference, so just go ahead and use whatever they're going to use. Now again, both of these are accurate, sine and cosine. They're both sinusoidal. The difference between when you use them is actually where the graph starts. So this blue graph right here, I actually have it in blue and you'll see that the equation or the graph starts at y equals 0. That's when you use the sine graph. You're going to start when y equals 0. And you're going to use this cosine graph when you're starting at the amplitudes. It could be either the upward or the positive or the negative amplitude. So you're going to use the cosine wave function when your wave starts at either plus or minus a. So let's go ahead and talk about the different variables that are inside this equation. We already know that the position in time, x and t. So what about the a, k, and omega? The a is just the amplitude. That's just really the maximum displacement in either direction of the graph, and we've seen that before. The k is something new called the wave number. You don't need to know conceptually what it is. All you need to know is the equation for it, which is 2πλ, the units for this k are going to be in radians per meter. Now the other equation is going to be, or sorry, the other variable is omega, which is the angular frequency, and we've seen that before. The angular frequency is just 2πt, the period. So notice that these two actually kind of look alike. 2πλ, 2πt. The angle of frequency also had another equation which is 2*pi times the frequency. That's really it. Those are your 3 variables. So let's go ahead and take a look at our problem here. So our problem we're going to whip a rope up and down to create a transverse wave. We're told some of the values. We know that the amplitude, this a here, is 0.5. We're also told that the wave speed of the rope is going to be 8 meters per second. Second, that's going to be v. We're also told that the wavelength of the wave that we're producing, so that's going to be lambda equals 0.32. Now we're also told at t equals 0, at the starting point, the end of the rope that we're going to hold is at the maximum upward displacement. So what it looks like is kind of like this. So you're holding this little rope like this at the maximum upward displacement and you're going to flick it up and down to create a transverse wave that's moving to the right like that. Now in the first part of the problem, in part a, we're going to write the wave function for this wave. So we're going to write an equation that's y of x and t. Now to do this we need to figure out whether we're using a sine or a cosine function, and remember that depends on where the graph is going to start. So remember, if we're starting at either the positive or negative amplitude, we're going to use the cosine function. So with this y of x and t, we're going to use a times the cosine of kx+−omegat. The next thing we have to do is actually figure out the sign of this function, whether it's kx plus or kx minus omega t. And to do that, we're just going to use this rule here. The direction of the wave determines the sign of kx plus or minus omega t. The rule is pretty simple. It's the direction is going to be opposite to the sign. So what this means is that your wave moves to the right like in the plus x direction, sign is going to be negative. It's going to be kx minus omega t. And then it's the opposite if you're moving to the left. Minus x, your sign is going to be plus, so it's going to be plus omega t. Alright? So what happens is we're moving to the right, our wave, so that means we're just going to use a minus sign. So it's k x minus omega t. The last thing we need to do is we actually have to figure out the values for a, k, and omega, so we can plug them back into our equation, and that's going to be the full wave function. So we actually already know what the amplitude is. That's what we were given, a equals 0.5. So we need to figure out k next. So this k value, remember, has has an equation. It's going to be 2 pi divided by the lambda, which is going to be 2 pi divided by our lambda value is 0.32. So pretty straightforward. If you go ahead and do this work this out, you're going to get 19.6. So now we have what k is. The last thing we need to do is just figure out what what omega is. So our omega equation, or omega, variable, remember, has 2 equations. We're going to have 2 pi divided by the period, or we have 2*pi times the frequency. Now we're not told in this problem anything about the period. All we know is the amplitude, the speed, and the wavelength. So we don't know the t, so we're not going to use this equation. We're going to use this 1, 2*pi times the frequency. Now unfortunately, we don't know what f is, but we can go figure it out. Remember that the only other place that f shows up is inside of the wave speed equation, v equals lambda f. Now we actually have 2 out of 3 variables. We know the wave speed, and we also have what the lambda is. So we can figure out the frequency. So our frequency is just going to be v over lambda. That's going to be 8 divided by 0.32, and you're going to get to 25 hertz. So that's the frequency of the wave that you're oscillating. Right? So now all we have to do is just plug this back into this f right here, and then we'll figure out omega. So omega is just 2 pi times 25, and if you go ahead and work this out, what you're going to get is a 157.1. So that's what we now plug into this variable here, omega. So now all we're all we're going to do is just pop each one of these variables back into our wave function equation, and then we're done. So our wave function equation is going to be 0.5 times the cosine. Now we have 19.6 for k. We have an x here for the position minus 157.1 times t, and this is our full wave function equation. Notice how when you write out the wave function, if you're just asked to write it out, you're going to plug in values for a, k, and omega, but not for x and t. This is a function here, so you're still going to have these inputs as x and t. Right? So you're still going to have those inputs there. All you have to plug in a, k, and omega. Now the last thing we have to do, that's part a, is just actually evaluate this wave function. We're going to figure out the displacement of a particle at x equals 0.4 and t equals 0.75. So basically what happens is that now that we have this wave function, we're just going to plug in some values and figure out the y value. So y when x is equal to 0.4 and t is equal to 0.75. You're basically just going to plug in these values for x and t inside of your equation. So this is going to be 0.5 times the cosine of 19.6 times 0.4 minus 157.1 times 0.75. So it's a lot of stuff to plug into your calculator. Make sure you do it carefully. And also make sure that your calculator is in radians mode. You cannot forget this. Make sure your calculator is in radians and not degrees, or or you're going to get the wrong answer. What you're going to get here is you're going to get a displacement of negative 0.5. So what this means here is that if your amplitude is 0.5 and negative 0.5, then at this position and time, you're going to have a particle that's right here at the bottom at the negative That's all that means. Alright? So that's it for this one guys. Let me know if you have any questions.

# Wave Functions - Online Tutor, Practice Problems & Exam Prep

### Intro to Wave Functions

#### Video transcript

A transverse harmonic wave moving to the left has a wavelength of 2.5 m and a wave speed of 12 m/s. The amplitude of the wave is 0.1 m. At x = 0 and t = 0, the displacement of the wave is y = 0. Write the wave function for this wave.

$y\left(x,t\right)=0.1\sin\left(2.51x+30.2t\right)$

$y\left(x,t\right)=0.1\cos\left(2.51x-30.2t\right)$

$y\left(x,t\right)=0.1\sin\left(2.5x+15.7t\right)$

$y\left(x,t\right)=0.1\sin\left(15.7x+188.5t\right)$

### Example 1

#### Video transcript

Welcome back everyone. So in this problem, we're given a wave function. It's given as y of x and t is equal to 6.5 times the cosine. Then we got some numbers in this equation. We've got 2π/28 x and then 2π/0.36 t. In this problem, we want to calculate the wavelength and frequency of the wave, and then eventually we want to figure out what direction the wave travels in. So let's go ahead and get started here.

Part A, we want to calculate the wavelength. We know that the general form of an equation or a wave function is going to be a sine of kx plus or minus omega t or cosine. In this case, we have an amplitude, we have a cosine, and the value that goes in front of the x is k. Our omega over here is the 2π/0.36. We want to calculate what the wavelength is. That's not k or omega, but we can get the wavelength by looking at those values. Remember that k is related to the wavelength, and omega is related to the frequency or the period.

So to find the wavelength, we really just focus on that wave number given to us in our wave function. We know that k=2π/λ and in our problem here, in our wave function, it's given as 2π/28. So lambda is equal to 28 meters. It's straightforward to extract the wavelength when it's written in this form.

Now, for Part B, we want to calculate the frequency. Instead of looking for the wavelength by looking at k, if we want the linear frequency f, we're going to have to look at omega. Omega is represented as 2π×f or 2π/T. In our equation, this 2πf was equal to just 2π/0.36. So f=1/0.36, and this ends up being 2.78 Hz.

For the last part of the problem, we determine the direction the wave travels in. We just have to look at the sign between x and t. If the form is kx plus omega t, it means that you are traveling to the left. So, the plus sign indicates that the direction this wave travels in is left.

That's really all there is to it, folks. Let me know if you have any questions. Thanks for watching, and I'll see you in the next video.

### Calculating Wave Speed from Wave Functions

#### Video transcript

Hey, guys. So in the last couple of videos, we saw the equation for the wave function. This y of x and t equation, which is either a sine or a cosine. Now in some problems like our example down here, we're going to be given this wave function equation and we'll be asked to calculate the wave speed. So I'm going to show you in this video how to calculate the wave speed directly by using the wave function and I'm going to show you a simple shortcut equation to do that. Let's check out our example here. We have a transverse wave that's traveling on a string, so if we want to calculate the wave speed, that's v, we actually have a couple of different equations that we can use.

Remember that for strings only we can use the special equation TensionMu. So I've got the square roots of my tension divided by mu. Let's take a look. I don't have the tension, I'm told nothing about the tension in this problem, and I also don't know the mass or the length of the string. So I'm kind of stuck here and I'm not going to use this equation for sure. So the only other equation to calculate v that I know is going to be this lambda frequency. Remember this applies to all kinds of waves. Now if you go through this equation, you'll notice that we don't have the lambda or the frequency. So it also looks like we're kind of stuck here.

However, remember that lambda is related to k and frequency is related to omega. So if we have our wave function equation here, remember this is the form ak aomega, then we can actually calculate this wave speed by using these two variables, k and omega. So I'm going to show you a shortcut equation to calculate v directly from omega and k, and this is how it works. Basically, we're just going to write 2 expressions for these two variables, and we're going to get something that includes omega and k. Remember that lambda is just related to k, so k=2pilambda. So if you rearrange for this you're going to get lambda=2pik. So I'm going to write this equation. I'm going to write lambda as 2pik and now I'm going to have another expression for frequency. Remember that omega is related to 2pifrequency. Right? That's omega. So if I can rearrange for frequency, f equals omega2pi. So now I'm going to plug that back in here. So omega2pi. So now if you write it this way and you multiply these two things, what you're going to see is that the 2pi's will cancel. And what you end up with is you end up with just omegak, and that's the shortcut equation. If you're given the wave function directly, then v, the wave speed is going to be lambda times frequency, but it's also going to be equal to omegak. So if you have the wave function, you could actually calculate omega or sorry. V directly from these two variables. So my omega remember is 6 and my wave number k is going to be 0.4. So if you plug this in, you're going to get 15 meters per second exactly for the wave speed. Now if you were to actually go and solve either lambda or frequency or any one of these values by using the long method, by doing all the calculations, you would end up also with 15 meters per second. Alright. So that's basically it. The last thing I want to point out is that sometimes this wave speed is also called the propagation velocity of the wave. So that's it for this one guys. Let me know if you have any questions.

### Example 2

#### Video transcript

Welcome back, everyone. So in this problem, we have a wave function that's given to us. We have y(x,t) equals 6 millimeters, and then we have some numbers here. Right? We have 5 and then the units for this radians per millimeters, and then 600 radians per second. Some of the general sort of form of a wave function is that the first number is the amplitude, and the units. This is going to be the k value. Right? The thing goes in front of our x, and this is going to be the ω value. Alright? So what we want to do is we want to figure out how long it takes for something to happen. How long it takes for a given particle on the string to travel between plus 6 and negative 6 millimeters. So I sort of want to visualize what's going on here. If you take a string and sort of whip it up and down like this, you're going to get some sort of a sine wave, and the particles in the string remember are moving up and down. Now in this case, the amplitude of this wave is 6 millimeters. So basically, what is what's going to happen is that the particles on this, on the string are going to sort of bubble up and down between positive 6 and negative 6 millimeters over time. We're asked to find how long it takes for it to do that. Right? So a given particle on the string, it's going to sort of bounce up and down between these two points, the crest and the trough, forever. That's always what's going to happen, and we're trying to figure out how long it takes for that to happen. So Δt. Alright? So how do we solve for this? Well, the basic relationship between Δt and velocity and displacement is that v=ΔxΔt. Right? So displacement over time. So if I solve for this equation here, I can rearrange this in Δt=Δxv. Alright? So just very simply here, if I want to figure out time, I need to figure out the displacements, what's the distance that these particles are traveling, divided by what's the wave speed or what's the transverse velocity of the particles that are bobbing up and down. Alright? So how do I figure this out? What's the displacement? Well, the displacement really is just going to be the distance between the top of the crest and the bottom of the trough, and it's really just the distance between plus 6 and minus 6. In other words, it's basically just double the amplitude. So this is really just going to be 2 times amplitude, which is just going to be 12 millimeters. Alright. Now I can also convert this to 0.012, but that's basically where the displacement is. Alright? So that's done. So all I really need now is I need to figure out now the velocity, the display the velocity of the particles that are moving up and down on the string. And we have a new equation for this, it's the transverse velocity or whatever, and basically, the equation for the velocity is it's going to be the, I'm sorry. It's going to be the, the displacement velocity of the particles, which is going to be ω/k. Alright? So if we do do ω/k, basically, I'm just going to take this, ω, which is 600, and the units for this are going to be really important. So notice how I have 600. This is going to be radians per second divided by, and this is going to be 5, and this is going to be radians per millimeter here. Alright. So this is important because what happens is the radians are going to cancel when you do this, and what you end up getting is you end up getting a 120 millimeters per second. So the units are going to be really important here. So this is what my velocity is. It's a 120 millimeters per second. Alright. So then how do I figure it now Δt? Well, really now I have everything I need to solve, because now I have Δx and I have v. So I'm just going to bring this down here, and my Δt is just going to be my Δx, the displacement, which is 12 millimeters, divided by and this is going to be a 120 millimeters per second. So again, the units are important here because if you got something like meters and millimeters, you're going to have to convert. But what we're going to see here is that millimeters will cancel, and you're just going to be left with seconds. And really what happens is this is actually just going to be 0.1 seconds. So in other words, it takes 1 tenth of a second or 0.1 seconds for a particle in the string to sort of bounce up and down between positive 6 and negative 6. That's, will always happen. Alright? So it's kind of a strange problem. We're really sort of pulling together a lot of different equations from from wave functions. Let me know if you have any questions. Thanks for watching.

### Transverse Velocity of Waves

#### Video transcript

Hey, guys. So in the last couple of videos, we saw the equations for the wave function, which are sine and cosine functions. Well, in some problems, you're going to be asked to calculate two different kinds of velocities. In the example that we're going to work out here, we have a wave function given to us. In part a, we're going to calculate the velocity of the wave, which is really just the propagation velocity, and we've seen that before. In part b, we're going to calculate something else entirely, which is called the transverse velocity. Now, these things sound very similar. The velocity of a transverse wave versus the transverse velocity of waves. They sound very similar, but they're actually very different ideas. In this video, I am going to show you the differences between these two velocities and how we calculate the transverse velocity of waves. Let's go ahead and check this out here. So we're going to start with what we know. Remember that the velocity of a transverse wave had another name, we call it the propagation velocity. Basically, this was just the velocity of the wave pattern that is moving left and right. So if you take a string and whip it up and down, as we have in our diagram over here, you're going to create a wave, and this wave pattern overall moves to the right like this, and that's what the propagation velocity is. Now, this is different from the transverse velocity of waves. They sound similar, but they're actually very different. Because what's happening here is that as you're whipping the string up and down, the particles that are on the string are also moving. They're moving up and down perpendicular to the direction of the wave. So that's what this transverse velocity means. Transverse just means perpendicular. It's the perpendicular velocity of the particles that are on the string or on the wave, and those particles move up and down. So that's the difference between these two velocities. The propagation velocity is the overall wave pattern moving to the left or right. The transverse velocity is going to be the velocity of the particles that are moving up and down. Alright? So before I actually get into the equation, I want to go ahead and start our problem now that we know the differences between these two. So in part a, we want to calculate the velocity of the wave, which is really just the propagation velocity, and we've seen how to do this before. If we have our wave function equation, we've actually seen this exact wave function before. Remember that this has the form \(a \times \cos(kx - \omega t)\). So we have the velocity of the wave, the propagation velocity, is really just going to be given by this equation right here, \(\lambda/ k\). Remember that was our shortcut equation. So we have \(v = \omega / k\). So our omega value is just 6, our k value is 0.4. If you go ahead and work this out, you're going to get 15 meters per second. So overall, this whole entire wave is moving to the right at 15 meters per second. What about the transverse velocity? That's what we're going to calculate in part b. Well, now I'm going to show you the equations for this. Remember that different textbooks will use different sorts of equations for the wave function. Some will use a sine and then some will use a cosine. I am actually going to give you both of the transverse velocity equations, basically based on which one your textbook actually uses. So if we're using a sine, your transverse velocity equation is also going to be a function of x and c, and it's going to look like this: \( \pm \omega a \times \cos(kx \pm \omega t)\). So we picked up another value of \(\omega\) outside of our cosine equation and we also have this \(\pm\) here. This \(\pm\) is actually related to this \(\pm\) that's inside of our \(kx \mp \omega t\). So basically, what happens is that the rule is that they have the same signs. If you have a plus sign here, you're going to have a plus sign outside of your equation. If you have a minus sign inside of your wave function equation, then you're going to pick up a minus sign when you take the transverse velocity. Alright? So, let's take a look at the other equation for cosine. It's going to look a little bit different. So this equation is going to look like this. It's going to be \(\mp \omega a\). Now we're going to have \(\sin(kx \pm \omega t)\). So the idea here is that this is a minus over a plus, which actually means that it's going to be the opposite sign of whatever you have inside of here. So if you have \(kx+\omega t\), then you're going to pick up a minus sign outside of your equation for the transverse velocity. If you have a minus sign, then it's going to be the opposites. And also, the other thing to realize is that the sine turns into a cosine and the cosine turns into a sine, so the equation of the trig function is always going to flip. Alright, so that's just a little bit about the transverse velocity equations. So let's go ahead and actually use them. We want to calculate the transverse velocity of a particle that's at some position and time. So we need to figure out which one of the equations we're going to use. So \(v_{\text{tx}}\), which one are we going to use? Well, we're starting off with a wave function that is of the form \(a \cos\), so we have a cosine here of \(kx\) and we have \(- \omega t\). So we're going to actually use this equation here to start off with, which means our equation is going to have this form right here. So the first thing we have to do is figure out whether we have a plus or a minus sign that's outside of our equation. Remember the rule. If we have a minus sign that's here, then what happens is we're going to have a plus sign inside of our transverse velocity equation. That's exactly what happens here. So we have a plus sign here. Next we have \(\omega\) and \(a\). So our \(\omega\) remember is just going to be 6, our \(a\) is going to be 3. Now we have the sine and this is going to be our \(k\), this is going to be 0.4, that doesn't change, \(- 6t\). Right? So what's goes inside the parenthesis doesn't actually change. So this is actually the format that we're going to use. So all we have to do now is just plug in the values. So \(v_{\text{tx}}\), so now we're going to plug in when \(x\) is equal to 0.75, and when \(t\) is equal to 0.2. So all you have to do now is just plug in the values. So we're going to have \(18 \times \sin(0.4 \times 0.75 - 6 \times 0 .2)\). So that's what happens when I plug in all the numbers. Remember to keep your calculator in radians, and what you'll get is \( -14.1 \text{ meters per second}\). So this is the perpendicular velocity of a particle at that specific time. It's going downwards at 14.1 meters per second. That's all that means. Alright. So that's how we use the wave function. The last thing I want to mention here is that the propagation velocity that will be calculated in part a is going to be constant at all points on the wave, whereas the transverse velocity is actually going to change with the position and the time. Remember, it depends on \(x\) and \(t\), so if you change those values, it's going to change the transverse velocity. So this just means that some particles in the wave could be going up and others are going down depending on where you look along the wave. Alright. So let's finish things off and talk about the last part here, which is calculating the maximum transverse velocity of the particles. So remember the idea was that when we had the wave function, this \(y\) equation here, the maximum displacement that we could possibly have was really just the amplitude, either the positive or negative one. It was basically just whatever is outside of your trig function, either sine or cosine. It's the same idea for the maximum transverse velocity. It's just going to be whatever is outside your sine and cosine, so that's just going to be \(\omega a\). So we want to calculate \(v_{\text{t max}}\) here, so this is just going to be \(\omega \times a\), and we actually know both of those values. This is just going to be \(6 \times 3\), so our maximum transverse velocity is 18 meters per second. Alright? So that's how you work out these problems. Let me know if you guys have any questions.

## Do you want more practice?

More sets### Your Physics tutor

- A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope ...
- A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope ...
- A sound wave is described by D (y,t) = (0.0200 mm) ✕ sin [(8.96 rad/m)y + (3140 rad/s)t + π/4 rad], where y is...
- Show that the displacement D(x,t) = cx² + dt², where c and d are constants, is a solution to the wave equation...
- Show that the displacement D(x,t) = ln(ax + bt), where a and b are constants, is a solution to the wave equati...
- (II) A transverse traveling wave on a cord is represented by D = 0.22 sin (5.6x + 34 t) where D and x are in m...
- (II) A transverse traveling wave on a cord is represented by D = 0.22 sin (5.6x + 34 t) where D and x are in m...
- (II) Consider the point x = 1.00 m on the cord of Example 15–6. Determine (a) the maximum acceleration of this...
- (I) A transverse wave on a wire is given by D (x, t) = 0.015 sin ( 35x - 1200 t) where D and x are in meters a...
- (II) Write the equation for the wave in Problem 29 traveling to the right, if its amplitude is 0.020 cm, and D...
- (III) A transverse wave on a cord is given by D( x, t) = 0.12 sin (3.0x - 15.0t) , where D and x are in meters...
- (II) A transverse wave with a frequency of 220 Hz and a wavelength of 10.0 cm is traveling along a cord. The m...
- (II) A transverse wave pulse travels to the right along a string with a speed v = 2.0 m/s. At the shape of th...
- Figure 15–44 shows the wave shape at two instants of time for a sinusoidal wave traveling to the right. What i...
- A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed v, ...
- A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed v, ...
- A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed v, ...
- A string that is under 50.0 N of tension has linear density 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm a...
- FIGURE P16.57 shows a snapshot graph of a wave traveling to the right along a string at 45 m/s. At this instan...
- FIGURE EX16.8 is a picture at t = 0 s of the particles in a medium as a longitudinal wave is passing through. ...