Wave Functions - Video Tutorials & Practice Problems

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Intro to Wave Functions

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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 gonna have to use something called the wave function to solve problems. So I'm gonna 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 gonna show you is that there's actually only three 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 uh graphs that go up and down like this. So they can be described by sine or cosine graphs. All right. 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 gonna have two 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. All right. So the, depending on the textbook that you're using, you might gonna see this, uh you might see this written in two different ways. So you might see this written as a sign of KX plus or minus omega T. And in some textbooks, you'll see this written as a cosine KX plus or minus omega T. Your professor might also have a preference to just go ahead and use whatever they are 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 in blue and you'll see that the equation or the graph starts at Y equals zero. That's when you use the sine graph, you're going to start when Y equals zero. 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, 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 knows the equation for it, which is two pi over lambda. The units for this K are going to be in radiance per meter. Now, the other equation is gonna be or sorry, the other variable is la omega which is the angular frequency. And we've seen that before the angular frequency is just two pi over t the period. So notice that these two actually kind of look alike two pi over lambda, two pi over T the angle of frequency also had another equation which is two pi times the frequency. That's really it, those are your three variables. So let's go ahead and take a look at our problem here. So our problem, we're gonna whip a rope up and down to create a transverse wave. And 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 gonna be 8 m per second that's gonna be V, we're also told the wavelength of the wave that we're producing. So that's gonna be lambda equals to 0.32. Now, we're also told at T equals zero at the starting point, the end of the rope that we're gonna hold is at the maximum upward displacements. What it looks like is kind of like this. So you're holding this little rope like this at the maximum or displacement and you're gonna 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 plus or minus omega T, the next thing we have to do is actually figure out the sign of this uh 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 gonna be opposite to the sign. So what this means is that your wave moves to the right, like in the plus X directions is gonna be negative. It's gonna be KX minus Omega T and then it's the opposite. If you're moving to the left minus X, your sign is gonna be plus. So it's gonna be plus Omega T. All right. 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 KX 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 the equation and that's gonna be the full wave function. So we actually already know what the amplitude is. That's we were given a equals 0.5. So we need to figure out K next. So this K value remember has this has an equation. It's going to be two pi divided by the Lambda, which is going to be two pi divided by our Lambda value is 0.32. So pretty straightforward, if you go ahead and do this and work this out, you're gonna get 19.6. So now we have what K is the last thing we need to do is just figure out what la what Omega is. So our Omega equation or Omega uh variable number has two equations. We're gonna have two pi divided by the period or we have two 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 gonna use this 12 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, is inside of the wave speed equation V equals lambda. F. Now, we actually have two out of the three 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 gonna be V over Lambda, that's gonna be eight divided by 0.32 and you're gonna get 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 La Omega. So Omega is just two pi times 25. And if you go ahead and work this out, which you're gonna get is 100 and 57.1. So that's what we now plug into this variable here, Omega. So now all we're all, we're gonna do is just pop each one of these variables back into our wavefunction equation. And then we're done. So our wavefunction 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 point one 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 why when X is equal to 0.4 and T is equal to 0.75 you're basically just gonna plug in these values for X and T inside of your equation. So this is gonna be 0.5 times the cosine of 19.6 times 0.4 minus 1 57.1 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 radiance mode. You cannot forget this, make sure your calculators in radiance are not degrees or you're gonna get the wrong answer. What you're going to get here is you're gonna 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 gonna have a particle that's right here at the bottom at the negative amplitude. That's all that means. All right. So that's it for this one guys. Let me know if you have any questions.

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Problem

Problem

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.

Welcome back everyone. So in this problem, we're given a wave function, it's given as a Y of X and T is equal to 6.5 times the cosine. And then we got some weird numbers in this, in this equation, we've got two pi over 28 X and then two pi over 0.36 T. What we want to do this problem is we want to actually first sort out a couple of different things. 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. So with part A, we want to calculate the wavelength. So let's just start start out with the wavelength there. Now 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. So in this case, we have an amplitude, we have a cosine and then our value that goes in front of the X is K. So that's basically what this thing is. And then our omega over here is the two pi over 0.36. Now, 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 because remember that K is related to the length the wavelength and omega is related to the frequency or the period. So really what we do is to find the wavelength in part A, we're really just going to focus in on that wave number that's given to us in our wave function. All right. So we know that two, that K is equal to two pi divided by Lambda. And in our problem here, this is in our, our wave function that's given to us as two pi over 28. So here's what happens when you get a wave number that's given to us. And oftentimes you're gonna see numbers like two pi over something really, what they're just trying to do is they're trying to represent this K where it's very easy to determine what the wavelength is if K is two pi over Lambda. But in our wave function, it's two pi over 28 then that just means that Lambda is equal to 28. So it's really straightforward to sort of extract that wavelength when it's written in this form. So really what happens here is that Lambda is equal to 28 and this is going to be in meters and that's all you have to do. So it's just uh it's better to sort of do this rather than having to sort of sort out a bunch of decimals and do a bunch of uh equations and calculations. All right. So let's take a look now at part B and part B, we want to do something very similar except instead of the sorry, instead of wavelength, we want to calculate the frequency. So what is F? All right. So, really similar here uh 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. So we're just gonna look at the equation for Omega to get the frequency. So Omega member is represented as two pi times the frequency or it's two pi divided by the period. Now, in our equation here, this two pi F uh really was equal to just two pi divided by 0.36. All right. So what happens here is if you compare these two things, really, what happens is that F must equal 1/2 0.36. So just by sort of comparing these things, uh really what happens is that your F is equal to just one over 0.36. And so if you actually just work this out with here, uh with the sense of being is 2.78 in Hertz. So that is the solution to that part of the problem. So we have 28 m and then a frequency of 2.78 Hertz. All right. So when it's, whenever it's written your constants with these two pies, it's basically just allowing you to calculate these things a lot faster. All right. For the last part of the problem, it's actually those are the simplest one, which directions of the wave travel in. Really, we just have to look at the sign uh uh between X and T. So remember if that for KX plus omega T, what this means is that you're going to be traveling to the left. So this plus sign means that the direction this wave travels in is left, right? So 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.

4

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Calculating Wave Speed from Wave Functions

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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 sign 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 this special equation square root of tension over mu. So I've got the square roots of my tension divided by mu. So let's take a look, I don't have the tension told nothing about the tension of 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 A K and Omega, 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 two expressions for these two variables. And then we're going to get something that includes Omega and K. Remember that Lambda is just related to the K. So K is equal to two pi over lambda. So if you rearrange for this, you're going to get Lambda is equal to two pi over K. So I'm going to write this equation, I'm going to write Lambda as two pi over K and now I'm going to have another expression for frequency. Remember that O Omega is related to two pi over frequency or sorry, two pi times the frequency, right? That's OMEGA. So if I can rearrange for frequency F equals Lambda over two pi or sorry Omega over two pi. So now I'm going to plug that back in here. So Omega over two pi. So now if you write it this way and you multiply these two things which you're going to see is that the two P will cancel. And what you end up with is you end up with just Omega over K 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 omega over K. So if you have the wave function, you can actually calculate omega or sorry V directly from these two variables. So my, my omega member is six and my wave number K is going to be 0.4. So if you plug this in, you're going to get 15 m 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 using, by you doing all the calculations, you would end up also with 15 m per second. All right. 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

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Example 2

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Welcome back everyone. So in this problem, we have a wave function that's given to us, we have Y of X and T equals six millimeters. And then we have some numbers here, right? We have five and then the the units for this radiance per millimeters and then 600 radiance 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 gonna be the K value, right? The thing goes in front of our X and this is gonna be the omega value. All right. 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 six and negative six millimeters. So I sort of want to like visualize what's going on here. If you take a string and sort of whip it up and down like this, you're gonna 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 six millimeters. So basically what is, what's gonna happen is that the particles on this uh on the string are gonna sort of bobble up and down between positive six and negative six millimeters over time. We're asked to find how long it takes for it to do that, right? So a given particle in the string that's gonna sort of bounce up and down between these two points that crest in the trough forever. That's always when it's gonna happen. And we're trying to figure out how long it takes for that to happen. So delta T, all right. So how do we solve for this? Well, the basic relationship between delta T and velocity and displacement is that V is equal to delta X over delta T, right? So displacement over time. So if I solve with this equation here, I can rearrange this and delta T equals delta X, the displacement divided by velocity. All right. So just very simply here, if I want to figure out time, I need to figure out the displacements, what's the distance of these particles are traveling divided by, what's the uh wave speed or what's the, the sort of trans velocity um of the particles that are bobbing up and down? All right. 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 six and minus six. In other words, it's basically just double the amplitude. So this is really just gonna be two times amplitude, which is just gonna be 12 millimeters. All right. Now, I can also convert this to 0.012. But that's basically where the displacement is. All right. 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. Um And basically the equation for the velocity is it's going to be the uh oh I'm sorry, it's gonna be the um the displacement velocity of the particles which is going to be omega divided by K. All right. So if we do do Omega divided by K, uh basically, I'm just gonna take this uh omega which is 600 the units for this are gonna be really important. So notice I have 600. This is gonna be radiance per second divided by and this is gonna be five and this is gonna be radiance per millimeter here. All right. So this is important because what happens is the radiant are gonna cancel when you do this. And what you end up getting is you end up getting 100 and 20 millimeters per second. So the units are gonna be really important here. So this is what my velocity is. It's 100 and 20 millimeters per second. All right. So then how do I figure it? Now? Delta T? Well, really, now I have everything I need to solve because now I have delta X and I have V. So I'm just gonna bring this down here and my delta T is just gonna be my delta X, the displacement which is 12 millimeters divided by, and this is gonna be 100 and 20 millimeters per second. So again, the units are important here because if you got something like meters and millimeters, you're gonna have to convert. But what we're gonna see here is that millimeters will cancel and you're just gonna be left with seconds. And really what happens is this is actually just gonna be 0.1 seconds. So in other words, it takes 1/10 of a second or 0.1 seconds or a particle in the string to sort of bounce up and down between positive six and negative six. That's will always happen. All right. So it's kind of a strange problem, but 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.

6

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Transverse Velocity of Waves

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Hey guys. So in the last couple of videos, we saw the equation for the wave function are a sine and a cosine functions. Well, in some problems, you're gonna be asked to calculate two different kinds of velocities. And the example that we're going to work out down here, we have a wave function that's given to us. And in part A, we're going to calculate the velocity of the wave which is really just a 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. So in this video, I'm gonna 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 just start off 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 you whip it up and down, like 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 gonna be the velocity of the particles that are moving up and down. All right. 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 cosine we have KX minus omega T. All right. So we have the of the wave, the propagation velocity is really just going to be given by this equation right here. Lambda over K. Remember that was our shortcut equation. So we have V equals sorry um Omega over K. So our mega value is just six RK value is 0.4. If you go ahead and work this out, you're going to get 15 m per second. So overall this whole entire wave is moving to the right at 15 m per second. What about the transverse velocity? That's we're going to calculate in part B. Well, now I'm going to show you the equations for this. Remember, the different textbooks will use different sort of equations for the wave function. Some will use a sign and then some will use a cosine. I'm actually going to give you both of the transverse velocity equations. Uh Basically based on which one that your textbook actually uses. So we're using a sine, your transverse velocity equation is also going to be a function of X and T. And it's gonna look like this. It's gonna look like plus or minus omega a times the cosine. This is gonna be a cosine here of KX plus or minus omega T. So we picked up another value of omega outside of our cosine equation. And we also have this plus or minus here. This plus or minus is actually related to this plus or minus, that's inside of our KX minus Omega plus or minus 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 gonna 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. All right. So um 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 minus plus omega A. Now we're going to have sine of KX plus or minus 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 plus 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 sign turns into a cosine and the cosine turns into a sign. So the, the equation of the trig function is always going to flip. All right. So that's just uh 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 VTXT, which one are we going to use? Well, we're starting off with a wave function that is of the form a cosine. So we have a cosine here of KX and we have minus 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, next, we have Omega and A. So our omega remember is just going to be six or A is going to be three. Now we have the sign and this is going to be RK, this is gonna be 0.4 that doesn't change minus six T, right. So what's, what's goes inside the parentheses 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 value. So VTXT uh So now we're gonna 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 gonna have 18 times the sign and then this, we're gonna have 0.4 times, 0.75 minus six times 0.2. So that's what happens when I plug in all the numbers. Remember to keep your calculator in radiance in which you'll get is negative 14.1 m per second. So this is the perpendicular velocity of a particle at that specific time. It's going downwards at 14.1 m per second. That's all that means. All right. So that's how we use the wave function. The last thing I want to mention here is that the propagation velocity 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 or 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. All right. 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, it was really just the amplitudes, 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 gonna be whatever is outside your sine and cosine. So that's just gonna be omega A. So the maximum transverse velocity you can have is Omega A. That's just the equation. So we want to calculate VT Max here. So this is just gonna be Omega A and we actually know both of those values. This is just gonna be six times three. So our maximum transverse velocity is 18 m per second. All right. So that's how you work out these problems. Let me know if you guys have any questions.