welcome to the next section and traveling waves in this section. We're going to be considering what happens when we have a piece of rope or a piece of string with the way of traveling through it. And let's say we look specifically at some small section Delta X of the string, but we'd see there is that we have a differential force. So we have one force pulling this way and we have another force pulling this way. So on our piece of strength, Okay, when we consider the components of these two forces, we're gonna have X and Y components for each one. Now we suppose that the tension in the X direction that should be constant because it's not moving in the X direction. It's only moving in the Y direction because of our displacement now in the X direction. That means that we should have some tension here in some tension here and notice after the wave passed. This passes. This is exactly how excess held stills because we have some tension in the X and some tension. The X in the opposite direction. So we're just gonna call this the tension of the road so then we have attention in the Y direction for each except at different positions. This is that we'll call this position X and we call this position X plus Delta X Okay, s so we know the two forces acting on the piece of string. So what does Newton's second law look like? We know that Net force in the Y direction is going to be equal to T Y X plus Delta X minus. Now, I'm going to take care of the direction here, so I don't need to worry about it at all later. So I know this tension is pulling to the left. Okay, Um, I know that F Net is going to be mass times acceleration in the UAE, which is T y X plus Delta X still minus T Wyatt X. I don't have a mass for this piece of string, But if I knew the total mass and the total length, I could find the linear mass density, so I could then right, the masses mu times Delta x multiplied by I'm going to use d squared the displacement with respect to time. Okay, because my displacement function here is what's causing it to go up and down now, I should be a little more specific here. Noticed that I've used a nod sign for the derivative. That's because we know that displacement is a function of two variables in this case. Why? And, er sorry, x and t because we're gonna be at a position X and at the time t and it's gonna cause a displacement in the why So d of X t is going to be equal to a time sign of K X minus omega T plus fi. And if we wanted to worry about the direction, we could say, this is in the J hat direction. Okay, so there's our displacement function. We're going to take the second derivative with respect to time because it's a function of two variables. We need to use this curvy D here in place of our standard d DT that we use. Okay, so we've changed our operator. This is what's called a partial derivative. It just means we're taking the derivative with respect to one variable on a function that is has multiple independent variables inside. Okay, so what can How come they were you right? T y here to be a little more useful. We'll notice that t y over tea is just the slope of the rope here. Okay, well, the slope of the rope is also going to be the derivative of the displacement with respect to X. Okay, because this is literally the slope of the displacement functions. So at X, this would be equal to T Wyatt X over tea and X plus Delta X. We would have this equal to t y at X plus Delta X, divided by t. All right, so using that little bit, we can then find the T y X can be replaced by tee times d d d x at x n t y At X plus Delta X can be equal to tee times d d d x at X plus Delta X Okay, so we have our tensions here and we can plug in So we have new Delta X The squared big D d T squared has to be equal to I can pull a t out of each of these terms and then I'm left with de de the X at X plus Delta X minus D d d x at x. Okay, notice here If I divide over Delta X, then I have a term that looks like this. I have d d d x at X plus Delta X minus D d d x at X all divided by Dell text which is essentially the derivative equation. Remember that The definition of a derivative look, something like this. And we have the limit as h approaches zero. Okay, so if we make Delta X very, very small, this actually would become the second derivative of the displacement function with respect tax. So I could rewrite this as do you square d d t squared. Then we're going to group are constants t over mu times d squared the the X squared. So this is an interesting result. What is it telling us? It's telling us that the relationship between these two derivatives is going to be this ratio of constant. What if I take those derivatives I would find Do you square DDT squared? Remember that it's going to be on the function d of x t. It was a sign of K X minus omega t plus fi. So if I take the second derivative of this with respect to time uh, sign to cosine cosine to sign. I get a negative there and then I pull two negatives out on the derivative. So that's three negatives. So I end up with negative Omega Squared, a sign of K X minus omega T plus fi. And that's going to be equal to t over mu. Now they're the second derivative with respect to X. I'm going to get a negative k squared times a sign of K X minus omega t plus. If I Okay, well, a lot of stuff just cancels including the minus signs, and we're left with Omega squared over K squared equals t over mute. Well, Omega. Okay, that's familiar looking. That is the wave speed, which means we find that V is equal to the square root of the tension in the rope divided by its linear mass density. We've found the wave speed as a function of the medium that the wave is traveling to something that I alluded to in the last section. Um and it's absolutely correct here, and in fact, we can rewrite this equation here to generically describe what is happening to generically described what is happening in any traveling wave through a medium. Okay. Where the displacement function is telling you about the medium and the wave. Speed will also tell you about the medium. Okay, so these are two very important equations. Here we have the speed of a wave traveling through a string, and here we have our generic wave equation that we will look at over the course of the next few videos.

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## Video Transcript

welcome to the next section and traveling waves in this section. We're going to be considering what happens when we have a piece of rope or a piece of string with the way of traveling through it. And let's say we look specifically at some small section Delta X of the string, but we'd see there is that we have a differential force. So we have one force pulling this way and we have another force pulling this way. So on our piece of strength, Okay, when we consider the components of these two forces, we're gonna have X and Y components for each one. Now we suppose that the tension in the X direction that should be constant because it's not moving in the X direction. It's only moving in the Y direction because of our displacement now in the X direction. That means that we should have some tension here in some tension here and notice after the wave passed. This passes. This is exactly how excess held stills because we have some tension in the X and some tension. The X in the opposite direction. So we're just gonna call this the tension of the road so then we have attention in the Y direction for each except at different positions. This is that we'll call this position X and we call this position X plus Delta X Okay, s so we know the two forces acting on the piece of string. So what does Newton's second law look like? We know that Net force in the Y direction is going to be equal to T Y X plus Delta X minus. Now, I'm going to take care of the direction here, so I don't need to worry about it at all later. So I know this tension is pulling to the left. Okay, Um, I know that F Net is going to be mass times acceleration in the UAE, which is T y X plus Delta X still minus T Wyatt X. I don't have a mass for this piece of string, But if I knew the total mass and the total length, I could find the linear mass density, so I could then right, the masses mu times Delta x multiplied by I'm going to use d squared the displacement with respect to time. Okay, because my displacement function here is what's causing it to go up and down now, I should be a little more specific here. Noticed that I've used a nod sign for the derivative. That's because we know that displacement is a function of two variables in this case. Why? And, er sorry, x and t because we're gonna be at a position X and at the time t and it's gonna cause a displacement in the why So d of X t is going to be equal to a time sign of K X minus omega T plus fi. And if we wanted to worry about the direction, we could say, this is in the J hat direction. Okay, so there's our displacement function. We're going to take the second derivative with respect to time because it's a function of two variables. We need to use this curvy D here in place of our standard d DT that we use. Okay, so we've changed our operator. This is what's called a partial derivative. It just means we're taking the derivative with respect to one variable on a function that is has multiple independent variables inside. Okay, so what can How come they were you right? T y here to be a little more useful. We'll notice that t y over tea is just the slope of the rope here. Okay, well, the slope of the rope is also going to be the derivative of the displacement with respect to X. Okay, because this is literally the slope of the displacement functions. So at X, this would be equal to T Wyatt X over tea and X plus Delta X. We would have this equal to t y at X plus Delta X, divided by t. All right, so using that little bit, we can then find the T y X can be replaced by tee times d d d x at x n t y At X plus Delta X can be equal to tee times d d d x at X plus Delta X Okay, so we have our tensions here and we can plug in So we have new Delta X The squared big D d T squared has to be equal to I can pull a t out of each of these terms and then I'm left with de de the X at X plus Delta X minus D d d x at x. Okay, notice here If I divide over Delta X, then I have a term that looks like this. I have d d d x at X plus Delta X minus D d d x at X all divided by Dell text which is essentially the derivative equation. Remember that The definition of a derivative look, something like this. And we have the limit as h approaches zero. Okay, so if we make Delta X very, very small, this actually would become the second derivative of the displacement function with respect tax. So I could rewrite this as do you square d d t squared. Then we're going to group are constants t over mu times d squared the the X squared. So this is an interesting result. What is it telling us? It's telling us that the relationship between these two derivatives is going to be this ratio of constant. What if I take those derivatives I would find Do you square DDT squared? Remember that it's going to be on the function d of x t. It was a sign of K X minus omega t plus fi. So if I take the second derivative of this with respect to time uh, sign to cosine cosine to sign. I get a negative there and then I pull two negatives out on the derivative. So that's three negatives. So I end up with negative Omega Squared, a sign of K X minus omega T plus fi. And that's going to be equal to t over mu. Now they're the second derivative with respect to X. I'm going to get a negative k squared times a sign of K X minus omega t plus. If I Okay, well, a lot of stuff just cancels including the minus signs, and we're left with Omega squared over K squared equals t over mute. Well, Omega. Okay, that's familiar looking. That is the wave speed, which means we find that V is equal to the square root of the tension in the rope divided by its linear mass density. We've found the wave speed as a function of the medium that the wave is traveling to something that I alluded to in the last section. Um and it's absolutely correct here, and in fact, we can rewrite this equation here to generically describe what is happening to generically described what is happening in any traveling wave through a medium. Okay. Where the displacement function is telling you about the medium and the wave. Speed will also tell you about the medium. Okay, so these are two very important equations. Here we have the speed of a wave traveling through a string, and here we have our generic wave equation that we will look at over the course of the next few videos.

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