Everyone, welcome back. So in the last couple of videos, I showed you the basics of standing waves, and I showed you what sort of what they look like on a graph. So imagine that you had a string set up between these two supports, and you had some kind of a frequency to set up a standing wave. It would look like this. You'd see a wave kind of like bouncing up and down. And then if you increase the frequency, you would end up with another standing wave with 2 loops where it would kind of look like this, and that would be a certain frequency. And then n=3, there'd be 3 loops and so on and so forth. And in those types of problems, we were actually already given what the fundamental frequencies are. Well, in this video, I'm gonna show you the equations for actually those transverse standing waves. How do we actually find out what frequency and wavelength and all those things are when they're not given to us? We're gonna see actually a couple of examples and a couple of equations. There's actually really only one equation that you have to learn, and then we'll do it a couple of examples together. Let's get started. Alright. So just remember that standing waves can only exist for special harmonic frequencies, and we have special values for the wavelength. Alright? I'm just gonna go ahead and actually give you these equations. The first frequency, which is n=1, remember that's called the fundamental frequency, is actually just given by v=v÷2l. We can actually figure out this out just by looking at the properties of the string and the wave itself. Alright? So, remember that the harmonic frequencies are just if we took fone1 and we just multiplied it by some integer n, where n could equal 1 or 2 or 3, depending on how many loops we were looking at. Right? So in other words, we're just going to stick an n in front of that equation. And so I basically could just take this equation and generalize it. This is nv÷2l. All right? Now we can take this equation and figure out what the wavelength is, and what you're actually gonna see is that it's 2l÷n. You can actually really quickly see this because remember that the relationship between speed, frequency, and wavelength is v=λf. So if you actually use this equation to solve for λ, what you're going to see is that the v will cancel out, and all you're left with is that this sort of, the equation is flipped from the fundamental frequency. All right? So these are really just the 3 equations you'll need to go ahead and solve problems. And if you actually look at it, this is the only one you really need to remember because from this one, you can actually figure out what this is and also what this is. So let's go ahead and just jump right into our first problem over here. We're going to take a look at this example where we have a 1.5 meter long string that's stretched between these two supports, So l=1.5. And we know what the speed of these waves is going to be. We know what the speed of waves is going to be 48 meters per second. So in this first problem here, we're gonna figure out what is the wavelength and frequency of the fundamental tone. How do we do that? Well, basically, the fundamental tone here is gonna be when n=1. So when n=1, we're just trying to figure out the λ, which is gonna be 1 over here. And We can actually just use our equation for now. Right? Or now for this. So this is just gonna be 2l÷n, in which n is just equal to 1. So what's the 2l? This is just gonna be 2 times the length of the string, 2×1.5, and you get 3 meters. Alright? So that's the first answer. Now you might be thinking of this number, and you might be wondering, how can the wavelength be 3 meters when the length of the string is only 1.5? And remember what the fundamental frequency looks like. That's sort of the most basic type of frequency. We can only set up a standing wave with one loop. Alright? And so basically, what's going to happen is that the loop is kind of just going to bob up and down like this. But remember, this sort of up pattern like this is only half of what a full wavelength is. A full wavelength goes up and down and up again. Alright? So this only represents one half of that wave. So that's why the full wavelength of this wave over here is actually 3 meters even though the string is only 1.5. Alright? So that actually perfectly makes sense because if you if you continued on, that would be a full wavelength. Alright. So that's 3 meters. So now let's take a look at f1, the fundamental frequency. You can actually figure out a couple of different ways. You could use this equation over here, or we could just stick to the 2 equations that directly give us f1. So this is actually just going to be v÷2l, and which this this is going to be 48 divided by 2 times 1.5. What What you're gonna get here is that this is 16 hertz. Alright? That's gonna be 16 hertz. So now let's take a look at the next equation over here, the next problem. The next problem says we're gonna take a look at the wavelength and frequency of the first overtone. What does that mean? Well, it turns out that there's actually 2 special words that will actually directly tell you what the value of n is. We've already taken a look at harmonic. Right? Harmonic just means basically, that's what the n value is. For example, the first harmonic is where n=1. So it's basically just telling you n=1. 2nd harmonic is when n=2. 3rd harmonic equals n=3, so on and so forth. So whatever the harmonic is, that's what n is. But the word overtone is something you might see, and it's kinda complicated or it's kinda just tricky because, really, what that means is an overtone is n tones over the fundamental frequency. So, for example, the second harmonic is when n=2, but that's also the first overtone because it's the first tone over the fundamental frequency. So n=2 is the second harmonic, but it's the first over tone. n=3 is the 3rd harmonic, but it's the second overtone, and so on and so forth. So when they're asking for the first overtone, really, what they're actually cluing you in on is that n=2. Alright? So we're just going to solve these exact same, same values, λ and f. Well, now we're just going to look at when n=2. Alright? This is going to be λ2. And again, I can just use this formula over here. This is going to be 2l÷2. In other words, that will cancel out and all we'll just be left with is l, and this is going to be 1.5 meters. So that's the wavelength here. Why does that make sense? Because remember, what n=2 is going to look like is it's gonna look like sort of a wave. It's gonna look like this. So it's gonna look it's gonna look like an up and down, which is gonna be a complete wave, and that's why the wavelength is equal to the length of the string. So that's going to be 1.5. And then finally, over here, we can see that f2 is just going to be remember, we can use nv÷2l, which case we're going to have 2×48÷2×1.5. We also could have just used n×fone because we already know what the fundamental frequency is. Either way, what you're actually gonna get here for an answer is that this is equal to 32 Hertz. Alright? So those are your answers for the, wavelength and frequencies of the fundamental and also the first overtone. Let me know if this makes sense, and thanks for watching. I'll see you in the next one.
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Standing Waves
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