What's up, everyone? So let's see if we can work this example problem out together. We're gonna put together a lot of information that we've seen from waves and standing waves in this problem. Let's break it down. So we have a string that's fixed at both ends. It's 8 meters long and has a mass of 0.2 kilograms. So in other words, I'm told that the length is 8 and \( m \) is 0.2. What I'm also told is that this is a string, and it's subjected to a tension of 100 newtons. Now we've seen tension before in these types of problems with waves and strings, and that's the variable \( F_t \). So in other words, the force of tension on the string is 100 newtons. So the first thing we want to figure out in part a is the speed of waves on the string. That's just the wave speed. So let's put these things together and see if we can look at part a over here. Now remember, we have a couple of equations for velocity, but for strings, particularly, we had one equation, which is the square root of tension over \( \mu \). That's what we're gonna take a look at here because that applies to strings only, in this case, in transverse waves. So let's take a look at this \( v \) equation. Now remember, the \( \mu \) is the linear mass density. And just so for your recollection, it's the linear mass density is really just the mass divided by the length. That's what that \( \mu \) value becomes. Alright? If I just go ahead and plug all this stuff into my calculator, this is really just gonna be \( 100 \) divided by, the mass, which is \( 0.2 \) divided by the length, which is \( 8 \). If you plug all that stuff into your calculator, what you should get is a velocity of \( 63.2 \) meters per second. Alright? So that's the first part. So figuring out just the speed of waves that, sort of behave or as they move along on this on the string here. Alright?
Let's move on now to the second part. The second part asks us to find what is the longest possible wavelength for a standing wave. So what does that mean? What's the longest possible wavelength? Well, remember that the first loop, the \( n = 1 \), is actually basically like it's only just one bump up like this. Right? Whereas the \( n = 2 \), that's where you actually have 2 loops. And remember, that's gonna look like a standing wave like this. It's gonna look something like this, and then it's gonna basically just fold back over itself. So you're gonna have 2 loops like this. The \( n = 3 \) is gonna look like 3 loops, so it's gonna look a bit like this \( 1, 2, \) and then \( 3 \). Alright? Which one is the longest possible wavelength? Well, as the ends get bigger and bigger, you have more and more loops, which means that the wavelengths actually get shorter. Notice how the wavelength for \( n = 1 \) involves one complete cycle; it is actually only \( 1/2 \) of the wavelength, so to calculate the wavelength for \( n = 1 \), you're actually not gonna use an equation. You'll actually just see that it actually has to be double the length of the string. So, \( \lambda_1 \) is going to be equal to \( 16 \) meters. This will always be true. The longest possible wavelength of a standing wave will always correspond to \( n = 1 \) always.
Now, we're going to calculate the frequency of that wave. There are a couple of different ways to do this, but one of the easiest ways is to use the equation \( f_1 = v / 2l \). Let's use our wave speed, which we already know as \( 63.2 \) meters per second, calculated earlier in the problem, and divide it by \( 2 \times 8 \), the length of the string. The fundamental frequency you'll calculate should be \( 7.9 \) hertz. Alright? So those are the 3 parts of this problem: figuring out the wave speed, figuring out the longest possible wavelength, and what the corresponding frequency will be, and that actually happens to be the fundamental frequency. Let me know if that made sense. Thanks for watching, and I'll see you in the next video.
