Okay, So in this question, where you're dealing with a yo yo, um and we're told that, you know, it is accelerating towards the ground. Um, someone you know, it starts playing with the audio, and it starts accelerate towards the ground with an acceleration of zero point 800 m per second squared. Um, and the time, the moment in time that we are concerned with is t equals 1.20 seconds. Um, just gonna repeat that. There we go. And we this is, you know, a problem in the waves chapter because, um, as the yo yo is rolling off the string, um, it rubs against the string and starts exciting transfers waves in the string as the length of the string increases. Um, and in part, a were supposed to show that, um, the rate of change in Lambda over time in the wavelength is 1.92 m per second. So what we're trying to show is that de Landau t t able to 1.92 m per second. Um, we're doing this specifically for the fundamental frequency of these transfers waves on the string. So, um, what? We can do here. The first thing that we should do that, you know, will be helpful is to figure out the relationship between, um the length of the string and time. So what is L As a function of time? And of course, this comes back to a kingdom attics equation. Um, and the one that's most pertinent here is one half a T squared, plus, uh, v t plus l initial. And so Ah, the thing is, our and we can save the nod as well. Um, but the thing is, the initial length of the string and the initial speed of the string are both zero. So the 2nd and 3rd turns dropout, um, so are l of tea becomes just one half a t squared and were given both the acceleration and the moment in time that we're concerned with. So, um, from there, we can take a look at how linda is related to l. So what is Lambda as a function of Oh, well, ah, the problem tells us that we can treat both ah, the end of the string at the person's hand as a note and the end of the string at the yo yo. As a note, which makes sense, those are both being held in place. So that means that our lambda is going to be equal to twice the length of the strength. So Lamda as a function of l is to l. Um, So now what we want to find is we want to find ah, Lambda or delenda DT. We have a function of, uh, Lambda in terms of El and we all function Ellington's t so we can write a function of lambda are a function of lemon in terms of l and L intensity. So we can write Let a function of Linda in terms of tea as, uh, a t squared because to L at Linda's to l two l times one half a. T or two times won't have 80 is just 80 square. Um, So our formula for Lambda over in terms of time is 80 squared. So if we take de lambda d t. And we just take the derivative of, uh, 80 squared, we'll end up with to a t, and then we want to evaluate this AT T equals 1.2 seconds. And so we can say that at the time that we care about, um, the rate of change over time, off the wavelength of these transverse waves equal to two times 0.800 times 1.20 And if you go ahead and plug that into your calculator, you will get that the Lambda DT is in fact, one point 92 m per second. So that's how you do part a, um, Part B is a yes or no question. Um, now we're dealing with the rate of change of the wavelength of the time but instead dealing with the fundamental frequency. Ah, and so the wavelength of the fund months over dealing with the wavelength of the second motor vibration So n equals two now. So, um, what we can do is we can bring back our function of L in terms of time that is still equal to one half a T squared. But our function of Lambda in terms of l is different now because and has changed. So in the second motor vibration, um, El Lambda is going to be equal to, um to l divided by n where n is now equal to two assistance and is equal to two. We can say that, uh, Lambda is just equal to help. So this is a change from the first part. Um, this part of the problem, um asks is the rate of change for the second harmonic the same as the rate of change. Um, in this part of problem, as if the rate of change for the second harmonic, it's the same as for the fundamental, and we can already see it. The answer is no. But we should go ahead and bring us to, um, its full conclusion. Um, and go ahead and say what the ah rate of change of the second harmonic is because it's not much more work at this point. So we have that lander is equal to El, which means that we to get a formula of Lambda in terms of time. It's just the same as l in terms of time. So Lamba is equal to one half a t squared, which means that if we do de lambda t t, then we'll have, um just a t for our our land, a derivative or our time derivative of Linda and A and T are still equal to zero point hate dear zero and 1.2. And if you evaluate that, you'll get that, um, for the second harmonic de lambda d T is equal instead to 0.96 m per second. And so our answer is very clearly, no, they're not the same. Um, And then as we move on for part C, we're basically being asked, Um, we're giving a slightly different situation where the mass of the ah yo yo is increased. Um, substantially. And it's increased in such a way that its mass distribution doesn't change. So, like its rotation doesn't change. And so the acceleration is still the same and were asked, um real this change the deal and a d t for the fundamental frequency. Um, and this question, ah, seems a little bit nebulous and weird. But it's actually, you know, you just need to think about what is our formula for de Lambda GT for the fundamental and that is to a T. So dealing DDT only depends on the acceleration and time, which means that it doesn't depend on mass, and so no, it will not change. Um, the the the Lambda DT will be the same as it is in part a even though we have the extra mass. So that's our answer for part C. And then Part D is very similar to part C. So here it's asking, you know, it's asking the same question a za part C. But this time we're dealing with the second harmonic instead of the fundamental. So it's asking Does deal end a DT change for the second harmonic? And once again we just look at our d lambda t t. But this time for the second harmonic wish we said was equal to 80. And once again, it only depends on acceleration and time. It does not depend on mass of changing the mass, but also not change the rate of change of the, uh, wavelength. And there we go. So we're all done, and that's what you do. This problem