Okay, so in this question, we have kind of a weird set up. We have a string that is attached to a surface at the bottom, and the top end is kind of attached to this metal cart that is able to roll around. And so it's the top is able to move freely. Um, so it's a little bit of a weird set up, but that's honestly the most difficult part of the problem. The rest of this is pretty simple stuff that you've dealt with throughout this chapter. Um, so the mass of this spring or string sorry, um, is equal to 5.50 g, which of course, is 0.550 kg. Um, the length of the string is 86 0.0 centimeters, which is 0.860 m. And we're also told that the we can assume that there is a constant tension in this string. Um, of 1.3 years, mutants, um and in part a were s to find the, uh, speed of transverse waves in this string as it moves. And so this is pretty, pretty straightforward part. Um, we just need to remember that, um the speed of, uh, wave in a string this under tension gonna be equal to the square root of the tension for us two divided by mu, which is Mu, of course, is just equal to theme ass of the string, divided by its length so that our form that just becomes, ah, tee times l over and And if you look at the variables that were given straight away to begin the problem, they are m l and T. So we have everything we need to do to solve this part of the problem. So we just go ahead and plug in everything. So our calculation is the square root of 1.30 times 0.860 divided by 0.55 zero. And if you could calculate that out, you should get Ah, the speed of the waves is 14.3 m per second. So that's your answer her A. And then in part b um were asked to find the noted tow anti no distance, um, for the three simplest modes of vibration. So that means for the first three modes of vibration, we want to find the distance between a node and it's a Jason anti. So to do that, it would be helpful to draw the the first three months of vibration. So we're told in the problem that the, uh and uh, that's attached to the platform of the bottom, um is going to have a note and then because the top end that's attached to the car can vibrate, Really, that's going to have an anti note, um, when it has a standing wave. So if we draw that the way that we draw like sound this thing, we live, we die like sound waves and other, uh, staying waves that we've done with this book the first node is gonna have, or the first mode of vibration is gonna have a note at one end and a note at the other end of the string, which gives us a diagram like this. And so the distance from Ah, a no to any Jason anti note is, of course, just the whole length of the string. Um, so for your first one, I'm going to number these Just 123 The first one, um, were the note Anti no distance is just, um, And then for the second motor vibration, we need to add a set. We have to add another node in between there. Um, so are, uh, drawing is going to look like this. And that's not quite to scale, obviously. But we do know that from a no to an 80 note and from an anti note back to a node, all those distances are going to be the same. So we have a note going to an anti note. That's one that's one. And then from 19 0, back to note and then from a note to an 18 So we actually have three lengths issued. Drew this accurately, these would actually be all the same size. Um, but that means that the note to anti no distance, um, from here to here is going to be one third of the length of the two. So l over three, and then we just do this again for the third motor vibration. And so we just add another note in there before we get to the end of the tube. So look this and I'll draw one more time. So we have a no to an anti note here and, you know, to a node, No, to an anti note anti no to a node and then notes the anti No again. So we have five lengths now. Five distances of a note in 18 0, that means that the distance the Noto anti no distance is 1/5 of the tube, so l over five. So for the first, the three simplest, most vibrations the distant node to anti no distances are l l over three and l over five, which then we just need to calculate out. So l, of course is still 86 10 centimeters. Um, and then l over three is equal to 28 0.7 centimeters, and l over five is equal to you. 17 0.2 centimeters. And I feel like all of that stuff explanatory. You just take 86. Divide by three in divide by five. So these are our answers for part B. I don't feel like it needs much more explanation than is there. All right. And then in part C, we need to find the frequencies for each of these three modes of vibration. And this part is also fairly straightforward. We're just doing you know the same calculation three times. Um so of course F is going to be equal to V divided by Lambda. And Lambda is going to be, in this case four times each of the links that we had in part B. Um, because the distance from a note to its adjacent anti note is always 1/4 off the full wavelength. So we just take these three distances 86 centimeters, 28.7 centimeters and 17.2 centimeters, and we multiply them by four to get the way of links for each of the three modes of vibration. So, um, if you want to, you can also calculate it directly from the length of the tube or not that you don't like this string. Um, And if you do that, you will just say that you will say that, you know, for the first motor vibration, Um, the wave like this going to be four times l ah, for the second one is going to be four thirds times l. And for the third one, it will be or fifths times l. And that's what I did when I was calculating this out so that is how our write it. So for F one, uh, the frequency of the natural frequency, the first motor vibration. You'll have 14.3 for your speed and then in the denominator you have for l is equal to the wavelength. So you have four times 0.860 liters. If you calculate that you should get 4.14 hurts where the first motor vibration. Um, And throughout this problem, it's important that you carry all the decimals until you get a final answer. Because as we're building on our previous work so much, if you make a rounding error, they will compound as you go through your work, and it will make your answers in part C a little bit off. Um, if you you know, round if you don't carry out the best in Ulster. So, um, that is the frequency of the first ah, motor vibration. And for F two, we have 14.3 Still, for the speed of the wave. Um, and then this time, uh, E Linda will be four thirds l, which means that if that's in the denominator, then there's gonna be, uh, you may take the reciprocal multiplying By taking typical, you can say that it's times three in the numerator and then four zero point 860 in the denominator. So we enough with just an extra three in the numerator and then your frequency for the second motor vibration is going to be at 12 point or hurts. And for the third one last but not least, F three will be equal to 14.3. And now the Lambda is, um, 4/5 times l. So we end up with a five and numerator because Lambda isn't a denominator. So and then you end up with four times 0.860 in the denominator one more time. And so frequently that out, then your frequency for F three should be 20.7, and then we're all done. There you go.