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Trialfrequency (Hz)Wavelength (cm)10.2520.4330.6240.81.751.01.6TrialWave velocity (cm/s)1121.231.241.3451.61/ Are the velocities for the different trials similar or...

Question

Trialfrequency (Hz)Wavelength (cm)10.2520.4330.6240.81.751.01.6TrialWave velocity (cm/s)1121.231.241.3451.61/ Are the velocities for the different trials similar ordissimilar? Did you expect them to be? Why or Whynot?2/ Calculate the mean wavevelocity for low tension (from the above data.)3/ Set the "Tension" to "High"and repeat the experiment, using whatever values for the frequencythat you deem reasonable.Copy this table in the answer box, and complete it using yourmeasurem

Trial frequency (Hz) Wavelength (cm) 1 0.2 5 2 0.4 3 3 0.6 2 4 0.8 1.7 5 1.0 1.6 Trial Wave velocity (cm/s) 1 1 2 1.2 3 1.2 4 1.34 5 1.6 1/ Are the velocities for the different trials similar or dissimilar? Did you expect them to be? Why or Why not? 2/ Calculate the mean wave velocity for low tension (from the above data.) 3/ Set the "Tension" to "High" and repeat the experiment, using whatever values for the frequency that you deem reasonable. Copy this table in the answer box, and complete it using your measurements from the simulator and your calculations Calculate the mean wave velocity (for high tension) Trial frequency (Hz) Wavelength (cm) Wave Velocity (cm/s) 1 2 3 4 5 4/ Did the mean velocity change when we changed the tension? Did you expect it to? Why or why not? 5/ Suppose the mass of the string in the simulation is 400g, estimate the tension force for both settings (low and high) from the data you have collected above. (Show your work. Do not forget to convert the units to base SI units before calculating the tension.)



Answers

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass $M$ from the free end of the string, producing tension $M g$ in the string. The oscillator produces transverse waves of frequency $f$ on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities $\mu .$ You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary $M ;$ then you measure the node-to-node distance $d$ for each standing-wave pattern and obtain the following data: $$ \begin{array}{l|lllll} \text { String } & \text { A } & \text { A } & \text { B } & \text { B } & \text { C } \\ \hline \mu(\mathrm{g} / \mathrm{cm}) & 0.0260 & 0.0260 & 0.0374 & 0.0374 & 0.0482 \\ M(\mathrm{~g}) & 559 & 249 & 365 & 207 & 262 \\ d(\mathrm{~cm}) & 48.1 & 31.9 & 32.0 & 24.2 & 23.8 \end{array} $$ (a) Explain why you obtain only certain values of $d$. (b) Graph $\mu d^{2}($ in $\mathrm{kg} \cdot \mathrm{m})$ versus $M($ in $\mathrm{kg}) .$ Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straightline fit to the data to determine the frequency $f$ of the waves produced on the string by the oscillator. Take $g=9.80 \mathrm{~m} / \mathrm{s}^{2}$. (d) For string A $(\mu=0.0260 \mathrm{~g} / \mathrm{cm}),$ what value of $M$ (in grams) would be required to produce a standing wave with a node-to-node distance of $24.0 \mathrm{~cm}$ ? Use the value of $f$ that you calculated in part (c).

In our little a box here which is an oscillator which produces ah waves on this string. Ah, frequency f and the basic laces attached to a string which is then attached to a pulley here fixed to a police and hanging from that police a mess em and the distance between the oscillator and the poli has linked l and ah, basically we want Thio uh, given a no to no distance So we're given a note to know distance D and recall that the note to know distance here this is a note and this is a note and this is d then d is equal to land over to this will be useful. So we're given D and ah, we want to explain why we only get certain values of D Well, so if you recall for ah for waves on a string we have l is equal to end Lambda over too. Ah and that means that Ah lambda here, which is equal Thio to d is equal thio to l over end. So you canceled these out and that means d is gonna be equal to l over end and n here are integers this is the important part 123 dot that that. So the reason we only get to wear certain values for D is because Ellis fixed and and his for introduced. So this is not a continuous function. This is a discreet function because and his integers So that's why we only get certain values for B for for D And so in part, we were asked to graph and what I've done is I've used some plotting software and ah, I have plotted these values for you here for you to look at and I will bring these onto the screen right now. So if we look at the data that were given in the textbook here and we plot ah mu d squared here which is muse the mass per unit length ah versus M, the massing in from a pulley. Here we get something on. What's important to see is that it follows. It looks like it follows like a linear trend. Everyone explain why the data follows. A linear trend follows a straight line, and the reason for this is this follows. And that's because if you look at what Amud Square is, you d squared is going to be equal. Thio mu times lame Just over two squared So it's a mule and a squared over four and ah, that's going to be cool through our equation, he's equal to root f over a meal we can rewrite. This has f lambda squared over to V squared and through vehicles limbed f We can write that as we know what the tension is. Ah, the tension is going to be mg I can read This is mg lambda squared divided by two lamb two squared f squared We're here. We've used f attention is mg and we've used V is equal and f so these cancel and then what you get is you get mg over to f squared. So I'm gonna write this as one. Ah, all right, this is G over to f squared times M So this here g is a constant the frequency from our oscillators constant. So this whole thing here is a constant. This is a variable here em It's our independent variable and this is our dependent variable mu t squared and we see that this is just of the form Why is equal to M X plus B where be is going to be zero and our slope M is this g over to f squared? So that's why it's linear is because we look at beauty squared versus M it. It's a linear function with varying. It's a linear function with a slope of G over to F squared. So if we use, um, some data software to dio a best fit line here, what we find is that the best fit line has slope and this is programming and I can't program in front of you to do this fit. So I'm just gonna have to tell you what that slope is. And that slope is going to be zero point 00 1088 meters. Ah! And so based on this slope, we want to determine the frequency f and we can determine that Ah, like this. We know that this is going to have to be equal to G over to f squared. So that means f is going to be the square root of 9.8 meters per second squared over two time 0.108 a meters here and that's equal to 67 0.1 hurts. Great. So finally, um, living on part D here. Ah, What we want to dio is ah for a string with mass per unit length is your 0.0 to 6 grams per centimeter We want to find ah m given that D is equal to 24 centimeters finding Ingrams So this is easy. We know that Mu t squared is equal to our slope times m And for m is gonna be mu t squared. Do whatever our slope and we've done this year. Ah, that's simply going to be 0.0 to six grams per centimeter times 24 centimeters squared. And now let's be careful of their units so transformed from meters, two centimeters, that gives us a grand value of 137.64 grams. All right,

Discussion belongs to the web. And shown in the figure. We have assigned a subtle way traveling to the right as speed. We even having linear mass density, millions and frequency effort and wavelength lambda one. And this string is attached to another string to having mewTwo equals to three million. And we have to calculate the frequency F two Wavelength λ two and speed veto. Okay, So we know that for away the speed where we still we it is equal to T by mu underwrote. So from here we can write that we to develop by even for the same tension, it will be equals to mutual developed by me one. Under route. Ok. I'm sorry. Newman. By mutual. Okay, this will be given by mutual because we is university proportional to this value. So even by mutual. So we have this relation. So from here we get the V2 equals two million, divided by mutual, which is three million underwrote manipulated by everyone. So from here we get two equals two. Riven by under rule three. So this is the way we steal in the second string. Okay, Now we know that the relation between frequency wavelength and uh frequency wavelength and the way we speak it is equal to London replied by frequency. So since wave is a property of uh since the frequency is property of waves, so it will remain same. So we can write directly that F two will be equal to F one. Okay, It will not change. So now if frequencies same then wavelength and uh we we speed both are related to this and we two is equal to this relation, then we can say that lambda to this will be also have this relationship. Lambda. Lambda two will be equals two lambda one back under three. So this become the relation for the lambda too.

So this is one of the problems that students love pretty much across the board because we don't actually use any numbers were just solving for this problem generically. Now, part is, just asked me to write the equation that relates the number of anti nodes, the length of a string, the tension in the string and the density of that string, which is one of the equations from the book that we were looking at. So for part, a that burst equation that we're looking at is that the frequency in this string is equal to the number of anti nodes divided by two times the length of the string, Um, multiplied by the square root of the tension in the string, divided by the density of that string. Okay, and that's all a is looking at now be is essentially asking what would happen? What would have to happen? Um, what would have to happen if the length of the string is double what would happen to the frequency? So when I'm looking at this year, um, if the length of the string is doubled, so are messing around with this part of the equation right here. How does it affect the frequency? Now, this is just a question of direct versus indirect relationships if its director in verse. Um, because the L. A. Is on the bottom of this fraction. It is an inverse relationship, meaning as the length of the string gets bigger, that frequency is going to get smaller, and any time we're checking for direct or inverse relationships, what really matters is if the factor that's being changed is raised to a power or square rooted now, L. A is not, which means if we double the length, the frequency then experiences the same factor but the opposite function. So instead of multiplying it by two, we would divide it by two as well. So if you double the length of a string, you will, um, if you double the length of a string, you will cut its frequency. And half. Now, see is where things start to get a bit interesting because C is asking us in terms of all this generic stuff, what would the tension have to be changed to in the string for the number of anti nodes to be one greater than it is if we keep the length of the string and the frequency of the string The same. So what that means is our frequency is staying the same. But instead of just and we have n plus one cause we're trying to increase it, the length of the string is the same. Um, the tension in our string is changing. So instead of ta, I'm gonna call that tension be for our new tension, and the material the string is made out of is staying the same as well. Now our next step is gonna be taking equation one up here that we solved and plugging it into equation, too, to get a new equation that kind of combined some of these things that we know and what I'm actually gonna do is I'm going to take equation one and use it to divide what's going on in equation too. So I'm going to divide equation to buy equation one. Now, the reason I'm doing this is because it really quickly shows what drops out because we have FAA on both of these. So they cancel out, and that leaves us with a one gonna write. Our final equation over here is equal to, um, n plus one stays there, but the to l. A gets dropped out on both sides because the length of the string isn't changing, It doesn't play a role in what the final account is. We have n plus one divided by N um, Same thing here. The T B and the TA are different, so I can't simplify them. But we do have this density on both sides of the square root of that density cancels out and we're left with times the square root of TB over t A. Now our goal here is to get t b by itself. So we need to do a little bit of rearranging. We can multiply by an and divide by n plus one to bring that over to the other side. Then we're going to square both sides to get rid of that square root and multiply by the initial tension. A. And that is what the tension in B ends up being. So there is our equation for any situation where we just want to add on one additional anti node to the problem. My clean up our workspace here a little bit because we're about to hit another fun part of this problem. Um, de is asking kind of the biggest question that we have here, which is saying, um, if the frequency has tripled and the length of the string is halved, so it's write down what we are looking for here. We know the frequency is being tripled and the length of the string is being cut in half. By what factors should the tension be changed so that twice as many anti nodes are produced? So we want to double the number of anti nodes as well. And the question here becomes what needs to happen to that, um, tension to make this happen. So when I write out my equation again because we want to see what we actually care about, how we get where we're going, we have f A equals and over to l times the square root of tea over you. So we are changing the frequency, the number of anti nodes and the length and the peace were looking at for How does it affect it is the tension is right here now because the tension is what our outcome is going to be. We want to rearrange this equation to have tea all by itself on one side. So I'm gonna multiply by two l to get to l A times the frequency, then divide by N to bring that over. And I'm going to square everything here as well to get rid of that square root and then multiply by that density function, and that is what our tension is going to be here. Now, let me tidy this up and rewrite it. So the tension is four times l squared, um, times F squared times mu divided by n squared. Now, if I'm looking at what's going on here, everything on the top here is directly related to the tension. Everything on the bottom here is inversely related to the tension in the string. So, um, from what we know, the frequency is being multiplied by three. And because the frequency is squared, we have a direct relationship of times three squared. So times three squared is what's going on with the frequency. Our next piece of information is that the length is being divided by two and the length is squirt as well. So it's divided by two squared, and then the number of nodes is being doubled because it's on the bottom, it's inversely related. Instead of multiplying it by two, we're going to divide it by two squared as well. Then we put all those terms together, we get nine divided by 16 and that is what will have to happen to the tension to make all three of those things come true for us.

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


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