Velocity of Transverse Waves (Strings)

Hey, guys, in this video, we're gonna talk about waves on strings. Okay? If you were to fix a wave at one end against something like a wall, grab the other end and just whip it up and down, you would see that you produce a bunch of waves traveling down the length of that string. All right, so this is what we want to talk about. Now, as we've said before, a wave on a string is a very common example of a trans verse wave. One who's oscillations are perpendicular to its propagation. Okay, we explain why this was trance first before, but if you whip this string up and down, you're gonna produce these wave pulses that travel towards the wall. Okay, These waves that travel towards the wall now a wave on a string is no different, really than any other wave. In a lot of respects, it has the same mathematical description. It has the same characteristics on amplitude of propagation, speed, wavelength, frequency, etcetera. But there are a couple things that are unique to each wave. The specific energy carried by that wave is going to be unique to the wave and the specific propagation. Speed of that wave is gonna be unique to the wave. Remember that propagation speed depends upon two things. The type of wave and the characteristics of the medium that the wave is in. Okay, so what we're gonna focus ourselves on in this video is the speed of this wave. Okay? And of course, we're talking about the propagation speed. All right. The propagation speed of the wave on a string is given by the following equation. Spirit of tea over mu. Okay, where t is the tension in the string and mu? Is the mass per unit length of that strength? Okay, let's do a quick example. Ah, 1.2 m string as attention of 100 Newtons. If it has a massive 5 kg, what is the frequency of a wave with a 12 centimeter wavelength on this street? Okay, well, we know we want to find the frequency of the wave, okay? And we're told the wavelength. Since we are dealing with both frequency and wavelength, we're probably going to be using the equation. V equals Lambda F. And we can easily solve for F by dividing Lambda over. So the only question is What's the the speed on the string? Well, we know that for waves on a string, the speed is just the tension divided by the mass per unit length. Okay, we know the tension. We know the length and we know the mass. So we know everything we need to solve this problem. This is gonna be 100 Newtons of tension divided by the mass per unit length. So it's 5 kg and mass and it's 1.2 m in length. So that whole thing is going to be about 4.9 meters per second. Okay, This means that the frequency is 012 m. It was given in centimeters, but I need to convert 2 m divided by 4.9 m per second, which equals hurts. All right. No big deal. That wraps up our discussion on our introduction. Toe waves on the strength. Thanks for watching guys

Hey, guys, let's do a quick example. An easy way to find the mass of a string is to produce a wave on it. A five centimeter string off unknown mass produces a wave with a centimeter wavelength that has a 70 hertz frequency. When it's tension is 10 Newtons Given this, what must the mass of the spring be? Sorry. The string. Be okay? Well, we're the only equation that we have so far that deals with the mass. On a string is the speed equation. Speed of waves on a string are given by the spirit of the tension divided by the mass per unit length. Okay, this is the tension divided by the mass per unit length. What we want is toe isolate that mass. Okay, so we can just rewrite This is the square root. Sorry. This squared of the speed is t over em. Over. L take em over. Ellen, multiply into the numerator. Take V squared divided into the denominator. This is M over. L is t over v squared And then lastly take l multiply it into the numerator and we have em. Is t l over b squared. So what? We need to know is V squared. Well, the relationship for speed off any wave is that it's equal to the wavelength times the frequency. We know that it's a 12 centimeter wavelength and a 70 hertz frequency. So this is gonna be a 0.12 You wanna in meters times 70 which is 84 m per second. Now we console for the mass, so the mass is gonna be t l over V squared. The tension is 10 Newtons. The length is five centimeters or m. And the speed we found was 84 and we that squared. And that comes out to 0007 kg. Or if I reduce that by a factor of 1000 that's 7 g. Either of these answers are perfectly fine. Alright, guys, that wraps up this problem. Thanks for watching

Hey, guys, let's do a quick example. A string is hung vertically from an anchor point. If you were to grasp the string at the bottom and shake it to produce a wave traveling up the string, what happens to the speed of the wave as it moves up the string? Okay, so we have some anger point here, a rope hanging. We grab that rope and we produce a wave on it by whipping it. So this is traveling upwards. What happens to the speed of that wave as it travels up the rope? Okay. A lot of people are probably gonna want to say it decreases right off the bat, but don't associate this with freefall motion. We know the speed of a wave on a string equals attention. Divided by the mass screen at length. The mass screen length is not changing, but the tension and the rope is okay. Tension and rope is always increasing the higher up the rope you get. Okay. The simple reason is this Imagine the rope was built up, was broken up into a bunch of chains. Chain links, right. How much tension is in the chain link? Second from the bottom. Well, all that's holding all that it's supporting is one little mg, which I'll say is the weight of a single link. So the tension is just one little mg. But what if we go? One, two, three, four, five up? Then there's four chains below it. So what's the weight? It's supporting four little MGI, so it's tension is four little MGI. Let's go. We have toe to the top of this chain link. It has one, two, three, four, five, six, seven, eight, nine. So the way that it's supporting is nine little mg. So the tension is nine little mg. So clearly the tension in the rope increases the higher you go up because there's mawr rope being supported, the higher you go up. Okay. So as the tension goes up, the speed goes up, so the speed of the wave actually increases as it goes up the rope. Okay, This is contrary to our initial sort of visceral guests that it should actually decrease on its way up like objects in free fall. Alright, guys, that wraps up this question. Thanks for watching

Hey, guys. In this video, we want to talk about the energy carried by waves on a string. All right, let's get to it. Waves on a string carry both kinetic and potential energy. They obviously carry kinetic because as a wave which has sorry as a wave on a string that string has masked as it propagates it has a velocity and has mass. So it has kinetic energy. But when the string is just flat static, no waves on it when you whip it, that whip stretches the string a little bit to produce a wave. That wave has some stretching this in that in that string. So it also carries some potential energy with it as well. Okay, The kinetic energy is due to the motion off the string. Sorry. Wrong color. Kinetic energy is due to the motion of the string and the potential energy due. The potential energy is due to stretching of the stream. Okay, This energy depends upon three things. It depends upon the frequency of the wave. It depends upon the amplitude of the wave and it depends upon the wavelength of the waves. So pretty much the three main characteristics of a wave. All right. Now it turns out that the kinetic energy and the potential energy are the same for a wave on a string. Okay, They are both given by the same equation given right here in our orange box, where the variables mu, Omega A and Lambda are the same as they have been so far. The total energy per unit wavelength is just gonna be the some of the kinetic energy and the potential energy per unit wavelength. What I mean by per unit wavelength is this is the energy carried by a single wavelength. Okay, the energy per unit wavelength is just double this, right? It's this plus itself. So it's one half mu Omega squared, a squared lambda. All right, let's do a quick example. A string 5 m in length carries a wave along its length. If the tension is 50 Newtons and the mass per unit length is kg per meter. How much kinetic energy does the string carry for away with the frequency of 50 hertz and an amplitude of one centimeter? So we want to find first the energy per wavelength. Then we'll find the total energy this is one half mu Omega squared a squared lambda. So we know Mu Mu was given to us the mass per unit length. Okay, the frequency is given to us so we can find the angular frequency pretty easily. The amplitude is given to us. The only thing that's not given to us is the wavelength. The wavelength. We do have to calculate using our wave equation. This means that the wavelength is f over V. The question is, what's the speed of the wave? Well, since this is a wave on a string with speed is just the tension divided by the mass premium length square rooted. The tension, as we're told, is 50 Newtons. The mass print length is 500. and this whole thing equals 10. Sorry, 100 meters per second. The answer in the square root is 10,000 square of 10. 0 is 100. So this means that the wavelength is hertz, right? The frequency divided by 100 m per second, which is Oh, sorry. This is backwards. I messed up on this equation right here. This equation is V divided by f not F divided by V. So This is 100 divided by 50 which is 2 m. Okay, now we have everything we need to know to solve. For the energy per unit wavelength, that's one half 0.5 was the mass per unit length. The angular frequency is just two pi times 50 hertz, right? Was the frequency squared times the amplitude 01 m squared. Everything needs to be an s I units times the wavelength which was m. And this is 0.49 jewels. Okay, but this is not what the question asked. The question asked how much energy is carried by the string. Now this is the energy for a single wavelength. So the question is, how many wavelengths does the string have? Well, the string is 5 m in length and if the wavelength is 2 m thin, the whole string carries 2.5 wavelengths. So the total energy he's going to be too 0.5 Sorry. 2.5 times the energy per unit wavelength, which is 1.22 jewels. Okay, now something else you could ask is how much power is carried by the string. Well, we know the energy per unit wavelength. And how long does it take for a wavelength to pass just the period. So if I do thought, if I divide the energy period of wavelength by the period, I get the power, okay. And what we have is we have this last Lambda divided by the period, which gives us the velocity. And this is the power period. This isn't the power per unit wavelength. This is the total power. If you, for instance, asked for two times the energy or the energy per two wavelengths, you'd have to divide it by two periods. So that tude canceled from the numerator and the denominator. This power is not dependent on number of wavelength. This is just the power. All right, So a 0.7 m string supports a wave of 0.5 m amplitude. If the mass of the string is 0.5 kg and the tension in the string is Newton's, what must the frequency of the wave be to carry 100 watts of power? Since we're dealing with power, we'll start there. Windows is one half mu Omega squared, eight squared times the okay. We know the tension in the string. We know the mass per unit. We know the mass of the string, and we know the length of the string, so we can easily find the speed right. The speed is just the square root of tea divided by mu, which is the square root of 15. Newton's divided by 05 kg, divided by seven. Right, That's the mass per unit length, and that is 14 meters per second. All right, now what we wanna know is the frequency, right? What must the frequency be? Toe have 100 watts of power. So we need to rearrange our power equation to solve for frequency. The frequency in this equation is actually the angular frequency, but we can always convert that toe linear frequency later. So this becomes omega squared is to pee. I multiply the two up the two from the denominator up, and everything else goes into the denominator mu. Yeah, a squared Z. Okay, so omega is the square root of two p over mu, a squared V, which is gonna be two times the powers 100 watts. The mass per unit length, as I said was the mass 5 kg divided by the length. 7 m, the amplitude. We're told this 70.5 m and the speed we found as 14.5. Okay, so this whole thing becomes in verse seconds. The last thing we need to do is we need to take this number, which is angular frequency and convert it into linear frequency. Don't forget that linear frequency is angular frequency divided by two pi. So this is 879 divided by two pi, which is about 280 hertz. All right, that wraps up our discussion into the energy carried by waves on a string. Thanks for watching guys.

Hey, guys, I hope you're able to solve this problem on your own. If not, here is a little bit of help. Waves air produced on a string by a paddle that can move up and down, shaking the friend of a string to produce a five centimeter amplitude wave. As shown in the figure below the string is a mass per unit length of half a kilogram per meter and attention of 100 Newtons. If the maximum power the paddle can deliver is 50 watts. What is the greatest frequency? The Sorry. What is the greatest frequency wave the string can support? Okay, so remember that power carried by a wave on a string is one half times mu omega squared a squared V. So if we want to know frequency, we'll solve for omega squared, multiply the two up to the numerator. Everything else goes to the nominator and then take the square root of both sides. Okay? We're told the maximum power we're told the mass prima length, we're told the amplitude. Okay, All we're missing is the speed of this wave. Now, this is a wave on a string, so the speed is just gonna be the square root of the tension divided by the mass period at length. We're told that the tensions 100 Newtons. So this is 100 divided by half a kilogram per meter and that speed is meters per second. Now we know everything we need to solve for that angular frequency. This is two times the power which we said was 50 watts divided by the mass screen at length, which is half a kilogram per meter. The amplitude was five centimeters or 05 m squared in the speed which was 14 m per second. And this whole thing is going to be 76 inverse seconds. But we're not done. Remember, we're looking for the greatest frequency that means linear frequencies. So we need one more step which says that the linear frequency is Omega divided by two pi which is 76 in for seconds over two pi, which is going to be 12 hurts. All right. And that wraps up this problem. Thanks for watching guys