Welcome to our second example video considering waves on strings. In this video, we're going to consider a string that is tied to a mess that is hanging. And we're going to examine how waves move on this string up here so that we can determine the mass over here. How are we going to do that? Well, first of all, notice that the mass we were to draw the free body diagram the masses What is actually giving us tension in our string? Because we find that if there's zero acceleration, f t minus M g equals zero. Therefore, the tension in the string is equal to mass times G. Well, we know that speed is related to tension like this. So if we give ourselves a few more numbers here, let's say that it has a length of 1 m up at the top. Notice that as theory wave comes to the pulley, it will reflect off it and come back and forth. So I won't actually care about all this segment down here. So we have 1 m long, and then we could also say that over the course of this meter that the rope has a Mass. That's very small. Will say it's just 1 g now. The reason I say it's 1 g is because we really want the mass of the string to not matter. Okay, so if we make that 1 g and assume that M is much larger than it, then we don't have to worry about it when drawing are free body diagram. Because even this is a half a gram, then it's not going to significantly influence the force of tension anywhere. Okay, so we have 1 g. We could then say that we pluck it and it goes back and forth 10 times over the course of six seconds. What do we do then? Well, we first of all, find speed. Wave speed is going to be 10 times. We have 2 m down and back over six seconds. We have mu is equal to 1 g that 0.1 kg divided by the length, which is 1 m. So there's our linear mass density, and then f t is mg. So when we saw for F T, it's V squared times mu, we can plug in m g. Do not confuse the mass that is hanging for the mass of the string We end up with M equals V squared mu, divided by G or in other words, up here. If we reduce this, we have 20 divided by six, so that could be 10 thirds meters per second. So we have 13th meters per second squared. I'm Sam U, which is 0.1 kg divided by 1 m, all divided by 9.8 m per second squared. Checking the units, then we have meters squared. We have 2 m here it will cancel and then we have second squared and second squared. Here they cancel. We're left with just kilograms, which is exactly what we want.

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## Video Transcript

Welcome to our second example video considering waves on strings. In this video, we're going to consider a string that is tied to a mess that is hanging. And we're going to examine how waves move on this string up here so that we can determine the mass over here. How are we going to do that? Well, first of all, notice that the mass we were to draw the free body diagram the masses What is actually giving us tension in our string? Because we find that if there's zero acceleration, f t minus M g equals zero. Therefore, the tension in the string is equal to mass times G. Well, we know that speed is related to tension like this. So if we give ourselves a few more numbers here, let's say that it has a length of 1 m up at the top. Notice that as theory wave comes to the pulley, it will reflect off it and come back and forth. So I won't actually care about all this segment down here. So we have 1 m long, and then we could also say that over the course of this meter that the rope has a Mass. That's very small. Will say it's just 1 g now. The reason I say it's 1 g is because we really want the mass of the string to not matter. Okay, so if we make that 1 g and assume that M is much larger than it, then we don't have to worry about it when drawing are free body diagram. Because even this is a half a gram, then it's not going to significantly influence the force of tension anywhere. Okay, so we have 1 g. We could then say that we pluck it and it goes back and forth 10 times over the course of six seconds. What do we do then? Well, we first of all, find speed. Wave speed is going to be 10 times. We have 2 m down and back over six seconds. We have mu is equal to 1 g that 0.1 kg divided by the length, which is 1 m. So there's our linear mass density, and then f t is mg. So when we saw for F T, it's V squared times mu, we can plug in m g. Do not confuse the mass that is hanging for the mass of the string We end up with M equals V squared mu, divided by G or in other words, up here. If we reduce this, we have 20 divided by six, so that could be 10 thirds meters per second. So we have 13th meters per second squared. I'm Sam U, which is 0.1 kg divided by 1 m, all divided by 9.8 m per second squared. Checking the units, then we have meters squared. We have 2 m here it will cancel and then we have second squared and second squared. Here they cancel. We're left with just kilograms, which is exactly what we want.

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