A fellow student with a mathematical bent tells you that the w...

Question

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is $y (x, t) = (2.30 mm) \cos [(16.98 r a d / m) x + (742 r a d / s) t]$ . Being more practical, you measure the rope to have a length of $1.35 m$ and a mass of $0.00338 kg$ . You are then asked to determine the following: (f) tension in the rope; (g) average power transmitted by the wave.

Accepted Answer

Hey, everyone in this problem, we have a string. OK. With mass 5 g in a length, 1.25 m. And it's tightly attached to a mechanical vibrator at one end and to a fixed support at the other end. OK. And the vibrator is gonna tra generate traveling waves in the string. OK. Which can be modeled by the function Y is equal to six millimeters times sine 1.5 pi radium per meter, X plus 90 pi radium per second T. And we're asked to find part one, the tension in the string and two, the average power transmitted by the traveling wave. OK. Let's start with part one. And we're asked to find the tension in the string. OK. And let's recall that we can relate the tension in a string to the speed of the wave through the following V is equal to the square root of T over mu, where T is a tension V is the speed of the wave and mu is the linear density. OK. So squaring both sides V squared is equal to T over mu. And so we can write the tension T is equal to V squared times mu. OK. And so again, the tension. T that's what we're looking for. We need to find V and we need to find mu hm. All right. So how can we find the, how can we find me? Let's start with the V. OK. We have an equation. General equation for a wave Y is equal to a sine KX plus omega T we want to find the speed V. OK. And V we know can be written as F times lambda. OK? Where F is the frequency and lambda is the wavelength. OK. Well, F and lambda aren't directly in Y. OK. We have K, we have Omega but they are related to the K and Omega value. Now let's recall how they're related. We know that Kay is equal to two pi over Lambda. OK. So binding K will allow us to find the wavelength lambda but we also know that Omega is equal to two pi F. OK. So finding Omega is gonna help us find the frequency F OK. Once we have F and lambda, we can find V. All right. So let's compare the equations. OK. The general equation, we have this K term inside the sign multiplied by the X. OK. So in our equation, it's gonna be this term here. So we get 1. pi OK. We have radiant per meter. On the right hand side, we have two pi and that's Radian. Can we have pie? We have unit radian divided by lambda. OK. We're gonna get LAMBDA is equal to two pi gradient divided by 1.5 pi radiant per meter, which is gonna give us Lambda is equal to 4/3 meters. OK. All right. So we've found our Lambda our wavelength. Now let's do the same for Omega. OK. All right. So looking at Omega comparing in our equation, it's going to be inside of the sign and multiplied by the T. So we get here this 90 pi radiant per second. OK. So we have 90 pi a radiant per second is equal to two high FK and the two pi has a unit of radiant. So when we divide, that's gonna cancel, we're gonna get F is equal to and the unit is one over second. OK? Or hurt. All right. So when we find V, we get F times lambda, which we now know, OK. So we have the frequency F which is 45 in verse second, OK? Times lambda or over 3 m. And this is gonna give us speed B about meters per second, right? All right. So we found V using the wave equation we were given, OK. We found Lambda, we found F that allow us to find B and now in order to find the tension, we also need to find me. OK. All right. So we have me and mu is the linear density. OK. So let's recall mu can be written as the mass per length. OK? Well, we know both of those things about our strength. We know it has a mass of 5 g. OK? Converting this into our standard unit of kilogram, we're gonna get 0.005 kg and it has a length 1.25 m. So divided by 1.25 m, this gives a linear density of 0. kilograms per meter. And so this is a me value we needed in order to find tension. OK. So now we can go back to our tension equation. Let's give ourselves a bit more room to work here. We can go back to our tension equation and we know that tension is equal to V squared. OK? So 60 m per second all squared times meal 0. kg per meter, which is gonna give a 10 of 14.4 newtons. OK? And that is the answer we wanted for part one. OK? And the tension in the string is 14.4 newtons. And how we got to the Newton? Well, we have meters squared per second squared, K times kilogram per meter. That's gonna give us kilogram meter per second squared, which is equivalent to a Newton. OK? All right. So that's part one. Now let's do part two. OK. So we're gonna go down, we're gonna start part two and part two wants us to find the average power transmitted by that traveling wave. OK. Let's recall that the average power pav can be written as one half the square root of tea T is attention omega squared A squared. OK. We know mu we found it in the question or in part one of the question, we know t we just found as well. We know omega because we found it in order to find the frequency and a, the amplitude we can get from the wave function that we've been given. OK. So we have all of these values. Let's just go ahead and plug them in. So we have one half, hey, the square root of me, 0. kilogram per meter times 10, 14.4 Newton. Yeah time omega 95. Hey Radian second. It's weird times B amplitude. OK? Now let's think about the amplitude. Let's go back to our wave function. OK? We see that the amplitude A is out in front of the equation K multiplying the whole side. So in our case, the amplitude A is going to be given by six millimeters. OK? And if we scroll down and keep on working six millimeters, we want to convert this into our standard unit. And so we get six times 10 to the negative 3 m all squared. And this is going to give an average power of 0.345 watts, OK, doing all of that calculation. And so the average power part two 0. watts. Hey, let's go back up to our answer choices. We found a tension of 14.4 newtons and an average power of 0.345 watts. That's gonna correspond with answer. A thanks everyone for watching. See you in the next video.

Key Concepts

Wave Function

Tension in the Rope

Average Power Transmitted by the Wave

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