String 1 in FIGURE P16.47 has linear density 2.0 g/m and string 2...

Question

String 1 in FIGURE P16.47 has linear density 2.0 g/m and string 2 has linear density . A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. What should the string lengths L₁ and L₂ be if the pulses are to reach the ends of the strings simultaneously?

Accepted Answer

Welcome back everyone. We are making observations about two wires with different linear mass densities. Now, we are told that the linear mass density of wire A is 0.15 kg per meter. And we are told that the linear mass density for wire B is 0.5 kg per meter. Now we are simultaneously going to send two waves through each of our wires. And they, we are told that they arrive simultaneously at the opposite end. And we need to calculate the required length for wire A and for wire B for this scenario to be true. Well, we have that wave speed is equal to the tension in the wire divided by our individual linear mass densities. But we also have that our wave speed is equal to the length of each wire divided by time from this. We are going to set up a couple of equations here. We're gonna say that va is equal to the tension in the wire divided by U of A. And we have that V of B is equal to the square root of TB over mu B. Now what I am going to do here is I'm going to solve. Sorry, this is actually T W because the tension is the same in each of our wires. So what I'm gonna do is I'm actually going to solve for our tensions. So what we have is that we have T W is equal to ma times va squared. And we also have that tension of W is equal to mu B times V B squared. What we can do is we can set these two equations equal to one another. We have mu A times va squared is equal to mu B times V B squared. Now, how can we simplify this even further? Well, I'm going to use the second half of our original equation here to set up two more equations. What we have is that VA is equal to the length of a divided by the time it takes for the wire to arrive to the opposite end. And we have that V of B is equal to the length of B divided by the time it takes to get to the opposite end. Now, remember both waves arrive simultaneously. So we have that the time time it takes for both waves to arrive is going to be equal. So now what I'm gonna do is I'm gonna take these equations for our wave speeds and I'm going to substitute it into our system of equations down here. Let me change colors real quick in order to do this. So what we have is that mu A times L A squared divided by T squared is equal to mu B times L B squared divided by T squared. Now, what I want to do here is I want to solve for the ratio between the length of A and the length of B. So here's how I go about doing that. I say that the length of A divided by the length of B is equal to, well, the times are going to cancel out here and I'll be able to move the linear mass densities over to the same side. What we get is that the length of A divided by the length of B is mu B divided by mu of A, which we have these values. So let's go ahead and calculate this ratio. We have 0.5 divided by 0.15. And what this gives us is 0.5774. So we have that the length of A is equal to 0.5774 times the length of B Wonderful. So where can we go from here? Well, I'm actually going to scroll down just a little bit here. And let's take another look at our diagram. We have that the length of both of our wires is going to be 3.5 m combined. So we have L A plus L B is equal to our total length of 3.5. Now, I'm going to substitute in L of A with our equality that we just established. What we get is 0.5774 times the length of B plus the length of B is equal to 3.5. What this gives us is that our length of B is equal to 3.5 divided by 1.5774, which gives us 2. m. Wonderful. Now that we have the length of B, we can go ahead and solve for the length of A, we know that the length of A is going to be the total length minus the length of B which is 2.22, which gives us 1.28 m. So now we have found both the length of B and the length of A which corresponds to our final answer. Choice of D. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

String 1 in FIGURE P16.47 has linear density 2.0 g/m and string 2 has linear density . A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. What should the string lengths L₁ and L₂ be if the pulses are to reach the ends of the strings simultaneously?

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