Hey, guys. So let's work this problem out together. In this problem, we're told that the wave speed for a certain string under tension is 20 meters per second. I'm going to call this v=20. I'm going to draw this out real quick. That's my sort of wave-like this, and this is sort of like my axis. Right? So now we're asked to calculate, well, if you cut the tension in half, so the tension is half of its original value, then what happens to the new wave speed? Let's take a look here because, obviously, we're going to be dealing with some kind of velocity or a wave speed equation, and we know this is going to be a string. So, because we have a string, we can always just use this formula over here. So we have that v=FTμ . Now notice how this problem has really no information other than 20 meters per second. And this is a classic kind of proportional reasoning question because they're asking what happens to something if something else gets cut in half. Right? If you change something, how does another variable respond? Here's what I'm going to do here. I'm going to say that vnew is really just going to be the square root of FTnew over the mass per unit length. The only thing we're told here is that the tension is cut in half, so we can assume that the mass per unit length will also stay the same. So here's what happens. This FTnew here, what they're asking or saying here, is that it's going to be one-half of the original tension. So what happens is this equation here, v=FTμ, is equal to 20. We actually know that already. But now what happens is vnew is going to be the new tension divided by μ. Okay? And what we can do here is we can basically just replace this FTnew with this expression, the one-half FT. So it's going to be the square root of one-half of the original tension divided by the mass per unit length, and what we can do here is we can basically just pull out the one-half that's inside of the square root, and this just becomes the square root of one-half times the square root of FT over μ, right? That's what happens when you sort of extract it, basically just splitting up the square roots. So notice how this piece right here, this equation, this FT over μ, is what shows up in our new equation, but now we just have an extra factor of the square root of one-half that's on the outside. That's what usually happens with these proportional reasoning type questions is that you can basically just extract out the same expression in the before case, but with just an extra constant that's out in front. So remember, this expression here is equal to 20. We don't know what all the variables are, but we just know that the whole thing equals 20. So really what happens is this just becomes the square root of one-half times 20, right, because this thing is equal to 20. So, in other words, the velocity, the new wave speed, is actually just going to be the square root of one-half times 20, which is equal to 14.1 meters per second. Now this should make some sense that the new wave speed is going to be less because, again, if you look at the equation here, if you reduce the tension then that means the speed is going to be reduced as well. It's not going to be half because of the square root that's inside here, but it is going to be reduced. That's it for this one, guys.
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Velocity of Transverse Waves
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