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Anderson Video - Spring Example- Max Speed and Acceleration

Professor Anderson
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>> Okay, let's try an example problem that might have relevance to your homework. That tends to wake everybody up a little bit. Let's do the following. Let's say that we have a spring with a block on it. And it looks like the following. It's on a frictionless table, so there's no coefficient of friction here, no static coefficient, no kinetic coefficient. And we're going to let the spring have the following value. It has units of 10 and in SI units what are the units for a spring? Well, it is Newtons per meter. Okay? You know that because when we multiply by meters, distance, we have to get back to Newtons. Okay. And let's stretch it out-- -- to 10 centimeters. And then we're going to let it go and 10 centimeters in SI units is, of course, 0.1 meters. Okay? And let's ask the following questions. What is the maximum speed of the block? And let's ask a follow up. What is the maximum acceleration of the block? All right. Do we have enough information here to figure this out? What do you guys think? What's your name on the front left over there? Sarah? Can somebody hand Sarah the mic? Sarah, let's have a little chat. Sarah, what do you think? Do we have enough information here to solve this problem? >> (student speaking) Yes? >> Maybe. >> (student speaking) No? >> Seems like we might need one more piece of information, right? Which is the mass of the block. If I put a really heavy mass on there, is it going to go back and forth as fast, Sarah? >> (student speaking) No. >> No. It's not. Right? So, we also need to know the mass of the block. And we'll say that the mass of the block is 2 kilograms. All right. We're just making up some arbitrary numbers here. So, now what do you think, Sarah, do we have enough information to solve this? >> (student speaking) Yeah. >> Yeah. I think we probably do. Doesn't seem like there's anything else here that is necessary. We know about the spring. We know how far we stretched it out. We know what the mass is. There's really nothing else in the problem. Right? So, what do you think we should do next? >> (student speaking) Put the numbers into the equation? >> Which equation? >> (student speaking) The last-- >> The last one we looked at. Okay. Excellent. The last one we looked at was this. X equals A cosine omega t plus phi, and let's say that we're just going to let phi equal 0. And so this just becomes A cosine omega t. That looks pretty good. But, we don't want X, we want V. We need to figure out what V max is. So, we know that V is related to X because V is just the derivative of X. And we saw what that was. V being the derivative of X means that we get negative A omega sine of omega t. So, if this is the velocity, what is the biggest speed that you can have, and let's just worry about the absolute magnitude of it. Sarah, what is-- what's the biggest V I could have? If this is our equation for V. How big can sine get? >> (student spekaing) 1. >> 1. Right? It goes from 1 to 0 to negative 1. So sine can only be up to 1. So the biggest speed that we can have is just the stuff that's multiplying it. It's just A times omega. Okay? But we know what omega is. Omega is the square root of k over m. And so we, in fact, have V max. It's A square root k over m. And now we can plug in some numbers and figure out exactly what that is. Okay. V max is a square root k over m. What did we say A was? 10 centimeters. Right? SI units, that's 0.1. What was our k? Anybody remember? What did we have for our k? Is it 10? >> (student speaking) 10, yeah. 10 Newtons per meter. We had a mass of 2 kilograms. And so now you can punch this in and see what you get. 0.1 times the square root of 5, you guys, one of you punch that into your calculator. Tell me what you get. And we'll approximate it right here. What's the square root of 5? Well the square root of 4 is 2. And we got to get a little bit more than that. So, maybe 2.3, and then I'm going to take a tenth of that, so I'm going to say this is 0.23. And if you run it in your calculator tell me what you get, exactly. >> (student speaking) .223. >> .223? Okay. So we were reasonably close with our guess. .223 is the real answer. All right. What about a max? And the units on that are, of course, meters per second. What about a max? Well, to get a max we have to go back to V, and we know that a is going to be a derivative of V. And we know what V is. It's right here. So when I take a derivative of this thing, what do I get? I get negative A omega squared. I have to pull out another omega from the argument. And then the sine goes back to cosine. And now, again, cosine can only go up to 1, and down to negative 1, so the maximum acceleration is just going to be A omega squared. Just worry about the magnitude of it, not the sign. And now look what happens. Omega is a square root of k over m, but we're going to square it. And so we get A k over m. Okay? And now we can plug in some of those numbers and calculate a max. A, we said, was 0.1. k, we said, was 10. m was 2. 0.1 times 10 is 1. We get 1 over 2, which is, of course, a half. And the units in SI units are meters per second squared. Okay? V max and a max, for our block on the spring, and in fact, we did have all the information we needed. All we needed was the amplitude, the spring constant, k, and the mass, m.