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1k9 Mase anacnao spring with stiltnoss 72 Nm. The damping constant tne mass and gne its damping Iaclor quasiperiod and quasileoutncythe systemN-secimM3ssMovedthe Ia...

Question

1k9 Mase anacnao spring with stiltnoss 72 Nm. The damping constant tne mass and gne its damping Iaclor quasiperiod and quasileoutncythe systemN-secimM3ssMovedthe Iaft of equilibnum and givenleftward velocity of 3 misec_ detemine the equation olmotionWhat I3 Ine equabonMobontYt) = (61) ~ sln (64) (Tpa Aad ansten Hrsina Ndicalsnaeded The damping lactorquasiponcosecondsTne quasilre quencyI2

1k9 Mase anacnao spring with stiltnoss 72 Nm. The damping constant tne mass and gne its damping Iaclor quasiperiod and quasileoutncy the system N-secim M3ss Moved the Iaft of equilibnum and given leftward velocity of 3 misec_ detemine the equation olmotion What I3 Ine equabon Mobont Yt) = (61) ~ sln (64) (Tpa Aad ansten Hrsina Ndicals naeded The damping lactor quasiponco seconds Tne quasilre quencyI2



Answers

A $$1 / 8-\mathrm{kg}$$ mass is attached to a spring with stiffness 16 N/m. The damping constant for the system is 2 N-sec/m. If the mass is moved $$3 / 4 \mathrm{m}$$
to the left of equilibrium and
given an initial leftward velocity of 2 m/sec, determine the equation of motion of the mass and give its damping factor, quasiperiod, and quasifrequency.

To start solving this problem, We're going to first find the auxiliary equation, which is gonna be people R zero to R squared. Plus three r plus two is equal to zero. Um, if we want to factor this, we can see that we have, um, factors of to our one and two, which does add up to three. So are factors are gonna be one and two, like so is equal to zero. So here we have to, uh, distinct real roots. Therefore, we are going to be over damped in this problem, so over. Damned what that looks like is, uh, let's look at it like this. Okay, um, and then we have our equation here. We start at one. Okay, So this is one here and then Well, our initial, um, our initial velocity zero. So we're just going to start and then going to go down like that. So this is our over damn system here. Okay? This just a case to zero without passing through the equilibrium. Okay, So, like that now, to find our solution. So this gonna take the form wire t is gonna b c one e to the negative t plus C two e to the negative to t So why of zero first is going to be able to C one plus C two is equal to one. Next we need to find why prime a t This is gonna be able to Negative, uh, negative. C one e to the negative t minus two C to eat the negative to T. Why prime of zero then is gonna be negative. C one minus two C two, which is even a zero. We can add these two equations so that we're gonna get minus C two is equal to one. Therefore, C two is gonna be able to negative one. Then we can plug this in on. Then if we get that C two is equal to negative one, then we're gonna get this is to so see one, it's gonna equal to two. So our final solution why of tea? This is going to be able to to eat to the negative T plus four Sorry, minus e to the negative to t like so

Okay, So for this problem, let's start by finding the auxiliary equations that's p of our is able to our square plus three R plus two as equal to zero. So if you notice, we can factor this because the factors of to our one and two, which add up to three So then we're gonna get our plus two and then artistry, or are plus one is equal to zero. So are two routes are negative one and negative too. Since these are both riel, then we're going to have on over damped system. So that's our classifications of this oscillation over damped. Now for the general solution. That's why of T is gonna be the C one e to the negative t then plus C two e to the negative to t so to solve for C one and C two, we're going to use our initial conditions here. So first we need to find why prime a t that's gonna be able to negative c one e to the negative t and then minus two c two e to the negative to t like So now we plug in zero, so why zero is equal to and then this becomes one. And this also becomes one. So we get C one. Plus C two is equal to one. Next, we're gonna have, um Why prime of zero. Okay, there's going to be native. See, one minus two. C two is equal to zero. So we get from here. Our substitution is C one is equal to two C two. So then we can plug that into here. So see, one is equal to two C two. Um Oh, sorry. Negative to see to You want us negative to see to. Then we plug this in here. So we get negative to see two plus C two is gonna be negative. C two is equal to one, so C two is equal to negative one. Then that means see one. When we plug this in here becomes positive too. See, one is in a few. Positive, too. So our solution is why of tea is equal to, um Why of tea is equal to soc. One is a to e to the negative t. And then we have minus e to the negative to t like so. Okay. Now, uh, for this system here to graph it, we notice first that we start at one. So we're going to start at one here and then we start going flat. Okay. Now, uh, to see if this crosses equilibrium, we're going to set this able to zero. So to eat the negative t minus e to the negative to t we're gonna attempt to solve for C if this 10 equals zero. Okay, so if we, um, move the e to the negative to t over to the other side, like so and then multiplied by each of the two t on both sides. So we get to eat two. The team is equal to one so e to the t equals 1/2. This does have a solution, but in the negative direction, So e to the t equals what 1/2 1 half is less than e. So that means that t is going to have to be negative for this to be true. So this does not cross the, uh, ex access in the domains. He greater than it? Zero. So again, we started flat, and then we just go toward zero like that

Okay, so let's start by finding the solution to this differential equation. So I will do that by first finding the auxiliary equation socially, B R squared plus four R plus seven. And set that equal to zero. So now, um, let's see if we can factor this. So, for example, seven, um, has on Lee the roots or only the factors one and seven, which does not add up to four. So therefore, we're gonna need to use quadratic formula. We're gonna get our signal to negative B and then plus or minus square root of B squared, which is 16 minus four times a, which is one time see, which is seven eso we get, uh, here. Uh, okay. All over two A, which is just too so out front. Here we have. Um that's gonna become just negative too, then plus or minus. Now we have 16 minus 14 here, so that's gonna be square root of to. So we get squared of 60 minus 14 Which is this positive too? So we'll have square root of 2/2. Ah, here. Oh, wait. Sorry, Actually, um, four times. Sorry. Four times seven is 28. So we actually have 16 minus 28. So 16 minus 28. That's going to be equal to, um, square to 12. So we're gonna have square root of 12 or negative 12/2. So we're gonna actually have so I square root of 12/2 square to 12 can become to square root of three over two. So we'll just get I square root of three. So that's gonna become I square root of three like so now, since we have imaginary roots, imaginary roots means that we're going to be under dense here, So under damped. Okay, So since we have an under dam system with Graf oven under dam system, it's going to look like an oscillation that goes to zero or with AMFA's Who that goes to zero. So we started, say to We're going ah, up first. So we started to go up and then let me do that in different color, actually, again started to then go up, and then we're going to all sleep till we get, um, our oscillations get smaller and smaller here. Now, for our actual solution, we're going to have first that we're gonna have the General solution. Wire T is gonna be able to see one e to the negative two teeth and then co sign, and then we're gonna have square root of three t Let me cried as t square root of three and then plus C to eat of the late of two t times Sign Ah, tee Times Square with a three. So then now Ah, actually, I'll rewrite it. A square to three t square root of three and then times t square root of three. I'm Steve. Okay, so now we need to use our, um, initial conditions. So if we plug in zero first into this sign of 00 to 0, and then we have C one e to the zero co sign a zero both of these Air one. So we gets get C one here. So why of zero? I think the C one is able to to so we can just plug into here, So that's just gonna be a hate to now we can. Or now we need to find the derivative. Why? Prime of tea. Okay, so we're going t to use product rule a couple times here. Okay. Right. Okay. So when we take this derivative first let's do Ah, first times derivative second. So that's gonna be equal to negative Two heats of the negative to t uh, and then times also square root of three. So we'll have to square to three and then sign square root of three t and then also will have minus four e to the negative to t and then co sign square to three t Then for the second half, we're gonna have a plus plus so first each of the NATO to t times driven of second. So then we're gonna need to add a square to three and then, um, cosigned square root of three t and then now ah, minus two c to easily native to t And then, um, leave that a sign square root of three t like So now the sign terms are just gonna go away when we plug in zero. So why prime of zero is equal to again. So these go to zero. This goes to zero and then Oh, sorry, this goes to zero and this goes to zero. So we're left with. This is negative four because this goes to one. This goes to one negative four and C two square root of three. So we have negative four plus seat to square root of three is equal to six. So we're gonna have C two is gonna be with you 10 divided by square root of three or 10 square root of 3/3. Like so. So our final solution. I'm gonna move this over here. Our final solution is gonna look like why of tea is equal to to eat a plate of two teeth, cosigned square root of three t and then plus 10 square root of 3/3 E to the negative to t time sign square root of three t like so. So that's our final, um, motion.

In this video, we're going to look at the spring that has a three kilogram mass attached and it has a 48 Newtons per meter stiffness. So que is 48 Newton per meter and the masses three kilograms. What this allows us to do is calculate the angular frequency omega, which is the square root of K over end, which is the square root of 48 over three spirit of 16 which is four. And we also know that initially it's displaced to the left, um, half a meter. And so we know that the original position at times zero is minus half minus because it's to the left and the initial velocity at time. Zero is two meters per second to the right. And so this is positive, too. Okay, Now we know that the differential equation is given by three. Why Double prime for three White oval, Prime plus 48. Why is equal to zero? Okay, there's no damn thing here. And so what we have here is based on the solution of this type of equation. The general solution is given by some constant C one times co sign of five t because Omega's five. I'm sorry for 40 plus some other constant times signed or 40. Okay, so if we use these initial conditions, then we can find C one and C too. So let's go ahead and do that. If we plug in zero into this equation, we get C one co sign. Zero is one plus zero co sign 00 is equal to negative half, Which means see, one is equal to negative half. And if we plug in, um, into the white crime, so why primary duty is four C one times negative sign of fourty plus four C two co sign of 40. Okay, so now if we plug in zero, we get co sign zero is one. So then what we're left with is plus four c two. And if we plug in zero for sign, we get zero. So zero times see, one is just zero. And we know that the initial velocity is too. And so then we have f C two is equal to 1/2. Okay, so see two equals 1/2 and see one equals negative on How so? Let's write that down. So we have our solution. Lie of t is equal to minus 1/2 co sign of fourty Let's one have sign of 40. Okay, now the next thing we want to do is we want to find the amplitude. The attitude is given by the square root off. See one squared plus C two squared. So for us, they're both half. So it's half squared, plus half squared, which is equal to 1/2 times the square root of two core. It's weird, too, over too. Okay, no, we know that the first time that this model reaches equilibrium is when the function equals zero. So, really, what we have to solve is we have to solve negative half co signed 40 plus 1/2 sign of 40 is equal to zero. If I rearrange this bye, adding co signed 14 to both sides, multiplying everything by two and dividing by coastline 40. We see that we have tension. 40 is equal to one. Okay, which means 40 is equal to high over for and that's the smallest angle. And so t is equal to pi over 16. So at pi over 16 seconds, which is approximately 0.2 seconds is the first time this mass reaches equilibrium. Okay,


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