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4.10.9Oueston Help "20 Mas? amtacheo scring hanging trom Ihe ceiling bnAreDu causing Ihe spring atrelch U745mupon contno Is 1 Neacm nelemne Ine steady-state so...

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4.10.9Oueston Help "20 Mas? amtacheo scring hanging trom Ihe ceiling bnAreDu causing Ihe spring atrelch U745mupon contno Is 1 Neacm nelemne Ine steady-state solulion for Ihe system_equilibrium; At timean exlemal fcrce of F(t)e3c32 6t N applied ! the syslem; The damping consrant for Ihe systemThe sleady staie solulicn Is Yl) =

4.10.9 Oueston Help " 20 Mas? amtacheo scring hanging trom Ihe ceiling bnAreDu causing Ihe spring atrelch U745mupon contno Is 1 Neacm nelemne Ine steady-state solulion for Ihe system_ equilibrium; At time an exlemal fcrce of F(t)e3c32 6t N applied ! the syslem; The damping consrant for Ihe system The sleady staie solulicn Is Yl) =



Answers

An 8-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 1.96 m upon coming to rest at equilibrium. At time t = 0, an external force $ F(t)=\cos 2 t N $ is applied to the system. The damping constant for the system is 3 N-sec/m. Determine the steady-state solution for the system.

So from the problem, we know that m iss equal. Sue one over a kilogram. Okay, You're close to 6 16 16 Newton over neater b equals to the damping is most to you in seconds over meter. So I'm just gonna let left to be the positive direction. So this gives us the following Formula 1/8. Why? Prime Prime Plus two. Why? Prime plus 16 y equals zero. And we're gonna call. Why? Prime Prive. Sorry. Why Prime equals game. Well, so what I'm proud would be equals to Kendall Square. We would have won over eight Camera square plus two camera plus 16. It was zero. Um, so this is a quadratic function. So many solving for camel one into which as equals to minus two plus or minus over a square root. How? For square root of four minus four times 1/8. Time. 16 over to over a equals oneness to plus or minus sign where it's of minus four over 1/4. This equals two minus eight. Plus or minus. Sign. Hey, I so from the camera function and we can no determined that our Y function has the following general formula so e power to the minus 80 times a co sign 80 waas Move! Sorry. 80 record You. Oh, So if I went to derive why, I would then have minus eight e power to the minus 80 a co sign a t plus B sine a t plus e power to the minus 80 times minus eight. A sign 80 plus eight Be co sign 80 and then we have that. Why is hero give it its eight most to 3/4? And why playing zero? The initial love Fort Velocity is a monster too. Why? Zero would be equals to a because they would be over fourth. Why prime zero? Now we pluck and zero to this formula which give us eight minus eight A plus eight b equals to two. So therefore we would have a equal speed over fourth in be equal to one. So remember previously we have where we have established our y t function. My t equals two e power to minus 80 me over four. Call sign A T. It's less sign 80. Combining these two together Give us a minus a T 5/4 sign a T plus arc Tanne three over force. So common factor. A power to the 85/4 time sign. 80 plus are tan iss, right? Written as the inverse of 10 three. Fourth plus high over two minus pi over two. So the sign can beat now, right as co sign. Cool. Sign a T minus. Hi. Over to minus 10. The inverse of 10. The fourth for our 10. So this is our final equation. From here, we can now determine the damping factor. Damn thing factor. Okay, this equal Soon E power to the minus 80 time 5/4 period for the period would be T equals two pi over omega and our omega equals to eight. This is our omega value. So this will give us pi over four. Our frequency. This one over tea. It was two for over pie.

Okay, so for this question, this is a differential equations question and given 8 kg mass, right moves attached to a spring. So here's our mass, and that's attached to a spring that's hanging from the ceiling. And it's allowed to come to arrest and were given that the spring constant, um, or given that spring, constant is equal to 40. Right? So, um, is about 80 kg a spring constant k 0 to 40 and were also given, uh oh, 40 Newtons per meter. And we're given that the dampening constant B is equal to three Newtons, uh, Newton seconds per, uh per meter. And then, uh, time to equal zero. Um so those are all the factors or constant stuff was given and were also given an equation, which is the force to sign of two tee times co sign of two teeth. Basically, from all this, we have to find the amplitude and frequency of the steady state solution of, uh, this entire system here. Right. Okay, so the first thing we can do is we basically have to identify that this is basically a trigger metric identity, so we can actually simplify the function. F f t all the way down to sign of 40. Yeah. So once we do that, this actually becomes a whole lot simpler problem. And then we can set up our differential equation by saying 88 de tu y over dx squared. Uh, so second, even plus three dy DT it was supposed to be DT squared dy DT plus 40. Why? And these are our constants that we had appeared mhm. And, um, this is all going to be equal to a sign of 40 because that is also part of the system. So now we have to find the steady state solution. Uh, when we do that, we actually use the undetermined coefficients method, uh, by putting the solution in terms of or in the form of the sine cosine form. So when we do that, we have y P is equal to a co sign of 40 plus b sign of 40. I think that's gonna be the general form that we get this into. And then we have to plug that into this equation here. So we have to find, uh, second the first derivative of, uh, the steady state solution, which is just the YP part right once again, Uh, so when we find that we'll get, uh, during the coast on this negative sign so negative for a sign of 40 because that would be the inside of here using chain rule and endured of signs co sign so I can be co sign up for a key. And then your second derivative will similarly be basically the same things during Sinus Co sign Negative 16 a co sign 40 plus and then the during all of this. So because there's coastline is negative signs that this will change to a negative 16 b times sine of 40 mhm tonight, you basically take all of these and you plug them back into this equation here. And when you do that, you'll basically end up with eight times this. So it'll get eight times negative 16 a co sign of 40 minus 16, the sign of 40. And then you have plus three times yeah, divided teacher at the first derivative negative for a sign of 40 plus for be co sign of 40. And then you have plus 40 times Why? So I'll be plus 40 times a co sign 40 plus B uh, sorry. Yahiko Center 14 plus B sign of 14. Mhm. And then all that equal to sign a 14. So once we have this huge equation, uh, we can start simplifying it, uh, down, uh, right so we can go and start by multiplying everything inside. Uh, so if you put eight inside here, you get negative. 1 28 a co sign of 40 minus 1 28 b sign of 40 plus or it actually minus minus 12. A sign of 40 um, plus 12 b coastline of 14. Right when you multiply those two in and then plus 12 b co sign of 40 plus 40. A co sign of 14 plus 40 b sign of 40. All of that equal design of 14. So now all you have to do is take your co signs and take your signs. Right. So you have coastline here? Uh, cause I'm here cause I'm here and then you have sign here, sign here and sign here. And you basically have to combine all of those together into one term and do that, Um uh, by basically taking, uh, coastline of 40 out of the equation like that factoring it out and then having all of your inside Constance a negative 1. 28 a. Right, so that takes care of that term. And then here you have plus 12 B that takes care of that term. And then here you have plus 40 A. So when you add that you actually get in your negative 88 a And then you add to that your sign of fourty term, uh, which you will actually also take out. And then you have negative 1 28 B minus 12 A. And then you add into that plus 40 b, and this actually becomes negative 88 b And then you said all that equal to sign of 40. Now you can see that there is no co sign term on this side of the equation. So this has to be equal to zero. And so what that means is negative. 88 A. Since this can't be equal to zero negative 88 a plus 12 b has equal to zero and then coefficient for this sign in the 40 here is just equal to one. Then that basically also says negative 88 B minus 12 a this part as we equal to one. Right. So now all you have to do is solve for both of them, and, uh, easy way to do that is, um, first saw for B, um, on this side. Right. So you'll get 12 b equals 88 a and then B equals 88 a over 12. Um, so you do that you can, uh, simplify this a little bit, Uh, 2. 22 a over three. If you divide both sides by four and then you can plug that into this equation here and you'll get negative. 88 times 22 a over three, minus 12 A equals one. And then when you multiply that in, you'll get 88 times. 22. Uh, so you'll get negative. 19. 36/3, a minus 12, A. Equals one. Mhm. And then, um uh, 12. You can basically rewrite that as, um, uh, 36/3. All right. And so that way you can combine both of these together. So you have 36/3, and then you multiply 3/19, 36 a minus 36 a three, right, and then you'll just combine these two and then you'll get, uh, a equals Negative 3/19. 72 when you do this. So now that you have a all you have to do is plug that back into this equation here or any of them, and, uh, plug that back in to find B, and we know B is equal to 88 a over 12. So if we plug that back in, uh, B equals 22 a over three, then that must be is equal to 22 times like there's 3/19 72 all of us, all of the three. Um, when you do that, you'll get basically negative. 11. These stuff will cancel out over 986 So that will be your value for B. So now that you have both A and B, uh, you can plug those into your y p solution here. When you do that, you'll get y p equals negative. 3/19 72 co sign of fourty right minus 11/9, 86 times sine of 40. So now that we have that, that is a steady state solution. Now we have to find the amplitude, and you can find the amplitude of this, uh, amplitude of a form in a co sign X plus B co sign X is equal to, uh, square root of A or amplitude. My smaller case a a squared plus B squared. So that will be this square plus the square, so that when you do that, you'll get negative 3/19 72 squared plus negative 11/9 86 square, and I'll give you the square root of 4 93 over 1972. Don't just take my word for it. Go ahead and solve for yourself so you can see the answer. But that will be your amplitude, and that's how you find it. And then also, the question also asked us to find the frequency. So are the angular frequency. Um, so in both cases, frequency is basically going to warn over tea and angular frequency is equal to, um, the angler frequencies to pi over t. So for both cases, you have to find t and T is given to you by two pi, uh, over four. Because in this case, in this case, in our case, specifically, uh, the period is two pi over four. Um, because because it's being multiplied here on the inside by four. Uh, so full period would be to pie. So you divide that by four and you get to power for the same thing as part of the two. So that way you frequency would be won over pi over two or two over a pie. And that means also your angular frequency is equal to, uh to pi over pi over two, which is the same thing as for and both of these are in four times one over a 2nd 14 hertz, basically. So your frequency is equal to two over pie and angular frequency is equal to four times a second to negative form, and those are your answers for this system, and that's basically how you solve it.

So in this case, we want Teoh were asked to look at a case where we have a block and spring that suspended from a ceiling here That's been oscillating. And I told the measure why? From the static equilibrium off the system. So that means that this in this case, where we measure why zero this spring is stretched and that stretched by an amount of whatever the weight of this block waas So what we can think about is that the the same the equal, the other stretched length of the middle of the block was here when the spring was stretched and then we have some static deflection there and then we're gonna measure our motion from that point here. So if we draw free by diagram of this block, we have mg acting and we have a force in the spring and the force of the spring is the change in length. So the change of length is whatever. Why is here plus the static deflection. So that's the change in length. That's the motion of this part of this end of the spring. The motion of this end is this. So we need to take the difference of those two. So we get this minus this. And that's the change at length of the spring totals. Changes link from when it was unstrap putting that taking, you know, we get em. Why double that minor equals M g. My minus f s. Which is then this. Okay, now, what we can see here is that the static deflection is actually mg over case. Okay, times a static reflection is mg. So that's just all this. All these two terms are basically a static force balance. So if we have statics, you write that better? Uh, okay, that's a static. That would be this equation of static for this system. This is just the static displacement from when the spring was on stretched. So why now? Measures the motion about that static equilibrium. And that's usually what we worry about in vibrations, where because in vibrations, constant forces don't cause systems the vibrate, they just cause the equilibrium position to change. And so usually we're just worried about the emotions about any liberal. So these terms Castle and we get this and we can either write it this way also, but we can take this and we can basically a strike on the solution from what we've had before in previous problems. So we have the homogeneous solution that has a frequency of the natural frequency two constants that we find from initial conditions. And then we have the particular solution. We're in this case f not equals K Delta. So f not over k is just dealt or not. And then we have one over the frequency ratio in the denominator times Sign over my nicotine, whatever this term was here goes into here. So if this was the co sign, this would just change to co sign. And so this is our final solution and we have again are forcing frequency. His part are part of the solution is part of this motion is oscillating with forcing frequency, and part is oscillating with natural frequency, and those things were constructively and constructively interfere with each other again and again. And it will look like a very complicated motion. But it's really only two frequencies combined

So for this problem, we've given that there is a mass that weighs £32 pops. And this is a mass that weighs £32. Uh, which is attached to a spring, uh, which is hanging basically from the ceiling. And it comes to arrest at its equilibrium position. And time is equal to t Uh, start Time T is equal to zero, and you have an external force, which is three co sign of 40 pounds That's applied to the system. Uh, and if the spring is constant, if the spring constant is 5 ft, uh, £5 per feet and the dampening content is £2 seconds per feet, then we have to find the steady state solutions. So that's the question. Fine said steady state solution. So we know that the masses £32 um, and we know that the external forces three times goes on the 14th. Uh, spring constant right is equal to five. Uh, dampening constant is equal to two. And these are just what we used to denote those. So we actually get an equation which is, uh, second derivative plus two times. Do I D. X plus five. Why were these guys. That's where these guys come into play the spring constant, the dampening, constant mhm. And you get all that equal to three co sign of 14 because that's the force that's being applied on the object of £32. Um, so now all we have to do is find a steady state solution. All right, so the steady state solution is divided by y p. Um, because this is the part that we're looking for when we say we're trying to find the steady state solution. Uh, we're not trying to find the lives of H or any of that other stuff. So I said B, we can say is equal to a co signer 14 plus B sign of 14. Um, we're using the undetermined coefficients method here because we're trying to find the solution in the same form as a given as a given forces being applied. So this is called the Undetermined Coefficients method, where we basically just have equation, which is in the same form as a solution that we're trying to find it in. So now, using this equation, um, we can try to find A and B, so our sins were given by P was fined also by private P White primer table B uh, dirt of coastline would be negative sign. So you have negative for a sign of 40. I mean, the dural Sinus co sign. So plus four b co sign of 40. Uh, we're also doing a product well, here. And then you take the second derivatives that as well. And then you get foot. Now you're 16. Uh, a co sign of 14 visitors. Science, Carson. And then during Christmas, negative signs left. Negative 16 b co sign a sign of 40. So now that you have those two, you can plug those directly into this equation here, Um uh, and then from there you can find A and B So we plugged y double prime into, uh, d squared wire over the X squared, and you'll get negative 16 times a co sign of four teeth, plus be a sign of 40. And here I just factored out the negative 16. Uh, so that, um, mhm. It's so that it's easier for us to dis visualize it and use it and stuff. Um, and then plus two times, uh, negative. Four a. So in this case, I will factor at the four. So we'll get eight times. All right, be co sign of 14 minus. All right, we'll take four hours to be an A minus a sign of 40. Mhm. And that's basically what we get for this, uh, this term here, Um, the next we have to add five plus five lives. So plus five times a coastline of 14 plus b sign of 14. And we're basically setting all this equal to let me actually move this over real quick. Have this here and then recently moving, um, basically setting all this equal to our initial thing here, which is three coasts on the 14th, so equal to three code sign of four teeth. So now that we have all of that, all we have to do is start condensing these and getting all of them into the forms are just signed and co sign. So over here, you can call this negative 16 a co sign of 40 minus 16 b co sign of fourty. And then over here we have this guy plus eight be co sign of four teeth minus eight, a sign of 40 and then over here I have plus five a co sign of 40. Yeah. And then plus five b sign four teeth. Of course. All that people at three co signed a four for Pete to know that we have that we can get to combining things. Um, mhm. So we have We'll have, uh we can take these coastline of forties here, right there, right there and right there, and can basically say, co sign of fourty, and then we'll have this thing in front of it, which has negative 16 8 plus a B plus five a. Right, because those are all the things that cost in the forties being multiplied to, uh, sorry here is supposed to be negative 16 b sign 14, and then you can add here and then say sign of 40 years. Well, and we'll basically do the same thing, will take the coefficients there. 16 b, uh, minus eight A. And then plus five B. And then all this is equal to three co sign of 14. Oh, I'm pretty sure you guys can tell exactly where this is heading, since there is no sign sign term on this side of the equation here. Basically the sign term will just cancel out should be equal to zero. Whatever's on the inside here will be zero. And that means there is no sign term. So all we're left with is if we combine these two negative 11 a plus eight b times co sign of four t is equal to three. Co sign of 14. Now there's close on. The four teas can cancel out, and all you have to do is find negative 11 a plus eight B equal to three. Um, but then, um, you The thing that you want to help help you out with this is since you have more than one variable, you know, that, uh, negative 16 B minus eight A plus five B or negative 11 B minus eight. A. Has to be equal to zero. So this will actually come into handy. Uh, once you substitute in, um, substitute in things for right here. Uh, so we can basically solve for a here we saw for a we'll get 1911 a equals three minus eight b and then a equals three minus eight B over negative 11. Um, so doing that, we're substituting for a and we're saying this is what is equal to, and we can actually just plug that straight into here and say Negative 11 B minus eight times three minus eight a. Oh, sorry, 3,000,008 B over negative 11. And we want all that. Two people, 20 And then when you basically, uh, multiply this inside, you'll get 8 16, even negative 24 plus 64 b and then all of this over negative 11. And then if you put both both sides of here over negative 11, well, you wouldn't really have to. Uh, next thing you do is move this 11 b over to the other side and you'll be left at negative 24 plus 64 B over negative 11 equals 11 B, and you multiply by negative 11 on both sides. Negative 24 plus 64 b is equal to negative 121 b and then you move this be to the other side and you'll be left with negative 24 equals negative 121 plus 64 be, which will give you negative 57 b and then finally, B is just equal to 24/57. Mhm. Um Oh, sorry, sir. Your remote you'll be subtracting so it will not be 57. Uh, you'll get a negative 1. 21 minus 64. Not plus it's for. So you have 185 years. So it'll be 24 over 185. And now that you have this value for B, you can go all the way back up to here or you say negative 11 a plus eight b mhm equals three. And you can basically just plug in your B plus eight times and you get busy with 24/1 85 equals three. And once you plug that in eight times 24 I will be 1 92. So that negative eight plus 1 92 over 1 85 equals three and subtract. So you have three minus 1 92/1 85 um, and out. And then during 9 11, 8 equals 363 over 1 85. Yeah, and finally, you divide both sides by negative 11. Yeah, and you'll be left with negative 33/1 85 for your value of faith. So now that you finally have both of those, you can plug those into your equation up here. Where was it? Uh, right here Your i p to come up with the solution that you want and do that y p equals a which is negative. 33 33/1 85 times co sign of four T plus 24/1 85 times sine of four t. And that is your steady state solutions for the system for this equation, so yeah.


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