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'Scud For * Dan0 € M {ki bh 2. A 10.0g mass hangs from a 25.0 cm spring with spring constant 5.00 Nlm; You pull down on the mass stretching (he spring 61...

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'Scud For * Dan0 € M {ki bh 2. A 10.0g mass hangs from a 25.0 cm spring with spring constant 5.00 Nlm; You pull down on the mass stretching (he spring 61 2.0 cm and mhemlet gos Qounting 8 oscillations in 20.0 seconds, solve for the following: period amplitude c) maximum speed of the glider Sk : 5,o0 Vlx d) maximum acceleration for the glider 0 Le e) kinetic energy at its equilibrium point when it is oscillating_ S,c0066 H2J5,0u

'Scud For * Dan0 € M {ki bh 2. A 10.0g mass hangs from a 25.0 cm spring with spring constant 5.00 Nlm; You pull down on the mass stretching (he spring 61 2.0 cm and mhemlet gos Qounting 8 oscillations in 20.0 seconds, solve for the following: period amplitude c) maximum speed of the glider Sk : 5,o0 Vlx d) maximum acceleration for the glider 0 Le e) kinetic energy at its equilibrium point when it is oscillating_ S,c0 066 H2 J5,0u



Answers

A 1.00 -kg glider attached to a spring with a force constant of $25.0 \mathrm{N} / \mathrm{m}$ oscillates on a frictionless, horizontal air track. At $t=0,$ the glider is released from rest at $x=-3.00$ $\mathrm{cm}$ (that is, the spring is compressed by $3.00 \mathrm{cm}$ ). Find (a) the period of the glider's motion, (b) the maximum values of its speed and acceleration, and (c) the position, velocity, and acceleration as functions of time.

We have a glider that has mass m people to one kilogram. It's attached to a spring that have this been constant k of 25 millimeters and then the spring is compressed. X is equal to minus three centimeters or divide by 100 to get minus 1000.0 three meters from the minus. Sign means that I moved to the left, the equilibrium position. That's enough from here that the amplitude since you're dealing with simple my motion is a positive value of X. So it's positive. 0.3 meters from Eman Cane. We confined angular frequency, which is the square root que over em looking in the numbers 25/1 Take the square root. So are angular. Frequency W is five Redd's for a second now, With all this, you can start now. So for part A, we want to find the period which is T is equal to to buy divided by W So just putting her numbers that kind divided by five. So we get for the period T is equal to 1.26 seconds. Now, Part B, you want to find the maximum velocity and maximum acceleration, So first be Max. The max was found by just English frequency W times A. We know the value. So I got a frequency, a spy, A s 0.3 amplitude. So maximum velocity is equal. 2.15 You just for a second. And then our I'm acceleration or a max. A max is equal to W square times a again. You know these numbers five squared times a day, so a max is 0.75 meters per second. It's where now, part seeing you want to find a position, velocity and acceleration. So we know that the general equation for a particle undergoing simple harmonic motion is given as X is equal to a damper to times co sign w the English frequency times T plus the space constants. So then the velocity just a derivative of X. So the inside Of course, saying wti and the derivative of that is just w visits the driven of his respect t on the outside coastline is minus sign so the velocity have minus okay, a w sign w t plus B. The acceleration is again the the derivative of the velocity with respect to t so get on the inside derivative of W ts w and derivative sign Just coastline to him. They still have minus okay. Have you squared? Co sign B T. Let's see. Remember, this is acceleration and here is the amplitude. Now let's look at the position X and evaluated AT T equals zero so we can find the face constant. So we have a co sign. W thank zero must be, of course, w 10 0 zero and then at T equals zero. This is equal to minus 0.3 So now we know by this point 03 is equal to a is positive 0.3 coastline the so divide both sides 5.3 You know, coastline being physical to minus 23 about of my 0.3 No, that's just becomes go Sign being is equal to minus one. To find b you take the inverse coastline in both sides. So we have the people to It was closing minus one. So then a base constant. It's fee is equal to just pie. So now I'm going back to x or position velocity and acceleration. Now you can just plug in on numbers. So for X, we have 0.3 coastline? No, I think that frequency is just fine. T plus the facial We face constant, which we just found That pie The velocity is minus 0.1 by sign by i t It's playing and then finally acceleration which will put us lower case A A is equal to minus 0.75 co sign my tea, plus why?

All right, this is Robert Call doing my interview question for numerator, This's from Sears. And so Mansky, Chapter fourteen, Number twenty nine, All right. And this problem. We have a glider that's oscillating back and forth. It's got a massive your point five kilograms attached to a spring that has a spring concert of foreigner fifty tons per meter and we're going to pull out two zero point zero four meters and then let it go. It is going to oscillate. So remember that means it's going to go from zero point zero four meters on this side. Too negative zero point three or four meters on this side. We're X equals zero is measured at the equilibrium point. So we have over here the useful equations. We have X AA position, speed and acceleration always functions of time. Here, remember, the ex of tea was solved for using Newton's second law, and then the other two came from taking derivatives of X. We also have ah, this variable omega here aren't w as it appears on your screen. So omega is your angular frequency. We found it to be the square root of K over m. When we solve for X fifty in the first place. And then we also have a R amplitude and tr time. Variable way also have this equation we get for free from the conservation of energy where we have speed as a function of position and that will be helpful for one of the questions that is asked is part of this. All right, so let's move on in the question itself. What it wants to know. First of all, for part, it wants to know what is the maximum speed of big wider here. And we also won't find out for part B. What is the speed at ah X equals negative zero point zero one five meters. See, we want the maximum acceleration. I have report me. We want acceleration at X equals negative zero point zero one five meters. Same position is for part B and then for a part e. We want the total mechanical energy. All these are pretty plugging, chug if we just go back. So for maximum speed we come back and we see that VI of tea is a function of sign. So that means remember, Sign can have a maximum value of one which means V can have a maximum value of omega times a sow. Remember that omega is the square root of Ko Ram, so the solution then will be the square root of four hundred fifty. That's K divided by n, which is you're a point five kilograms and all of that is going to be multiplied by our amplitude of point zero four meters. That gives us an answer of one point two meters per second. So let's record that here we have ah, maximum B is equal to one point two meters per second. So for part B, we expect to get something less than that. Notice that it's asking for the speed at a negative position. But because we're looking at the speed is a function of position. You see here that the exes squared, which means it doesn't matter if we're on the negative side of equilibrium or the positive side of equilibrium. The speed will be the same. Now the direction might be different. It could be going out or in on either side, but the speed will be the same. So we're just going to go ahead and plug in our negative zero point zero one five meters into the axe. In this equation, that's going to be the square root of here We have Omega squared, so we that's going to cancel out the zero point five we have K divided by AM's that's foreign or fifty, divided by m multiplied by the AMP. Lou amplitude point zero four squared minus point zero one five squared and the answer in this case is going to be one point one one meters per second. So that's reasonable because it's less than the last one. So we find that the AT X is equal to negative. Zero point zero one five meters is equal to one point one one readers per second. All right. For the maximum acceleration, we can do something very similar to what we did in part a. Where we go town to acceleration is a function of time. We see that the amplitude of this trick in a metric function is going to be omega squared times a. So it's going to be former fifty, divided by point five times is point zero four and we get a maximum acceleration of thirty six meters per second squared. So maximum acceleration here is equal to thirty six meters per second squared now that might seem large, but generally accelerations will pure larger in a problem like this. And now we want the acceleration at a particular position. Now you might be tempted to come to our speed is a function of position and take some sort of derivative. But that's unnecessary. With much simpler equation we can refer to, which is Newton's second law. We know that the only force here is the spring force. We hooks law, and we know from Newton's second Law that f equals M A. So this is also equal to mass times acceleration. So looking at that weaken, very quick re solve for what the acceleration is at any position. In this case, we have negative K times X divided by AM. So we're going to type it and notice that in this case the negative will matter because it's going to give us a sign on the acceleration. So we're gonna have a negative foreigner fifteen times and negative point zero one five, all divided by point five, which is our mass. So putting that and we get thirteen point five meters per second squared Finally, we have the total mechanical energy now total mechanical energy. Remember, there's a mechanical energy should be constant here. It's defined as falling. It's mechanical Energy is equal to potential energy plus kinetic energy. And that should be a constant in this case because we have no non conservative forces acting here. It's just the conservative spring force that matters, so mechanical energy should be conserved. But we see that we also have some points of interest. We have the point of maximum extension and compression, where we have only potential energy of one half k squared, and we also have the point equilibrium. Where are one half K X squared, which is the potential energy for spring is going to be zero Cause X equals zero and it'LL all be inside one half em. The maximum squared so I can say, is equal to one half a a squared or that's also equal to one half and ReMax squared himto. If I type in either of these answers, I should be able to get my maximum are my constant mechanical energy. My told a mechanical energy would join that being zero point three six. Jules, I hope this video was helpful and

All right. So there's two immediate things we can find from this graph, the first of which is the periods of follow along for one complete cycle. I see it goes from 10.1 seconds. 2.3 seconds. That gives us a period of 0.2 seconds. The next thing I see is the max acceleration. That's a X max is 12 meters per second squared. Okay, Party. We start with the, uh, the equation for the period, which is to pie mrk route em over. K rearranges to stall for em. So this is, um, t squared over four high squared times k plugging everything we know. We know T wing. Okay, we get a mass of 0.25 three kilograms, part B. So the acceleration during periodic motion is given by negative K X over em. Um, which also equals negative Omega X. But we're just going to use the CAC's over m part. So this equals negative. Okay. So you could take a step back here. We want to solve for X max. Okay, that happens at a Max. So x max, which is the amplitude I rearranged equation will equal a max times the mass over K. And we just care about the the absolute value of X max. So this equals, um 0.121 meters and depart. See up here. So they're storing force at maximum amplitude. That's X equals a equals negative K A or K A. You know, if we just care about the magnitude in this equals 3.3 Nunes.

Let us write down two important equations that will help us solve the problem. We know that the total energy of the simple harmonic oscillator is equal to its kinetic energy. That's its potential. Oh, you know his kinetic energy. It's equal to 1/2 empty squared, and its potential is equal to 1/2 K. Hey, squirt. Okay, Expert Sorry about the K Express. They also know that the total energy of simple harmonic oscillator is equal to 1/2 K a squared, whereas the impotent also noon Second law for simple harmonic oscillator gives us that negative K X, which is the restoring linear force Bye hopes Law is equal to mass times acceleration. And so this top equation relates ex with the with the amplitude and the bottom equation relates acts with the acceleration. And so in party, the velocity is a maximum whenever x zero. So we're gonna plug in X is equal to zero in this top equation and sulfur B, and that will be our maximum velocity. Plugging that end gives 1/2 M v. Mac's squared equals 1/2 k squared. Everything in here is known except for the velocity, which is what we're solving for, and Sophie Max is equal to a square root Kayla Graham. And then we can just plug in the constant to give us. And this is equal to 1.2 meters per second. And that's answer to party for part B. We want to sell for the loss of you. One of her ex is equal to negative 10.15 meters. And so for this we're going to use this equation again. Except we're not going to plug in X equals zero. We're gonna plug in that X is equal this and we're going to solve her V again. When we do that, we get thie is equal to plus or minus square root of K over him, Times square root of a squared, minus X squared. When we plug in the values that the problem gives us, we get plus or minus 1.11 years for second. The reason we have plus or minus is because at one point the mass is going to be moving to the right at 1.11 years for second, and then it will slow down and we'll start exploring this way and then when it passes this point, it's going to move the same speed, but in the opposite direction. That's what the minuses indicated here when reality is talking about the speed. So we just say that the speed is 1.11 meters per second. Since he doesn't care about the direction we can read the mind Son and Parsi. We need the max acceleration, and we know the max acceleration occurs wherever the object is not moving and X is equal to. Plus, you might say this is a standard fact of simple harmonic motion is that X is equal to plus on my essay. The acceleration is the max and so going back to the first page here we can use this equation, except we're gonna plug in that X is equal day and sulfur What little a's So a max is equal to K over him, and now we can plug in the values that throne gives us. And when we do that, we have 36 meters per second squared, and that's the answer to Percy Parte. De wants the acceleration whenever X is equal to negative 0.15 And so for this we can just take the same equation that we just used. Except instead of replacing X with a, we're going to a place of Native 0.15 Other than that, it's exactly the same. And so when he put that and you get 13.5 meters per second squared party wants the total energy in this. And we know based on here that he is equal to 1/2 k examples of squared. So just rewrite that. It's 1/2 K a squared and everything here we know so we can split it, and that's what we get, and that's that.


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