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[-/1 Points]DETAILSSERPSE1O 15.3.0P.013.MY NOTESPRACTICE ANOTHEplastic ball attached to sprng Moves simple harmonic motion The amplitude of the ball's motion 1...

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[-/1 Points]DETAILSSERPSE1O 15.3.0P.013.MY NOTESPRACTICE ANOTHEplastic ball attached to sprng Moves simple harmonic motion The amplitude of the ball's motion 11.0 cm and the spring constant 5.00 Nlm When the ball halfway between equilibrium position and its maximum displacement from equilibrium, ItS spccd 21.7 cmfs What the mass the ball (in k9)?What the periodoscillation (In s)?What the maximum acceleration of the ball? (Enter the magnitude In m/s? . m/szNeed Help?

[-/1 Points] DETAILS SERPSE1O 15.3.0P.013. MY NOTES PRACTICE ANOTHE plastic ball attached to sprng Moves simple harmonic motion The amplitude of the ball's motion 11.0 cm and the spring constant 5.00 Nlm When the ball halfway between equilibrium position and its maximum displacement from equilibrium, ItS spccd 21.7 cmfs What the mass the ball (in k9)? What the period oscillation (In s)? What the maximum acceleration of the ball? (Enter the magnitude In m/s? . m/sz Need Help?



Answers

A $100 \mathrm{g}$ ball attached to a spring with spring constant $2.5 \mathrm{N} / \mathrm{m}$ oscillates horizontally on a frictionless table. Its velocity is $20 \mathrm{cm} / \mathrm{s}$ when $x=-5.0 \mathrm{cm}$ a. What is the amplitude of oscillation? b. What is the ball's maximum acceleration? c. What is the ball's position when the acceleration is maximum? d. What is the speed of the ball when $x=3.0 \mathrm{cm} ?$

Hi, everyone. This is the problem based on concept of frequency, of oscillation, of Lord listening. If I spring office brings question K loading by the body and then its frequency is one upon goodbye route of Cuba. I am? No. In the first quarter, it is given a spring question of the spring is 2.5 Newtons per meter mast loaded? Having the value. I'm one kg your frequency, You will get one upon to five. Okay, 2.5151 So it is 3.796 hearse. Be part. All right. Sorry. Amplitude of oscillation. You will calculate. Mhm xD square. Blessed beauty square upon group F for displacement minus five centimeter velocity is given from the centimeter per second and frequency We have measured 12796 The on solving it application. You will get 6.4 centimeters. Yeah, Pardon me. Mhm applying contribution of energy. Okay. Uh mhm for initial position. Five centimeter and final position. Three centimeters. Behalf and the I square half K X. I split half and we finally squared plus half. Okay, accept no substitutes. The volume to calculate your final moss is quite What be well is 20 centimeters. That is fine too. He is 2.5 and 2.5 He put off. Okay? Yeah, I must point what? We have to spend half 0.5, right? And two. Why? Angina? Three. It's square. Okay on solving it. Final velocity. You will get from P 8. 20 centimeters per second there, so thanks. Part of

So we know that, uh, total energy off this equation in six jos. And we also know that initially will to the potential energy times, the kinetic energy. Um, so to make this problem looks Morrison Co Also, since we're looking for the amply two of off the vibration, which is not related to the kinetic energy we should be, um, assuming that we should be looking at the whole system at the point when the potential energy it's at this maximum, which means we have no kinetic energy at this point. So give us equals two. You're Max plus Zero, which is sex shows. So we know that the massive on potential energy, which is when the displacement this at this greatest would equal to six jos and the equation of kinetic energy is one have k times this place Manasse. And since we have the we know that the system is at its maximum displacement at this time, we should say that it's at the office at a so okay times, I swear. That's the math on this basement. Um and then the only thing we need to do is to calculate a so only tons both sides by 2 12 Joe's who goes to okay Square a goes to a square, rode off love, choose bye bye k and became plugging k here, which is 1200 new temper meters. And in the end, that calculation should give us 0.1 meter and as the displacement, we're just the massive displacement. Also, the amplitude of vibration off the system. We'll be on the variable we're looking at is the maximum speed, um in order to so that we need to take another assumption, which is, um, we need to look at the system when the kinetic energy said this greatest so that we don't have to worry about the potential energy. So now we have the whole of the energy of the host system because to six Jos Andi close to okay. Kinetic energy of this massive alone plus zero. So now the potential energy, Cyril. So now we have, um, the kinetic energy equation 1/2 and be square. And now the V set. It's Massimo also because of the mass close to six shows, because K mass and sterile Nico to six shows the total energy of the system. So now we can say that we can solve this thing. This equation in the similar way off, how it'd part a. So the times too. But both sides and the square the mouth squared was 2 12 shows the Mass. It's very close to 12 shows over an and the mouse equals two square root off. 12 chose and have this, um, we knew they were conversion here because and be close to 55 friends. But the only Univision use here is kilograms. So it's gonna equal to 55 times 10 to narrative three. And the guy was arriving 5.5 times. 10 tra now, too Kilograms. So I want five times 10 to the amount of two kilograms, and I should give us to answer off something around 14.77 meters per second. That's the massive bomb speed. Now we can take it will get part C um, which is pretty much asked us to combined part A and part B and um to find something that's when the precision ISS at house, this pasta bay over to and what street would be. So you know yourself that we put out the energy equation again. So eco's too potential energy, you plus canali energy que. And now, since we have, we know that we have a displacement. But not it. It's Max among. So we can be rid off either you or K between you to incorporate both of the equations to in order self the problem. So we need to ride it this way. You is close to 1/2 times. Okay, Spring, constant elsewhere, displacement and also the K kinetic energy close to 1/2 times. Um, over V Square. We know that the system of the equation is still the same six shows, and we know that a spring constant, um, let's give it this way so that it's less writing. But we know that both spring constant and asked this basement and we also know the mass. So is Ashley is pretty easy and straightforward to solve this equation because the only thing we don't know here is the velocity. Then, um, every times both sides by two. 12 Joe's times okay, asks Plus and these were and I'll give us The square was too tough, Jos. Minus chaos. Where and we would just be and Reese were about any Soviet. Just be 12 shows minus chaos. Where to buy? But I am and a swear route. And now we can plug in all of the numbers we have. So 12 shows already have the number here. Okay, Close to 1200 Newtons per meter times asked where we know that it's a square. Bye bye. Four and then we know em is, in fact 15 times tend to announce it to Kim Kilograms. And since we know that and he should be 0.1 from a hard day of the question when one meter from part A, we should also plaque that in and in the end, this giant thing should give us something about 12 point 7 to 9792 meters per second. And that should be the final velocity of the question. Um, and the question I was asked us what assumptions were made in order to solve the problem. And we can care clearly see that there's no energy lose who need to consider here the potential energy and kinetic energy just transformed from one another. And we don't consider things like friction or like air friction or anything like that. The, um, the only two energy here we're considering. It's Joe's potential energy and kinetic energy

First of all, let's do part a. Well. The total energy of the system could be described is E. Is equal to potential energy bliss. Kinetic energy. Well, when the Supreme makes maximum displacement, then this'll equation can be written is he is equal to potential energy. Well, in this case, when the messes it extreme point, kinetic energy is equal to zero. So the systems total energy is equal to one divided by two times k times a square. He raised the amplitude. It's maximum displacement and it becomes two times e divided by the supreme constant Well, we plug in numbers to multiply Bay six divided by 1200 so amplitude is equal to see the 0.1 m. See the 0.1? Yeah, right. Well, let's do part B Well at mean point total energy off the system is equal to kinetic energy. It mean point potential energy zero. So total energy is equal toe half and we square. Here is the maximum speed since masses, it is at the mean point and solving for maximum speed. Right? Maximum speed is equal to root two times e divided by m. Let's plug in numbers two times six divided by am, that is, 55 times tend to devour minus 3 kg, which is equal to 14 0.77 meter per second. Great. Now, let's do Parsi well in order to find out the speed when the ball is a position X is equal to was a divided by two. That's half of the amplitude we use. He is equal to one divided by two three times X square. This one divided my to him. We square right now, solving for me. We is equal to the route two times e minus Key times X square divided by divided by M right, Let's plug in numbers, then we is equal to route. Do multiply by six minus 1200. Multiply by a square divided by fall divided by AM, which is 55 times 10 to the power minus three. So we is equal to we is equal to 12 point 792 Made the 1st 2nd no data amplitude is 0.1 meter, so it is a good 12.792 m per second. Well, the most important assumption that we made in order to solve the problem is that we used the principle of conservation of energy. We also assumed that the total energy of the system transformed to a potential energy it acts, is equal to the maximum X is equal to maximum displacement and transformed to a kinetic energy it equilibrium position. Well, we can see that the speed when the ball is a position X is equal to sorry. The X is equal to plus a divided by two. That's half of the amplitude is less than the maximum speed. So the speed it this displacement is less than the maximum speed and that Maxim's because part of the energy is in the form of potential energy and paid born.

All right. 85. We have swinging. Pendulum got to swinging pendulums. Really? This one's going to swing into this one, and then they're going to stick together. Um, and we want to find the frequency. And the angular displacement after they have stuck together are right. So how do what do we know? What do we know? We know that this object is being released from here, so it has gravitational potential energy, and then it will swing into this object here. So when it gets down here, that energy will be converted to kinetic energy, and then it will hit that. So we will end up with a collision in which momentum is conserved, and then it will swing up together until it's, um, energy is no longer kinetic, would have kinetic energy when it starts going again and, uh, is all changed back into gravitational potential energy. So will be changed back into. So with the collusion, we're gonna change in kinetic energy, and then it will be changed back into gravitational potential energy. Okay, So what do we know? Do we know? Yes, we do. Gravitational potential energy is M g H. And I believe we know all those converted to kinetic energy, which is one half m b squared. All right, so, um, the mass of us 2 kg gravity is 9.81 The height is given at 10 centimeters, 0.10 m. One half It is the object coming down before the collision. So too. And the velocity is what we need to know. Right? Mass on both sides. Get rid of that. So you don't have to do that in the computation 9.81 times 0.1 multiplied by to take the square root. I got 1.401 point 40 m per second for the velocity before the collision. All right, so then we have a collision. And so momentum is conserved. We'll go with mass one velocity one just before it hits the other ball, plus mass to velocity to. And then after they collide, they're sticking together. So combine the masses and find the velocity at which the combined thing starts to swing up to the left. Right. So we have the mass of the first ball. It is moving at 1.4 m per second before it hits the other one mass of the other ball was 3 kg. Yes, and it was not moving before it hits the other one. Then they hit and they stick together. So two plus three is five, and it leaves with a velocity of Well, we don't know yet. This is zero. Okay, so the velocity that is leaving with, um is, uh, two times 1.4 divided by 50.56 meters per second. Okay, so then that's what it leaves with. Um, that's at the bottom of the swing. What? The velocity is combined, and then the combined mass moves up until it runs out off energy. Yep. Right. So we're back, Thio. The combined thing is leaving and has kinetic energy. And then that kinetic energy will eventually run out and be all converted into gravitational potential energy. Right. So this is when they are together. Um, at the beginning of their together movement, the velocities 0.56 when they reach a certain height and they stopped moving. So they combined mass gravity. And we're looking for the height. Perfect. All right. We could get rid of the fives to make our computation life's easier. we would have 0.56 square Divide by two. Divide by 9.81 and I got a height of 0.16 0.16 Uh, meters is thehyperfix are right off the ground. Now what do we have to figure out the frequency? I'm not sure he needed any of the stuff for, but that's fine. So the frequency is the reciprocal of the periods Omega, divided by two pi and omega is the square root of G over l for simple harmonic motion divided by two pi. Same thing is multiplying by 1/2 pi. All right, so we have that gravity is 9.81 and the pendulum length who was 0.5. So divide those. Take the square root, then divide by two, then divided by pi. If you don't put parentheses around it, it won't divide by the pie. So be careful of that frequency is 0.705 hurts. Okay, so it asked for the frequency, so that's an answer. And then it asked for the data. All right, so we have thio. So we have it down here, and then it swings up here and steps. The height difference is 0.16 Okay, let's see what kind of triangles we could make here. Right here. We're looking for this data. This is a right angle. This is the length of the pendulum. So 0.5 Yes. Okay. And then this is the, um Oh, the light. And so So this is not 0.5. That that whole thing is 0.5. Subtract the point. Oh, 16 And we get 0.484 So this is the length to here is 0.484 Great. There's data. This is a high pot news. This is adjacent. So Cotto, uh, that iss There you go. So we have the cosine of the angle is equal to the adjacent, divided by the high pot news. So divide them. Take the cosine inverse. So it looks like a my calculator. Um, and then you will get an angle of 14.5 degrees. So the angle of that of data there will be 14.5 degrees


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