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Block mast attached spring horkzont frictionless surface Tne positlon seconds) described bY the qubuon x(t) m cos(8 t). a) at what tlme the velocity maximum? b) wha...

Question

Block mast attached spring horkzont frictionless surface Tne positlon seconds) described bY the qubuon x(t) m cos(8 t). a) at what tlme the velocity maximum? b) what the amplitude oscillation when the mass doubles? what the maximum velocity of the block Ms d) a: what time does the block have position 42m? what energy : of the system at 1.0 5?the mass at any time

block mast attached spring horkzont frictionless surface Tne positlon seconds) described bY the qubuon x(t) m cos(8 t). a) at what tlme the velocity maximum? b) what the amplitude oscillation when the mass doubles? what the maximum velocity of the block Ms d) a: what time does the block have position 42m? what energy : of the system at 1.0 5? the mass at any time



Answers

A simple harmonic oscillator consists of an 0.80 $\mathrm{kg}$ block attached to a spring $(k=200 \mathrm{N} / \mathrm{m})$ . The block slides on a horizontal frictionless surface about the equilibrium point $x=0$ with a total mechanical energy of 4.0 $\mathrm{J}$ (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in 10 s? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at $x=0.15 \mathrm{m} ?$

Hello and welcome to this video solution of liberate. Here we are given a block of mass to Katie's executes as hmm while attached to a horizontal spring, having spring constant. 200 newtons permitted. So let me write down the given parameters. So we have the mass of the block equal to two kgs and the spring constant equal to 200 newton permitted The maximum speed of the block as it slides. The horizontal fictional selfish is 3m/s of fellow city maximum is equal to three. It was per second From the lady here to calculate the approval two blocks motion. Right so far from this calculator, angular speed or the angular frequency of oscillation, which is equal to the overall cable. Am right. That is equal to if you plug in. The values of this is 200 and here we have to But this is equal to 10 ingredients per second. Right? And from the expression of velocity V f mega, a right and well assorted three equal to omega is 10 and a Right. So here from here we have a amplitude is equal 2.3 m. So this is, I'm sorry. Yeah. Mhm. Extra to calculate the magnitude of the maximum acceleration, the maximum acceleration B is equal to a equal to omega square A. Right? So here we have Omega Square root, mean square times is .3. This gives us 30 m/s. Second square part C. A magnitude of the minimum acceleration. Now look for a minimum acceleration we can consider the the, particularly on the opposite end. So here we have minus omega square Right? This gives us -30. We just per second squared, but if you consider the magnitude of minimum uh magnitude of acceleration, which is minimum, okay then that will be equal to zero. Right? Because at x equal to zero we have got an acceleration a minimum. We call to zero. It can be both. The question didn't specify what is whether it's because absolute value or in general acceleration considering it has affected So -30. If you consider that it does affect it and if you consider the modelers of acceleration, that is the absolute value of acceleration, then you will have it at him. Now let me let's go to party here is asked. How long does the block takes too complex seven cycle of its motion Time period of motion is equal to two by by women. Right. This gives us point to the point 62 weeks serious And the total time to make seven oscillations will be 17. This is equal to seven times of 676-8 which is 4.4 seconds. I hope this is clear to you and have a very good rest of the day. Thank you.

Okay, so for this question, I'm not going to tell you the exact the answer off. But I'll teach you how to do it. And you can you circle the herd to find things so we know that without express the position Oh, the motion as X equal to the ampule Times co sign angular frequency times time asked the initial face five. Right. So the velocity is the derivative off position, So we have d x d t They have negative. Owe me a P MP you time sign on the dotted. Similarly, the exploration is DVD. So you have negative Omega Square. A close eye on me. Right? So for question A is asking for the force countered. Okay, so we know that a force equal to negative k x and you go, Jimmy right in this question, we know the mass. We know the acceleration, and we knew the position we can define. Okay, go to navigate m A over X. You can also, dear, if this equation from this re questions, So if we take Yushin One and the Christian three here, or use a divide by axe, you can get exactly the same results here only the only difference is that here you have kay and am But here, you're gonna have Omega Square. So you Because we know that Omega is secret. You square it, children. Right? So was dizzy reissue by exactly the same answer. So once we find Kay, we can also use this. The mission to find the anger for the ceiling. This for questioning a question. Be mpv off the motion here. We can use energy conservation. All right, so we know that the total energy is secret to the Maxima Kinetic energy, which is half a a square. All right, So are we already know Kay from questioning. And is he good to you at any moment? The kinetic energy X squared? That's a potential energy. Sorry. This is a petition urging the X square plus neck energy envies good. We know the mass. We know the velocity when Okay, when you acts, we can use this inquisition to find what stamping. So in this case and figure this secret you Sharon a scare. I X script us and over. Okay, please. Right. That's him. Questions. See what's the Okay. So, for questions, see, what is the maximum speed So if we look at the expression of the velocity here, so when sign would you want it will have its Maxwell speak, which is so the max on a vehicle too omega I'm standing. We already know the amplitude down here you can find similarly for D the maximum acceleration. It's gonna be yes, only us clear times. Damn you. And you can plug in the number with your catheter and find financing.

For this problem. There is a two kilogram block that is undergoing simple harmonic motion when attached to a horizontal spring. This spring has a spring constant of 200 Newtons per meter, and the maximum speed of of the block as it flies is three meters per second. We were asked to figure out what is the amplitude of the blocks motion. The magnitude of its maximum acceleration, the magnitude of its minimum acceleration and how long it takes for the block to complete seven cycles. So first thing we're going to do is find our amplitude. Um, we have a velocity so we can use velocity equals amplitude, times omega. And in this case, we're actually going to swap out omega for Omega equals the square root of the spring constant divided by the mass um, which equals the square root of 200 divided by two, which gives us a value of 10 rads per second. Um, now we are going Thio Ah, plug that 10 rounds back in here and we have our velocity or velocity is 3.0 meters per second, equals r x m times are 10 rads and we can solve for X m, which will be X M equals three divided by 10 which equals zero point 30 meters. So our amplitude here is 0.30 meters. All right, Next we're going to find our our acceleration. So our max acceleration, acceleration, Max equals our amplitude times Omega squared. We already found that our omega is 10 rads per second and we just found that our amplitude is 0.3 meters. So our acceleration max equals 0.3 times 10 squared. So we get an answer of 30 meters per second. So our maximum acceleration is 30 meters per second as faras minimum acceleration goes. That's gonna be either when the spring is fully extended, um, to its absolute end or fully compressed, In which case, um, at both points, it will be zero meters per second. All right, last part is to find how much time for seven cycles. Um and we have the equation. Period equals to pi times the square root of mass divided by are constant. We have our mass we know it's two kilograms. We have our cake, our spring constant. So we have t equals to pie times the square root of two divided by 200 which is going to give us in a, um, time a period of 0.6 to eight seconds, and then we want seven cycles, so we're gonna multiply that by seven, which gives us 4.4 seconds total, so it takes 4.4 seconds for seven cycles of this oscillation to occur.

All right, question 32. 3 significant figures, by the way. Probably right all that out. But the mass is given at 0.3 kg, and the position is given at a positive 0.240 m. The acceleration at that spot is given as negative 12 0.0 m per second squared, and the velocity is given at that spot as 4.0 m per second. All right, we want to calculate the spring constant. Well, the acceleration is dependent upon negative. There we go. Equals negative omega squared times the position. Omega is the square root of K over em. So Omega squared is K over him. All right. I think we know that we know the acceleration, which is a negative 12. We don't know. Okay, we do know the mass. We do know the position. Multiply. Let's see by 0.3, divided by 0.24 multiplied by a negative and the spring constant will be equal to 15.0. Mm, Yeah, 15.0. Okay, that's K. So what they asked for first, Then we know the velocity is, um, the square root Omega square root a county over em times the square root of the amplitude squared minus the position squared. All right, so the velocity is for okay, It's 15 and 0.3 a squared, minus 0.2 for squared. All right, Um uh, so I take 15 divided by the 150.3. Take the square root of that for divided by that. And I have 0.566 equals square root of the argument. Here we go square both sides squaring a square root in versus of each other and just gets rid of that leaves the argument the way it is. Um, 0.24 squared is 0.576 which will add to the 0.32 which is a squared. Take the square root and a is 061 for meters. So the amplitude is 0614 Alright, What else? All we want the maximum velocity, which is Omega times. The amplitude. So Maiga's that squirt a k over em The amplitude 0614 You're not square rooting that. So do that. 1st 15 divided by the 150.3 square root. Then multiply by the 0.614 to give 4.34 m per second and the acceleration maxes omega squared. It's okay over em. Times the amplitude. So 15 divided by 0.3 multiplied by 0.614 and you will get 30.7 m per second squared for the acceleration.


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