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12 points) A 301 +/- 2 gram block hangs from & spring with a spring constant 0f 9.90 -/- 0.04 Nln (about the same &s the springs in our lab.) The block is t...

Question

12 points) A 301 +/- 2 gram block hangs from & spring with a spring constant 0f 9.90 -/- 0.04 Nln (about the same &s the springs in our lab.) The block is then pulled dowrr to a point where the spring is 35.0 cm longer than its unstretched length (before the mass is hung on it); and then released Draw the problem; and create a general equation for the vertical position of the block as & function of time; Y(t) Identify all the variables in the problem What is the expected frequency o

12 points) A 301 +/- 2 gram block hangs from & spring with a spring constant 0f 9.90 -/- 0.04 Nln (about the same &s the springs in our lab.) The block is then pulled dowrr to a point where the spring is 35.0 cm longer than its unstretched length (before the mass is hung on it); and then released Draw the problem; and create a general equation for the vertical position of the block as & function of time; Y(t) Identify all the variables in the problem What is the expected frequency of oscillation of the spring? How certain are You of this value? b) What is the position and velocity of the block (magnitude & direction!) a t = 1.50 seconds after release? expected position AND velocity on graph of position in time Show these values match the



Answers

A 2.0 kg block is attached to the end of a spring with a spring constant of 350 $\mathrm{N} / \mathrm{m}$ and forced to oscillate by an applied force $F=$ $(15 \mathrm{N}) \sin \left(\omega_{d} t\right),$ where $\omega_{d}=35$ rad/s. The damping constant is $b=$ $15 \mathrm{kg} / \mathrm{s} . \mathrm{At} t=0$ , the block is at rest with the spring at its rest length. (a) Use numerical integration to plot the displacement of the block for the first 1.0 s. Use the motion near the end of the 1.0 s interval to estimate the amplitude, period, and angular frequency. Repeat the calculation for $(\mathrm{b}) \omega_{d}=\sqrt{k / m}$ and $(\mathrm{c}) \omega_{d}=20 \mathrm{rad} / \mathrm{s}$

Hello and welcome to this video solution of numerous. Here we are given a spring mass system where a force if he's acting on the mosque. So let me write down the given parameters. So we have K. The spring constant equal 200 newton were permitted. Right? And if we have got a mass M equals true one kg and the force acting on it is equal to 10 new comes party. Here we have to calculate the amount of compression in the spring. So it is pretty clear from the spring the equation spring equation we have difficult cakes. Right? The forces 10 that is given. And here X. K equal 200 X. Right. So here we have X equal 2.1 meters. So this is the compression in the spring part of the here you have to calculate that some of the potential and kinetic energy of the system so equal to the potential energy in the spring is half K. X squared. And with the kinetic energy of the masses of M V squared here we have half time skis 100 times 0.1 square plus half the mosses. Quarter one kgs. And the velocity has given us four m plus two m per second. So to square. Right, if you calculate this, you will get 2.5 jewels. This is part next part C. Time period equal to 25 Duke over us in baking. This is equal to two pi The masses one and G is 100 trade seconds. So this is equal Hi everybody five seconds party, they have to calculator, amplitude of motion. Great. So here we can say the total energy is conserved. Okay, that means let's say we we take the amplitude as a. Okay, yeah, this means in extreme position, the compression of the supreme is equal to so that's a delta X equal to a plus X. Right now we actually start with the conservation of energy. So we have the total energy stored right at the extreme end it will be informal potential which is equal to half key is square right? Actually half A plus X. All squared right now this will be equal to this energy that is due to this compression and due to the velocity. Right? So here we have E plus the amount of work done by the force. So if dot exe right now we can say that half key. Here we need to calculate plus X. We know exist the compression which is 0.1 whole squared here he is 2.5 the forces 10 newtons times 0.1 great. Okay, here we have cake or 200. Really this X will be this is actually the right because due to this fourth body moved full employment. So this is it. This gives us Christie it's squared. Thus lost. Nx plus 0.5 he called you good point fight plus 10 X. Sorry can E right now solving this quadratic equation you will get X equal to or sorry equal to them to do the equal to 0.2 m. So this is the solution of party next year party. Now you have to calculate the potential and the extreme meant. So we we have already depend on the expression half K X plus a whole square. Right now you plug in the values the initial values you will get this equal to 4.5 jewels if we have the potential energy at the right extreme end is given us. So obviously because at the right and we have to force fight. So here we are half. Okay, X plus a whole square. Find us fifth kinds of where? Because today is the distance between two extremes. So this gives us if you calculate this will be cool too 0.5 jews. Now we have to tell the that we have to see from this that the difference different values in the part, B, E and F. Do not violate. Law of conservation of energy has work done by the external forces. There is work done by the external force. That force is equal attending constraint, so that doesn't violate the conservation of I hope this is clear to you and have a very good rest of the day. Thank you.

Okay, so we're asked to find our general solution for part A. But it might be better to solve for Part B first, because we need to know our frequency. Another Susan. See. But are W Valium? So let's start with far be I'll leave party for leader So we know that R M is equal to nine. That's nine x double primer plus Okay, which was 0.25 extra Secret's out. Now it's the wide by nine, you get one over six squared exit. So so when you know that, it'll be easy. But someone over six so t is equal to two pi over one of the six, which is 12 fine and in our physical to one over 12 pie. Okay, so we know our frequency is equal to one over six. So our general form of our solution is gonna be X is equal to a sign of one over six times T shirts t over six. Let's be coastline. Uh, see over six. Okay, you know, our nation value that is equal to zero, and that's given or not. That's equal 2.75 So we have a sign of so plus be co sign. I'm so is eagle 2.75 So we get that B is equal 2.75 Now we know our velocity That's equal to the derivative of our functions that we get. V is equal to one over six. Co sign vehicle sign of tea over six, minus one over six. Be sign of sea over six. Now let's plug in our initial value for our velocity. You get that one over six A is equal to zero. So in a a a sickle today. Okay, it's our general solution is it's ready to brand X o. T. Is equal to a which was ill and B which is 00.75 co sign of sea over six. Okay, now we're asked to grab our exit e function That's just gonna look like a co sign of lung function. So we're gonna have something like this. Get and for party for asks, you solve what X is equal to were asked to find when the first time exited deserve Oh, okay. So that's when t over six is basically so we have this fortune. So we need to know what this is. All so coastline and zero when we're pi over to sort of see over 16 sequel part over too. Someone produced, we get that T is equal to six by over two, which is 35

And this problem a block weighing 10 new twins attached at the lower end to a spring that has a spring constant of 200 new ones per meter. Um, and then it's attached to the ceiling, and the block is also leaving vertically and has a kinetic energy of two jewels. As it passes through the point where the spring is unstructured, we're being asked to find the period of oscillation and then, using the law of Conservation of energy, determine the maximum distance the block moves both above and below. The point at which the spring is stretched is un stretched. And then what is the amplitude of the oscillation? And finally, what is the maximum kinetic energy off the block as it oscillates, They're starting with the period before we're going to use the equation. Period equals to pie times, the square root of mass divided by spring constant. Um, and here we have Thio realize that we are not actually given the mass were given the weight. So when we plugged these values in, we're actually going to plug in to pi times, the square root of 10 divided by 9.8 over our spring constant of 200. And when we solve for that, we get that our period is 0.45 seconds. So we have an answer for a period 0.45 seconds by our next task is to figure out, um, the maximum distance the block moves above and below the point at which the spring is unstructured. Um, And in order to do that, first, to figure out what are what is the distance between the UN stretched length and the equilibrium lang with the mass attached. So to find that distance, we're going to use mass times gravity divided by our spring constant. Um and so our mass times gravity, that is our weight. So that's our 10 Newtons divided by our 200 which is our spring constant. And we get that that distance is 0.5 meters Now, during simple harmonic motion, Um, we like to establish X equals zero at the equilibrium length um, the middle of the oscillation and to read the total energy without any gravity terms. So we're going to write Total energy equals R kinetic energy, plus our potential energy and our potential energy equals 1/2 Okay, X squared. Uh, And so when the block is passing through the unstructured position, we're going to have that our energy equals our kinetic energy. Plus 1/2 Okay, times are 0.5 squared. Um, and we have values for K e E r. A kinetic energy is 2.0 plus 1/2 times, 200 times, 0.5 squared. So we get that our energy actually equals 2.25 jewels. Now, when the object is at the top most point or the bottom most point of the oscillation, all of the energy is going to be potential energy. So that means. And since it's a lot of conservation, if our total energy is 2.25 jewels than at the top of the bottom, it's also gonna be 2.25 Jules. And we can make that equal to our potential energy, which is 1/2 k x m squared on and substituting. And we have 2.25 equals 1/2 times 200 x m squared and solving for X m amplitude. We get that it 0.15 meters. Um, but that doesn't necessarily mean that that is how far it is from the unstrapped or stretched from the stretch position. So we need to take it this value into consideration and add or subtract the 0.5 meters value we found above to get the top most in bottom, most position. So our cop most position, the highest position is actually going to be 0.15 meters minus 0.5 meters we found above. So the highest point is gonna be 0.1 meters and then the lowest we're actually gonna add 0.15 plus 0.5 So the lowest position, it's going to be 0.2 meters. Um, so those are gonna be our to our max. These two values here go up here for our max positions, and then it's far is our, um our amplitude. We found that here. So our amplitude is zero point foot 0.15 meters and then for our maximum kinetic energy is gonna be equal to the maximum potential energy. And we found that that is actually 2.25 jewels

Hello and welcome to this video solution of numerous. This question is based on the principles of simple harmonic motion. So this situation based on a spring my system and we have got the mass of the object as two kg. So let me tabulate the given parameters right away. We have mass M equal to two cages. Now it is acted upon by an horizontal force of 20 newtons That is required to hold the object at rest when it's pulled .2 m from the equilibrium position. So there is a extension of this mask which is equal to 0.2 m from the equilibrium point. And this is held by a force that is equal to Np Newton's. Now here we have to calculate the force consent of the spring. So party forced consent has given us F equal to from the basic equation for situation of spring, which is difficult to cakes. Now here we have forced us 20, which is equal to K. We have to calculate time. Success. It opened to great. From here we have two equal to 100 newtons per meter. So this is party part B. We have to calculate the frequency of oscillation. So omega is the angular frequency of oscillation which is calculated as but overall cable. Yeah, if you put this value of K and then you will get uh route 50 dads two radiance first again, omega is actually if you know maybe I called Jeff. So if you want to convert this into read in hearts, then you have to calculate if So if we will get to pay by omega, omega by to pay Which is 1.13 heads. So there's an answer to part B. Now let's move on to part C. Here we have to calculate the maximum velocity we have v max equal to oh my God, Now Omega, we already know it's route 50 Times is 0.2. The one for me. Days later this gives us meters per second per second. So this gives us 1.14 meters per second. There's a maximum velocity of population part D we have to calculate. Okay, so here it is asking that where is this maximum speed? Oprah Okay, so maximum speed walkers have the equilibrium point. That is that X equal to zero. We have caught maximum speed max speed or velocity whatever. It's a maximum velocity question. E now you two for fire. Okay. Now the question is to calibrate the maximum acceleration. So acceleration of maximum acceleration is equal to omega square Mhm. So it's equal to 50 times. Point to this is equal to 10 m past. Against where? Okay, so now we have to find out the that the distance where this equilibrium where this acceleration is maximum. Right? So the acceleration is maximum attics equal to plus my nancy. That is at the extreme ends. He's X equal to plus minus of 0.00.2. We do straight we have to we have in these positions we have caught we have the maximum accelerations fine. Now we have part G. Okay so we have to calculate the total energy of oscillation of the system. So total energy is given us half. Okay is square. This is equal to half times How much we got GSK 800 Times 0.2 Squared this is jules which just gives us regions. Next we have got part age. Now it is given that what is the speed of oscillation? When the position is 1/3 of the maximum value, That means we have X equal to a way through Which is equal to 0.2 by three m. The speed is given by an expression equal to omega route over over a squared minus X squared. Now you have to plug in the values of A and X. Here and omega, we know this gives us the velocity At x equal to a by three. Call to 1.33 m. again. Similarly, we have to calculate the acceleration i at that same point I acceleration equal to omega's critics here. You have to plug in the value of omega. Yes, it took 50 So 50 times of and so Excess. He had 2.0.2 x three, two by three meters per second squared. This gives us 3.33. I hope this is clear to you and have a very good rest of the they.


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