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A spring that is 20 cm long is hung from the ceiling 800 mass is attached to the bottom of the spring and the new length of the spring is 34.6 cm: The mass then pul...

Question

A spring that is 20 cm long is hung from the ceiling 800 mass is attached to the bottom of the spring and the new length of the spring is 34.6 cm: The mass then pulled downward until the length of the spring is 41.0 cm and released_ What is the spring constant k? What is the amplitude of oscillation?What is the period T?What is the maximum velocity Vmar of the mass during an oscillation? What is the maximum acceleration &ma during= an oscillation? What is the frequency_

A spring that is 20 cm long is hung from the ceiling 800 mass is attached to the bottom of the spring and the new length of the spring is 34.6 cm: The mass then pulled downward until the length of the spring is 41.0 cm and released_ What is the spring constant k? What is the amplitude of oscillation? What is the period T? What is the maximum velocity Vmar of the mass during an oscillation? What is the maximum acceleration &ma during= an oscillation? What is the frequency_



Answers

A spring with spring constant $k$ is suspended vertically from a support and a mass $m$ is attached. The mass is held at the point where the spring is not stretched. Then the mass is released and begins to oscillate. The lowest point in the oscillation is $20 \mathrm{cm}$ below the point where the mass was released. What is the oscillation frequency?

In this problem, it is given a spring is suspended from the ceiling. Suppose this is the screen and it it's suspending, suspended from the ceiling and the Mars Suppose them it is given here 0.250 k g is connected to this spring and here it is, given that it is stretch the spring by 20 centimeters before coming to rest at its lowest one supports After the stretching the spring this block comes here and these distances 20 centimeter. That means this block is doing harmonic motion about this mean point and we can we know that that means position key X Spring force will be equal to the gravitational force. Your gravitational pull which is equal to M G. So from here we can get the value of K, which is mg by ex mgs 0.25 into G s, 9.81 divided by X, which is 0.10 So the law off they will be equal toe 25 Newton. Well, meter, this is the spring constant for this question now inquired to be, we have to find the amplitude so we can see here in this diagram. This is amplitude and it is half off the total distance traveled, so he will be equal to 25 to which is equal to 10 centimeter. Knowing about three or part C in part C. What is the frequency of the installation? We know that formula for frequency frequency is equal to one upon to five on the roof cave. I am putting the value off K and M the value off case 25. The value off Emmys 0.25 sweet will be around 1.6 hers.

I I prevent. This is the problem. Basement concept of time. Period of loaded spring given by two pi root off and upon King I scream is suspended from it vertically. Yes. Mm. Foods can be calculated as the mission of every school to PX minus. Empties Card to Jerusalem. Case car two mg by X mass of suspending particles. 125 g is nine pointed an extension in the Spanish point when it's okay. Skardu 24.5 Newton parameter. This is the answer of party. Inspired to be We have to fight. Yeah, the applicant and amplitude of oscillation with me Fight centimetres. A total distance prevalent in Vienna solution. Mm. 10 centimeter. Simple frequency of oscillation is receive the color of time period, that is man upon to wrote off. Okay. Opponent Kate is given from deep 4.5 upon mass. That is 0.25 So it is 1.6 months, That's all. Thanks for watching it. Just a moment. Hard size have tried very clearly. This is a moment, please. That's all. Thanks for watching it

So this question belongs to the oscillation in which we have given the spring constant K. This is equal to 1200 newton per meter. And it is mounted on the horizontal table and Hamas, M is given as three kg. And uh we have given the displacement, it is equal to 2.0 centimeters. This is equal to 0.2 m. Okay, so we have to calculate here the frequency for the part A. Okay, so calculating frequency to frequency from the frequency is given by one by two pi under root of K. Buy em So we have all the values. So this is equal to one by two by underwrote of K, which is 1200 M, which is three. So from here frequency comes out to be 3.18 hertz. So this becomes the answer for this problem. Okay. For the part eight. Now moving to the part second. Okay. Which is uh in which we have to calculate the maximum acceleration, so acceleration maximum. this is equal to omega square marker being completed. A and omega it is equal to uh under root of K by m. So and holy square A. So this is equal to okay, bye M. Player by eight. Okay, so substituting the value. So maximum acceleration will be equals two K, which is 1200 amplitude which is 0.2 divided by mass and which is three kg. So from here, maximum acceleration comes out to be 8.0 m per second squared. So this becomes the answer for the second part. Okay, now moving to the third part, in which we have to calculate the maximum speed. So maximum speed. We it is equal to omega A. So omega this is under root of K by M. R b, amplitude A. So substituting values under root of 1200. They were by three more player B amplitude 0.2 m. So from here, after solving maximum speed, we this will be equals to 0.4 m per second. So this will be the answer for the third part of the problem. Okay, thank you.

So we're given an object that undergo simple harmonic motion with an amplitude of 12 centimeters, which we might want to use meters, sometimes to on the acceleration is given by 108 centimetres per second squared restaurant A whole bunch of things that first goes to find the period trying to find the period need to relate acceleration and angular accelerant, acceleration and the imp itude. And we know that Omega's a equals a omega squared. So omega is the square root of X l maximum acceleration over that. If you plug that in you actually with three parts, right? So now we're just gonna kind of go down the list. We know that the period is two pi over Omega, right? We're just two. Pi over three is about 2.1 seconds. It doesn't depend on the mass. The linear frequency is one over that. So it's about 0.48 hurts doesn't depend on the mass. The maximum speed that this object will go is a Omega A is 12 centimeters and omega is three. So that's 36 centimetres per second. Still doesn't mean of the mass, uh, next rest to find the energy that's given by 1/2 a k A squared, um, or you could do 1/2 him V squared. Either one watch. You get it. In this case, we haven't found K yet, so I'm gonna use the 2nd 1 We have v here that we can just plug in. Um, and if you do it appropriately in s I where I'll use meters per second, I actually find that this is 0.65 times the mass right, which we leave is a variable here. And then finally be spring constant, which you could get from equating the energy here or just rearranging our angular frequency one. And so you end up with nine times in, right. And so you will notice The only dependence is on em are right here. Right? The energy and spring constant. Both are linearly proportional, but otherwise, there's no other, uh, ems anywhere. Right? Nothing else depends on it.


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