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The position of a mass oscillating on a spring is given by € (8.2cm) cos [2nt / (0.56s)] -...

Question

The position of a mass oscillating on a spring is given by € (8.2cm) cos [2nt / (0.56s)] -

The position of a mass oscillating on a spring is given by € (8.2cm) cos [2nt / (0.56s)] -



Answers

The position of a mass oscillating on a spring is given by
$x=(7.8 \mathrm{cm}) \cos [2 \pi t /(0.68 \mathrm{s})] .$ (a) What is the frequency of this
motion? (b) When is the mass first at the position $x=-7.8 \mathrm{cm} ?$

So the party weaken Identify the period using the general function. We know that the general function would be to but rather would be co sign of two pi divided by the period t multiplied by the time T and the function is equaling to co sign of to pie over 0.38 seconds multiplied by t. And so therefore, we can say that the period T is equaling 0.38 seconds for part B. The 1st 0 crossing for a co sign function occurs at 1/4 of the period. So we can say that t equaling 1/4 of the period 0.38 seconds. This would be equaling 2.95 seconds. That is the end of the solution. Thank you for watching.

Okay, So when the Connecticut Angie, you could show the potential energy. So we know that the total energy could you I don't have the energy. Bless attention. All right, So when these two are equal, so it's equal to two times the potential energy, which is to as Yes. Hey, ex squid, right? The X is what we're looking for, and we know that the total energy is equal to the Maximo potential energy. Right? So this isn't a half day a script, all right? So we could pay Big s owed half. And after Castle, you wouldn't know that x equal to hey over two. It's simple.

So the question was asking us, uh, how far is this mass front equally been? Position off the spring when the last potential ng's equal for the Canadian, Which means he's asking us to find out the displacement when the kinetic energy is equal to a less the elastic potential energy. Okay, we just won't have any square. You won't have k X square. We don't have these for the Omega Times square of a score. Months acts where we used to speed here will make us the angle of frequency is the M V two X. It is the displacement we're looking for and we know that Omega I wish that the angular frequency is equal to school square off there. Force constant divided by the mass which is k o. M here, okay, and we plugging back to the the equation. So we use this square and KLM substitute for omega. We get that bees and simply equal to the square root K over m Times Square, a square minus expert. And which means that the square is equal to K. M times a square minus X square. Okay, and we plug him back two originally, inclusion here we get it 1/2 and we swear we just won't have em times. Okay, well, and times a square minus X square equal to won't have. Okay, X where? And you see your m and M. Cancel out we get won't have. Okay, a square minus X square equal to we'll have chaos square in this case, you see won't have K and won't have k cancel also, which means that a square minus X squared is equal to X squared. So a square is equal to to act square. Okay, so we know that since a scores you go to to ask where so a is able to square root of two times X So displacement we're looking for is equal to a over square over to or if you want to, uh, make it look a bit a little bit nicer, which is square to over a over to. Okay, so that's the answer for this question. Thank you.

All right. We're told that they were given a position function, and we're told that Omega and A are going to be Constance that being said and go ahead and fine arts review, And that's going to be equal to a times omega, no sign of Omega T and that is equal to our velocity. I also find our second derivative. That's double, and that is going to be equal to a Omega squared times. Negative sign of negatives. Aaron for the negative out front sign. Oh my God d All right, and that is equal to or exoneration. Now we can show that our acceleration Is that a max moving out of the lawsuit? Zero in our velocity, that zero where accelerations. And that's because the one off the acceleration are always positive numbers. That being said, if we look at this graph, the red function is our velocity, and the blue function is our acceleration. Any time our acceleration is that zero our losses at a maximum at any time. Our velocity is at zero, as I've already marked. These acceleration is the next month that shows that anytime you have a lost season, zero your explorations in person. By the way, this graph was created plugging in one for an Omega. The reason for that is they're both constant. So they're not actually going to change. Will be cops found throughout the whole problem. Problem. Just change. What? The graph. It looks like a little bit.


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