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$\bullet$ $\bullet$ One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.355 $\mathrm{kg}$ . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 1.80 $\mathrm{m}$ and $1.06 \mathrm{m},$ and the period of the oscillation is $2.15 \mathrm{s},$ find (a) the force constant of the spring, (b) the maximum speed of the glider, and (c) the magnitude of the maximum acceleration of the glider.

a) 3.03 $\mathrm{N} / \mathrm{m}$b) 1.08 $\mathrm{m} / \mathrm{s}$c) $=3.16 \mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Lectures

04:16

In mathematics, a proof is…

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$\bullet$ A 0.500 kg glide…

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A 1.00 -kg glider attached…

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On a frictionless, horizon…

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A $1.00-\mathrm{kg}$ glide…

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A 175 -g glider ona horizo…

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A 175-g glider on a horizo…

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A $175-$ glider on a horiz…

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A glider of mass 0.150 $\m…

Suppose you attach the obj…

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A 2.00 -kg object is attac…

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A $2.00-\mathrm{kg}$ objec…

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An ideal spring has a spri…

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A massless spring with spr…

03:03

The period of motion of an…

Okay, So first question was asking us what is the force constant. While four constant K is equal to mbabic us where another massive school 0.35 And we can determine Omega as well Omega, which is the anger frequency secretly to high frequency and frequencies. Well, over time, here's there's two pi wall over time here I remember in question the time Paris given, which is t equal to two point one C 15 second and this will give us Omega is even true Pie times one over. So going one fire second, which just about 2.9 to to radiant her second. If we plug it back to the K equation, you get K is equal to and which is zero point 355 kilogram times 2.922 radian for second, the whole thing square. And this will give us the force. Constant is about 3.3 new tongue per meter. Okay, so the second question was asking us What is the Maximus be all the glider? Well, the Maximus be is equal to a omega. We just empty to you has the angular frequency and how home in determining amplitude member in question. You were saying that the glad er is doing an ace. The simple hormone emotion okay, and was keep bouncing between the one point a mirror and the 1.6 meter, which means that the amplitude is simply equal to one point. Hey, zero meter minus 1.6 Meaner. Do you have a right to Okay, I come out with such a question. Well, let's take a look. You know, single homily motion like this, right? We used lying here, which means that it was Oh, this is 1.8 on the top, and this is 1.6. So, between there, this is the two times off the amputee. Okay, So in order to find out an empty we need to use 1.8 minus 1.6 and you want to Okay, so this will give us the amplitude is about 0.37 meter. So the max mus be is equal to 0.37 meter times The angular frequency, which is 2.922 radiant for a second. This Well, give us the maximus be is 1.0 a. Neither per second. Okay, lets go. Get a question at sea, which I believe the last question but his problem. So the because I see you was asking about what he was asking us. What is the maximum acceleration so well, we know maximum acceleration. It's equals who? A We'll make a square. She's We know A and Omega us so we can just plug in a valley with just 0.37 meter times Omega square, which is 2.922 radiant for secon square. And this will give us 3.16 meter for seconds. Where and these are the answer for the question. Thank you.

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