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$\bullet$ $\bullet$ An apple weighs 1.00 $\mathrm{N}$ . When you hang it from the end of a long spring of force constant 1.50 $\mathrm{N} / \mathrm{m}$ and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

$L=2.67 \mathrm{m}$

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

University of Sheffield

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All right. So the question was asking us What is the are stretched length off this spring? Well, first, let's determine the bouncing frequency, which is the frequency in a simple harmonic motion. Where does he go to anger the frequency Do you want to? And we know the anger of frequencies able to square Okay over him, which is force constant. Over mass. We substitute squaring Kale in for the Omega. We got frequency over simple harmony. Motion. Everyone is equal to the square root K or M. Do you want to buy? Okay, we know the force constants given us 1.50 Newtown per meter. Mass of the object, which is a massive apple, is you go to force of habit with just one new tongue to ever not going any meter per second square. Okay? No, If you're plugging, we'll get square A 1.5 new tom per meter divided by 1/9 point A. You never for a second square okay, over to buy moves in the crowd a year. Oh, we're to buy. This will give us the frequency is equal to about 0.61 hers. Okay. And in a question. It was saying that the frequency oldest simple pendulum is half the bounce frequency. So the frequency of a simple pendulum you see, with the 1/2 off the balance frequency this will give us I want to buy two was just 0.61 of hers. You have I to which is equal to 0.305 hers. Okay. And we know that the frequency off a pencil in you seek equal to the reciprocal off the time here. Was this already Just one over to buy square. Uh oh, gee, because t is equal to square. I stayed with the to fight hand square. Well, ology okay. And we're looking for to determine the value for the stretch of land here, which is al here. Okay, so you would do sound implication here. We'll get ah, you see simplification. So we get to pie. Times Square well over G is equal to one over. Have to and square well over G is equal to while we're at two times 1/2 of it, sometimes one over to buy. OK, And I would get we can get a l is just simply equal to well over after a square times G over four by it's where. Okay. And were you planning the value we get 1/2. I mean, I'm sorry. One over after two square was just one over 0.305 hers square. Okay, times gravitational constant. We just 9.8, you know, for second square, over full by square. And this will give us Yeah. Stress on stretch land over the spring is equal to 2.67 neither. Okay, And that's the answer for this question. Thank you.

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