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$\bullet$ $\bullet$ A mass is oscillating with amplitude $A$ at the end of a spring. How far (in terms of $A$ ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

$x=\frac{A}{\sqrt{2}}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

Cornell University

University of Michigan - Ann Arbor

Hope College

Lectures

04:12

In physics, potential ener…

02:18

In physics, an oscillation…

01:23

A $450-\mathrm{g}$ mass on…

01:41

A spring with spring const…

03:03

The period of motion of an…

03:50

$\bullet$ A mass of 0.20 $…

01:11

Mass-Spring System The fre…

03:04

A block of mass M is conne…

07:21

A block of mass $M$ is co…

01:35

CE A mass on a spring osci…

09:28

Suppose you have a 0.750-k…

08:14

A force of $3 \mathrm{N}$ …

02:55

A pendulum of length $L$ a…

02:03

A mass $m$ is connected to…

04:15

$\bullet$ $\bullet$ An obj…

01:22

A 1.6 -kg mass attached to…

04:16

Suppose you attach the obj…

00:44

A 0.49 -kg mass attached t…

05:33

A 3-kg mass is attached to…

03:00

Suppose you have a $0.750-…

01:48

The period of oscillation …

02:23

A massless spring with spr…

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.

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