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Finding x when Elastic Potential Energy equals Kinetic Energy 
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#1
Jan2510, 09:05 PM

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Here's the problem I've been working on:
"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?" Now I know that Us = 0.5*k*x^2 and K = 0.5*m*v^2. I first thought that the answer would be .5 (half of A) since that would be halfway between fully stretched and at equilibrium, then I thought maybe at equilibrium, but neither worked. I also then tried to set the two equations equal to one another, but can't go anywhere with this. Any ideas? This is a fairly simple question that is completely stumping me... 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Jan2510, 09:19 PM

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#3
Jan2510, 09:29 PM

P: 2

Ok, so I tried setting 0.5kx^2 = 0.5mv^2, and end up with x = sqrt(mv/k), which tells me nothing because I don't know v, m, or k. I tried using conservation of mechanical energy by saying MEi = 0.5kxi^2 + 0 (which means there's only potential energy in the system) and MEf = 0.5kxf^2 + 0.5mvf^2, and setting MEi = MEf, but still got nowhere... grr...



#4
Jan2610, 05:25 AM

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Finding x when Elastic Potential Energy equals Kinetic Energy
When you use mechanical energy conservation, you say that the sum of kinetic plus potential energy at one point is the same as kinetic plus potential energy at another point. I am sure you know this but you are not applying it correctly.
When the mechanical energy is all in the potential energy form at displacement x = A, you have ME = (1/2)kA^{2} where A is the amplitude. When the kinetic energy is equal to the potential energy at displacement x < A ME = (1/2)kx^{2} + (1/2)mv^{2} Now you set KE = PE in the last expression and demand that the ME at displacement x be the same as the ME at x = A. That's mechanical energy conservation. 


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