Finding x when Elastic Potential Energy equals Kinetic Energy

rkelley7891

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

kuruman

That's what you should do. How far did you get? Try again. Don't forget that mechanical energy is conserved, which means that the ME at the point where potential energy is equal to the kinetic energy, the total ME is the same as at the point where the kinetic energy is zero, i.e. the point where all the mechanical energy is in the form of potential energy.

rkelley7891

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...

kuruman

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)kA² where A is the amplitude.

When the kinetic energy is equal to the potential energy at displacement x < A

ME = (1/2)kx² + (1/2)mv²

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.