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Oscillator with equal amounts of kinetic and potential energy

  • Thread starter Dwrigh08
  • Start date
  • #1
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Homework Statement



If you have a simple harmonic oscillator based on a spring of force constant 5.79 N/m, with an attached mass of 0.828 kg, and the oscillator is initially displaced 3.5 cm from equilibrium, at what distance from the equilibrium point (in cm) will the oscillator have equal amounts of kinetic and potential energy?


Homework Equations



I am using Kinetic energy=1/2 mass x omega^2 and Potential energy= 1/2 spring constant x displacement^2 and omega= root of (spring constant/ mass)

The Attempt at a Solution



My first thought was to set the kinetic energy and pitential equal to each other and solve for the displacement. When I try to do this I end up with 1. My next thought was to take into account the initial displacement and subtract that from 1. Neither work for me. I feel like i am missing an obvious step. Any help would be appreciated.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Can you wright down the expressions for kinetic energy and potential energy for a SHM in terms of k, amplitude and displacement?
 
  • #3
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I know that total energy= 1/2 k A^2. I tried to solve for total energy and then using E= K+U and rewriting E= 1/2 m w^2 + 1/2 k x^2. I come out with an energy from the first equation and then try to solve for x using the second equation. I dont think that is the right approach.
 
  • #4
rl.bhat
Homework Helper
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In SHM, at any instant, the kinetic energy = 1/2*m*w^2(A^2-x^2) = 1/2*k*(A^2-x^2)and potential energy = 1/2*k*x^2
Equate KE = PE and solve for x.
 

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