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Energy of simple harmonic oscillator

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle moves along x-axis subject to a force toward the origin proportional to -kx. Find kinetic (K) and potential (P) energy as functions of time t, and show that total energy is contant.

    2. Relevant equations

    K = (1/2)m*v^2
    P = (1/2)k*x^2
    E = K+P

    x = Asin(wt + [tex]\tau[/tex])

    v = dx/dt = wA(cos(wt + [tex]\tau[/tex])

    3. The attempt at a solution

    K = (1/2)m*v^2 = (1/2)m*w^2A^2(cos^2(wt + [tex]\tau[/tex])

    P = (1/2)k*x^2 = (1/2)k*A^2(sin^2(wt + [tex]\tau[/tex]))

    But when I add these to get the total energy, the terms with t do not cancel, and so the total energy is not constant. I can only imagine then that I've done something wrong in the above, very basic steps. Any suggestions would be appreciated. Thanks.
     
  2. jcsd
  3. Nov 19, 2008 #2
    Another equation that comes in handy is w^2 = k/m. Rearranging this you get k = m*w^2 which can be substituted into the first equation, to get K = (1/2)m*v^2 = (1/2)k*A^2(cos^2(wt + [tex]\tau[/tex] )).

    When adding K+P, the (1/2)k*A^2 (which is present in both terms) can now be taken as a common factor and using the trigonometric identity cos^2(x) + sin^2(x) = 1 shows the energy is constant.
     
  4. Nov 19, 2008 #3
    Of course, that seems so obvious now! Thank you very much!
     
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