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Finding displacement from the equilibrium

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Give the following answer in terms of the amplitude, A. At what displacement from equilibrium is the energy of a Simple Harmonic Oscillator two-thirds KE and one-third PE?

    2. Relevant equations
    I would guess
    E=.5(k)(x)^2 + .5(m)(v)^2
    If one is 2/3, would I change the 1/2 in it's particular equation to 2/3? What would I do from there?

    3. The attempt at a solution
    So far I have no idea where to even begin.
  2. jcsd
  3. Nov 6, 2009 #2

    Let me see if I can lead you in the right direction. You've got a couple of formulas there for the kinetic energy and the potential energy-- these can be defined further for simple harmonic motion, since the functions x(t) and v(t) are known.

    Once you have these formulas, which contain the amplitude, you can set the 2/3 / 1/3 ratio and find x.

    Have a look at the wikipedia page-- specifically, scroll down to Dynamics of simple harmonic motion and then, most importantly, Energy of simple harmonic motion.

    http://en.wikipedia.org/wiki/Simple_harmonic_motion" [Broken]

    Post again if you're still stuck. Good luck!
    Last edited by a moderator: May 4, 2017
  4. Nov 6, 2009 #3
    Hello and thank you for the answer!

    Unfortunately, I'm still a bit stuck. There is an example in the book that is the exact same question but the energy is split 50/50 instead of 1/3 and 2/3. They got an answer of +/-.707.

    How they got to that answer is completely beyond me.
  5. Nov 6, 2009 #4
    I'm sorry, this problem is actually easier than I thought it was.

    EDIT TO ADD: I just want to add, that by the above, I don't mean to imply that the problem is easy or hard or whatever. Only that you don't have to go into a whole lot of detail with the sin/cos stuff that I originally thought you did to arrive at the answer.

    You have given the following equations in your first post:

    [tex]U = \frac{1}{2}kx^2 [/tex]
    [tex]K = \frac{1}{2}mv^2 [/tex]

    [tex]E = U + K = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 [/tex]

    Now if only your total energy formula was expressed in a form that had the Amplitude, you would have an equation that you can use to determine x or v (since you know the ratio you want).

    Fortunately, you can do just that, using the equations from the wiki page I mentioned. The important result, for you, is that:

    [tex]E = U + K = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kA^2[/tex]
    Last edited: Nov 6, 2009
  6. Nov 6, 2009 #5
    Okay I understand that, but how do I turn the .5(k)(A)^2 into an answer?
  7. Nov 6, 2009 #6
    I've given you the missing piece you need. What have you tried?

    You have a formula for the potential energy. You have a formula for the total energy. You have a ratio of the total energy to the potential energy, for this particular problem.

    How can you put it all together?

    The reason I ask is that going any further will amount to basically giving you the answer.
  8. Nov 6, 2009 #7
    Every time I do it I get 1/3*A... but that isn't right.
  9. Nov 6, 2009 #8


    User Avatar
    Homework Helper

    Displacement in a SHM is given by
    x = Asinωt
    Velocity will be
    v = dx/dt = Aωcos(ωt) = ωsqrt(A^2 - x^2) ans E = 1/2*m*(A^2 - x^2)---(1)
    Velocity is maximum when x = 0. So vmax = Aω and Emax = 1/2*m*vmax^2 = 1/2*m*ω^2*A^2 ----(2)
    Ratio of E/Eamx is given. From equation 1and 2 find x.
  10. Nov 6, 2009 #9
    Okay this time I ended up with x=1.5A

    Sound anywhere near correct?
  11. Nov 7, 2009 #10


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    Homework Helper

    E = 2/3*Emax
    E/Emax = 2/3
    (A^2 - x^2)/A^2 = 2/3
    Solve for x.
  12. Nov 7, 2009 #11
    No. You should know this is incorrect physically. Why?

    That said, rl.bhat's way is the complicated way to solve this. If you use the potential energy term, which is already in [itex]x^2[/itex], you don't need to calculate the velocity to get at x. For example, in the example you said where they want to know at what x the energy is split 50/50, you know 50% of the total energy is equal to the potential energy:

    [tex]\frac{1}{2}(\frac{1}{2}kA^2) = \frac{1}{2}kx^2[/tex]

    [tex]\rightarrow \frac{1}{2}A^2 = x^2 [/tex]

    [tex]\rightarrow \frac{1}{\sqrt2}A = x [/tex]

    [tex]\rightarrow \pm .707A \simeq x [/tex]

    You can use this method for any proportion of U and K.
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