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Displacement x of simple harmonic oscillation

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    The amplitude of simple harmonic oscillation is A = 10 cm. Find a displacement x when K = 1/6 U. Here K is a kinetic energy and U is a potential energy.


    2. Relevant equations
    [tex]KE = \frac{1}{2} k A^2[/tex]
    [tex]U = \frac{1}{2} k x^2[/tex]

    3. The attempt at a solution

    I'm not sure the correct method of doing this problem but here is what I have attempted based on an example from my notes:

    [tex]KE = \frac{1}{6} U[/tex]
    [tex]\frac{1}{2} k A^2 = \frac{1}{6} \frac{1}{2} k x^2[/tex]
    [tex]A^2 = \frac{1}{6} x^2[/tex]
    [tex](0.1)^2 = \frac{1}{6} x^2[/tex]
    [tex]x \approx 0.24 m[/tex]
     
  2. jcsd
  3. Nov 20, 2011 #2

    vela

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    Your formula for kinetic energy isn't correct.
     
  4. Nov 20, 2011 #3
    You also need another equation (or law) regarding energy.
     
  5. Nov 21, 2011 #4
    This may be it, I think:

    [tex]K = \frac{1}{6} U[/tex]
    [tex]U = \frac{1}{2} kx^2[/tex]
    [tex]KE = U + K = U + \frac{1}{6} U = \frac{7}{6} U = \frac {7}{6} \frac {1}{2} kx^2[/tex]
    [tex]KE = \frac {1}{2} kA^2[/tex]
    [tex] \frac{1}{2} kA^2 = \frac {7}{6} \frac {1}{2} kx^2[/tex]
    [tex]A^2 = \frac{7}{6}x^2[/tex]
    [tex](0.1)^2 = \frac {7}{6}x^2[/tex]
    [tex]x \approx 0.0926 m [/tex]
     
  6. Nov 21, 2011 #5

    vela

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    Looks good.

    By the way, what does KE stand for? In my first post, I mistakenly thought you were referring to the kinetic energy as KE is a common abbreviation for it.
     
  7. Nov 21, 2011 #6
    I was using KE as the kinetic energy of the spring... wasn't sure if energy in a spring should be referred to as kinetic energy or just energy.

    Should it just be E for energy?
     
  8. Nov 22, 2011 #7

    vela

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    You're using the approximation that the spring is massless, so it has no kinetic energy. It only has potential energy. Your KE is the total energy of the system.
     
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