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Homework Help: Simple harmonic motion and energy - yet another answer key disagreement?

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm.
    When it is at x = x1, its kinetic energy is K = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = –1/2xm, the kinetic and potential energies are:

    A) K = 5J and U = 3J
    B) K = 5J and U = –3J
    C) K = 8J and U = 0
    D) K = 0 and U = 8J
    E) K = 0 and U = –8J

    2. Relevant equations

    [tex]
    E = 1/2kA^2
    [/tex]

    [tex]1/2kA^2 = 1/2mv^2 + 1/2kx^2[/tex]




    3. The attempt at a solution

    [tex]
    1/2kA^2 = 8[/tex]

    [tex]1/2mv^2 + 1/2kx^2^ = 8[/tex]

    [tex]A = 4/\sqrt{k}[/tex]

    [tex]x = -1/2(4/\sqrt{k}) = -2/\sqrt{k}[/tex]

    [tex]1/2mv^2 + 1/2k(-2/\sqrt{k})^2^ = 8[/tex]

    [tex]1/2mv^2 + 2 = 8[/tex]

    [tex]K = 6 J[/tex]

    [tex]U = 8 - 6 = 2 J

    [/tex]


    As you see, this is not one of the choices. Am I doing something wrong?
     
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 24, 2008 #2
    Sorry about the bad formatting earlier; I have fixed it now.
     
  4. Nov 24, 2008 #3
    Looking at it, I guess by process of elimination it has to A because energy isn't negative, thus B, C, and E are ruled out. And it cannot have full kinetic energy between equilibrium and amplitude, so D is ruled out.

    However, it disagrees with my calculations.
     
  5. Nov 24, 2008 #4

    LowlyPion

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    Homework Helper

    Don't you know what the total energy in the system is at x1?

    Don't you also know at x = 0 is where U = 0? And also at Xm is where K = 0, so ... which answer meets these requirements?

    Edit: Ooops. Looks like you figured it out.
     
  6. Nov 24, 2008 #5

    gabbagabbahey

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    Homework Helper
    Gold Member

    It looks to me like your original solution is fine, and the answer key is incorrect....

    The total energy is 8J, so [tex]\frac{1}{2}kA^2=8 \, \text{J}[/tex]. And so at [tex]x=\frac{-x_m}{2}=\frac{-A}{2}[/tex] , the potential energy is:

    [tex]U(x)=\frac{1}{2}kx^2=\frac{1}{2}k \left( \frac{-A}{2} \right)^2=\frac{1}{4} \left( \frac{1}{2}kA^2 \right)= \frac{1}{4}(8 \, \text{J})=2 \, \text{J}[/tex]
     
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