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SHM - Horizontal Spring Question

  1. Dec 16, 2007 #1
    1. A 41g object is attached to a horizontal spring with a spring constant of 15N/m and released from rest with an amplitude of 22.4cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? Answer in units of m/s.



    2. Relevant equations
    W=Square Root of (K/m)
    1m=100cm
    V^2+(wx)^2=(wA)^2
    w= omega
    A= Amplitude


    3. The attempt at a solution
    K= 15N/m
    m= 41kg
    A=22.4cm=0.224m
    W=sq root of K/m = 0.605

    V^2=(wA)^2
    V = 0.0677
     
    Last edited: Dec 16, 2007
  2. jcsd
  3. Dec 16, 2007 #2
    Halfway to the equilibrium position would be half the amplitude.
     
  4. Nov 18, 2008 #3
    I have the same problem. I set it up two different ways.
    v^2=k/mA^2-K/mx^2, then square root
    A-full amplitude
    X-half the amplitude
    or
    v=[(k/m)(A^2*X^2)]^(-1/2)

    but none worked.

    I'm pretty sure I'm setting up the amplitude stuff wrong.
     
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