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Simple harmonic motion of two horizontal springs

  1. Sep 12, 2004 #1
    Figure 1 shows a mass m in equilibrium between two stretched helical springs on a smooth horizontal surface. k_1 and k_2 are the force constants of the springs and e_1 , e_2 are their respective equilibrium extensions.
    The mass m is displaced X to one side ( X is smaller than e_1 and e_2 ), and then released. how you show that the acceleration (a) of the mass when its displacement from the equilibrium position is x obey the following equation :
    a = - [ (k_1) + (k_2) ]*x / m

    Please show me step by step !! thanks you.

    Attached Files:

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  2. jcsd
  3. Sep 12, 2004 #2


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    The force is
    [tex]F(x) = k_1 (x - x_{1E}) + k_2 (x - x_{2E})[/itex]
    Simply note that F(0) = 0 and your desired result follows.
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