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Math Help - mass between two springs equation help

  1. #1
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    mass between two springs equation help

    can some one please help with the equation of motion, I have attached the question. its confusing me cause there are two springs attached to one mass,

    what I get is

     m \frac{d^{2}x}{dt^{2}} = -k_{1}(x-x_{0}) + k_{2} (l-x_{0})

    but than when I solve for the equilibrium position I get

     x = \frac{k_{1}-k_{2})x_{0} + k_{2}l}{k_{1}}

    but the correct answer is

     x = \frac{k_{1}-k_{2})x_{0} + k_{2}l}{k_{1} + k_{2}}

    I dont understand how it is k_{1} + k_{2}

    so my this equation  -k_{1}(x-x_{0}) + k_{2} (l-x_{0}) = 0  must be wrong

    can anyone suggest how to get the correct equation for the equilibrium position?



    thanks
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  2. #2
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    Re: mass between two springs equation help

    Let F_1 be the force exerted by the spring with Hooke's constant k_1 and F_2 be the force exerted by the spring with Hooke's constant k_2. At x=x_0, you know F_1=0. At x = l-x_0, you know F_2 = 0, so F_1 = -(x-x_0)k_1. For F_2, when x<l-x_0, F_2>0 and when x>l-x_0, F_2<0, so F_2 = -(x-(l-x_0))k_2 = -(x-l+x_0)k_2

    Then, F_1+F_2=0 when -(x-x_0)k_1-(x-l+x_0)k_2=0, so (k_1+k_2)x=(l-x_0)k_2+x_0k_1. Then the forces are equal when

    x = \dfrac{x_0k_1+(l-x_0)k_2}{k_1+k_2} = \dfrac{(k_1-k_2)x_0+k_2l}{k_1+k_2} just as expected.
    Thanks from Tweety
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