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Mechanics, Potential Energy of a Spring (Lagrangian Mechanics) watch

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    I'm trying to solve this question but I'm failing at the first step.

    I'm supposed to get the Kinetic Energy into the form \frac{1}{2} \dot{q} . M \dot{q} and the Potential energy into the form \frac{1}{2} q . Kq where M and K are dxd matrices.

    I can do it for the k.e. but not for the p.e. as I don't know how to work out the p.e. of a spring. My notes say "The potential energy of each spring is k/2 times the square of the displacement from the natural length." - I don't know what that means.

    Could someone show me how to work out the p.e.
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    (Original post by primaverasnap)



    I'm trying to solve this question but I'm failing at the first step.

    I'm supposed to get the Kinetic Energy into the form \frac{1}{2} \dot{q} . M \dot{q} and the Potential energy into the form \frac{1}{2} q . Kq where M and K are dxd matrices.

    I can do it for the k.e. but not for the p.e. as I don't know how to work out the p.e. of a spring. My notes say "The potential energy of each spring is k/2 times the square of the displacement from the natural length." - I don't know what that means.

    Could someone show me how to work out the p.e.
    Well, you know hooke's law right?:
    F=kx
    integrating both sides w.r.t x gives us the work done i.e the potential energy in the spring:
    WD=\int Fdx=\int kxdx=\frac{1}{2}kx^2+C
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    So would the p.e. = k/2 (q1^2 + (q2 - q1)^2 + (3a - q2)^2) ?
 
 
 
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Updated: November 27, 2011

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