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    • Thread Starter

    A light spring AB, having natural length a and modulus of elastocoty 3amn^2, lies straight and at its natural length at rest on a horizontal table. A particle of mass m is attached to the end A.
    The end B is then moved in a straight line in the direction AB with constant speed V. The resulting motion of the particle is resisted by a force of magnitude 4mnv where v is the speed of the particle. If x is the extension of the spring at time t , show that


    B is moving at V m/s
    A is moving at v m/s

    Therefore extension is x = (V-v) m every sec,

    i.e. dx/dt = (V-v) m/s

    Tension in spring is T where

    T = λx/l
    T = (3amn²).x/a where λ = 3amn² and l = a
    T = 3mn²x

    Accelerating force on particle at A is given by T-F where F = 4mnv,

    T-F = mdv/dt
    3mn²x - 4mnv = mdv/dt
    3n²x - 4nv = dv/dt

    But dx/dt = V-v
    v = V - dx/dt
    dv/dt = -d²x/dt²

    substituting for v and dv/dt,

    3n²x - 4n(V - dx/dt) = -d²x/dt²
    d²x/dt² + 4n.dx/dt + 3n²x = 4nV
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    • Thread Starter

    thank you
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Updated: June 30, 2004

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