A particle with a mass of 0.660 kg is attached to a horizontal spring with a force constant of 23.76 N/m. At the moment t = 0, the particle has its maximum speed of 15 m/s and is moving to the left. (Assume that the positive direction is to the right.)

(a) Determine the particle's equation of motion, specifying its position as a function of time. (Use the following as necessary: t.)
x = _______

(b) Where in the motion is the potential energy three times the kinetic energy?
± ______ m

(c) Find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m. (Be sure to enter the minimum time and not the total time elapsed.)
_______ s

(d) Find the length of a simple pendulum with the same period.
________ m

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  1. m a = m d^2x/dt^2 = -k x

    .66 d^2x/dt^2 = -23.7 x

    let x = A sin (wt-phi)
    where w = 2 pi f
    then v = dx/dt= A w cos(wt-phi)
    and a = d^2x/dt^2 = -Aw^2 sin (wt-phi)
    = - w^2 x
    -m w^2 x = - k x
    w^2 = k/m
    w = sqrt(k/m) = sqrt(23.76/.66)
    = 6 radians/s
    so in the end
    x = A sin (6 t - phi)
    v = 6 A cos(6 t- phi)
    BUT the max v = 15 at t = 0
    cos is max of 1 at 6t-phi = 0
    so - phi = 0 so phi = 0
    x = A sin (6 t)
    v = 6 A cos (6 t)
    when t = 0, cos (6t) = 1
    15 = 6 A
    A = 15/6

    so now
    x = (15/6) sin 6 t
    v = 15 cos 6 t

    OK? I think you can take it from there
    like(1/2) k x^2 = (1/2) m v^2

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