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An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.

What is the object's velocity when its potential energy is 2/3E?

the answer is

±√(k/m)(A/√3)
can you explain the √3 thank you

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1 answer

  1. let's say

    x = A sin w t
    v = A w cos w t
    a = -A w^2 sin wt = -w^2 x

    w = sqrt k/m
    but we do not really need to know that
    max potential energy = (1/2)k max x^2
    = (1/2) k A^2
    max kinetic energy = (1/2) m v^2
    but max v^2 = A^2 w^2
    so
    max ke = (1/2) m A^2 w^2
    the max ke = max pe
    so
    (1/2) k A^2 = (1/2) m A^2 w^2
    so
    w = sqrt(k/m) whew, that's settled
    when it is stopped, all the energy is potential
    so
    E = (1/2) k A^2
    (2/3) E = (1/3) k A^2 = potential energy
    so where are we at then?
    (1/2) k x^2 = (1/3) k A^2

    x^2 = (2/3) A^2
    x = (2/3)^.5 A
    sin w t = (2/3)^.5
    sin^2 wt = (2/3)
    so
    cos^2 w t = 1 - 2/3 = 1/3
    remember v = A w cos w t
    so
    v^2 = A^2 w^2 cos^2 wt
    v = +/- A w sqrt (1/3)
    since w = sqrt (k/m)
    we agree

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