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

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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answered by Damon
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