A 200g mass is attached to a spring and allowed to execute simple harmonic motion with a period of 0.25seconds. If the total energy of the system is 2.0 joules, find the force constant of the spring and the amplitude of the motion.

Answer 1

F = -k x
x = A sin w t
v = Aw cos wt
a = -Aw^2 sin wt = -w^2 x
but
F = m a
-k x = m(-w^2 x)
so
w^2 =k/m
(2pi)^2/T^2 = k/m
k = m (2pi)^2/T^2 = 0.200* (2pi)^2/0.25^2
k = 126 Newtons/meter
now back to get A
when at max x, energy = (1/2) k x^2
= (1/2) k A^2
so
(1/2) k A^2 = 2 Joules
we know k, solve for A