F = k x
.5 * 9.81 = k * 1.5 * 10^-2
k = 3.27* 10^2 = 327 Newtons/meter
{{{{to find period ----you might know f = (1/2pi)sqrt (k/m) but anyway:
if x = A sin wt where w = 2 pi f
v = Aw cos w t
a = -Aw*2 sin wt = -w^2 x but F = m a
so
- k x = - m w^2 x
w^2 = k/m
w = sqrt(k/m) = 2 pi f }}}
so
f = (1/2 pi) sqrt (k/m)
f = (1/2pi) sqrt (327 / .5) = (1/6.28) sqrt(654) = 4.07 Hz
T = 1/4.07 = 0.246 seconds (part a)
I think you can get the maximum potential energy from my note below. That is the max kinetic energy as well. However you can also get (1/2) m v^2 at x = 0 from my derivation above
potential energy of spring = (1/2) k x^2
A mass of 0.5kg is attached to one end of a helical spring and produced an extension amplitudeof 2.5cm.The mass is now set into vertical oscillation of 10mm.Determine (a)the period of oscillation(b)the maximum kinetic energy of the mass(C)the potential energy of the spring when the mass is 8mm below the center of oscillation (g=10ms-2)
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