Posted by Molly on Sunday, November 21, 2010 at 6:30pm.
A mass of 1 kg is hanging from a spring with a spring constant of 3 N/m. At a distance of 0.2 m below the equilibrium it has a velocity of 1 m/s in the upward direction. What is the amplitude of the oscillation?

college physics  Damon, Sunday, November 21, 2010 at 7:17pm
F = k y = 3 y
when y = .2, F = .6 N up
w = sqrt (k/m) = 1.73 radians/s
y = A sin w t
v = w A cos w t
a = w^2 A sin w t = w^2 y
at time t
v = w A cos w t = 1
y = A sin w t = .2
a = F/m = .6/m = .6
so
1.73 A cos wt = 1
A cos wt = .577
and
A sin wt = .2
A sin wt/A cos wt = .2/.577 =  .346
tan w t = .346
w t = 19.1 degrees = .333 radians
sin wt = .327
A = .2/.327 = .611
check
.611 cos 19.1 = .577 right
v = .577 (w) = .577*1.73 = .998 right
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