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
1.73 A cos wt = 1
A cos wt = .577
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
.611 cos -19.1 = .577 right
v = .577 (w) = .577*1.73 = .998 right
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