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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