Posted by Anon on Tuesday, December 7, 2010 at 7:43pm.
At t = 0, a 785 g mass at rest on the end of a horizontal spring (k = 125 N/m) is struck by a hammer, which gives the mass an initial speed of 2.70 m/s.
(a) Determine the period of the motion.
T= .4979201367s
Determine the frequency of the motion.
f= 2.008354 Hz
(b) Determine the amplitude.
(c) Determine the maximum acceleration.
(d) Determine the position as a function of time.
(e) Determine the total energy.

Physics  Damon, Tuesday, December 7, 2010 at 7:55pm
m = 0.785 kg
k = 125 N/m
w = sqrt (k/m) = 2 pi f
T = 1/f
KE at center of motion = total energy = (1/2) mv^2
KE = E = (1/2)(m )(2.70)^2
so
U at max swing amplitude A = E = (1/2) k A^2
so
A^2 = 2 E/k
max acceleration = w^2 A
x = A sin w t
we already did total energy because I chose to find amplitude that way.