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.

1 answer

m = 0.785 kg
k = 125 N/m

w = sqrt (k/m) = 2 pi f
T = 1/E 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.