A mass-spring system oscillates on a horizontal frictionless surface with an amplitude of 3.35 cm. If the spring constant is 222 N/m and the mass is 0.580 kg, determine the mechanical energy of the system. Determine the maximum speed of the mass. Determine the maximum acceleration.

I already figured out the mechanical energy, which was .125J. Need help with the rest. Thanks!

1 answer

omega = w = 2 pi f = sqrt(k/m)
w = sqrt (222/.58)
= 19.56 radians/sso

x = .0335 sin 19.56 t

v = .0335(19.56) cos 19.56 t

a = -.0335(19.56)^2 sin 19.56 t
= -19.56^2 x

max v when cos 19.56 t = 1
and x = 0 so all energy is kinetic

max v = .655 m/s
(1/2) m v^2 = .125 J agree with you

max a when sin19.56 t = -1
or
.0335(19.56)^2 = 12.8 m/s^2