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
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
1 answer