omega = sqrt (k/m)
= sqrt (12.6/.0107) = 34.3 radians/s
x = .378 cos (omega t)
v = -.378(34.3) sin (omega t)
when x = .378/2 cos(omega t) = .5
then omega t = 60 degrees = pi/3 radians
then sin (omega t) = .866
(I am ignoring signs here, we need speed not velocity)
v = .378 * 34.3 * .866
= 11.2 m/s
A 10.7 g mass is attached to a horizontal spring with a spring constant of 12.6 N/m and released from rest with an amplitude of 37.8 cm.
What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless?
1 answer