Posted by noah on .
A 1.25kg mass stretches a vertical spring 0.215m.If the spring is stretches an additional 0.130m and released,
(a)how long does it take to reach the equilibrium position again?
(b)Write an equation giving its position y as a function of time t.
(c)What will be its maximum speed and maximum acceleration?

college physic SHM 
drwls,
The spring constant is
k = Mg/0.215m = 57.0 N/m
The frequency of oscillation is
f = [1/(2 pi)]sqrt(k/m) = 1.075 Hz
The angular frequency is
w = sqrt(k/m) = 6.75 rad/s
The period is 1/f = 0.93 seconds
(a) 1/4 period = 0.2325 s is the time needed to return to the equilibrium position
(b) If y is measured from the equilibrium position
y = 0.13 cos wt = 0.14 cos6.75t
(c) Max speed = w*(amplitude)= 0.13 w
Max acceleration = w^2*(amplitude)