# college physic SHM

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?

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1. 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)

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