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March 4, 2015

March 4, 2015

Posted by **noah** on Tuesday, July 26, 2011 at 11:22pm.

(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**, Wednesday, July 27, 2011 at 1:50amThe 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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