Okay so we got some spring stuff: A mass weighing 20 pounds stretches a spring 6

Question

Okay so we got some spring stuff:

A mass weighing 20 pounds stretches a spring 6 inches. The mass initially released from rest from a point 4 inches below the equilibrium position.

(a) Find the x of the center of mass at times t=pi/12, pi/8, pi/6, pi/4 and 9pi/32 s. (Use g=32 ft/s^2)

(b) What is the velocity of the mass when t=(3pi)/16 ?

(c) What direction is the mass heading in that instant.

Note: It isn't completely necessary to give me answers for each value of t in part (a), but please give one so I can tell if you are correctly setting things up.

Explanation / Answer

Since weight = mg = 20 lb

mass = 20/32

k*6/12 = 20

k = 10

a) The equation for simple harmonic motion of spring mass system = A*cos(wt + C)

w = sqrt(k/m) = sqrt(32/2) = 4

At t = 0, x = -4, dx/dt = 0

A*cos(C) = -4

A*2.316*sin(C) = 0

C = pi, A = 4

x = -4*cos(4t)

Therefore position of mass at pi/12 is -2, at pi/8 is 0, at pi/6 is 2, at pi/4 is 4, at 9pi/32 is -3.6955

inches

b) Velocity = dx/dt = 16*sin(4t)

At t=3pi/16 v=16*sin(3pi/4) = 16/sqrt(2) = 8*1.414 = 11.31

c) upwards

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