A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 8 inches below the equilibrium position. Find the equation of motion. (Use g

x (t) = -0.66 * cos (9.8 * t)

Step-by-step explanation:

We have that Newton's second law for a system is:

Knowing m the united mass and k the spring constant  

m * (d ^ 2x) / (dt ^ 2) = -k * x

where x (t) is the displacement from the equilibrium position. The equation can be expressed like this:

 (d ^ 2x) / (dt ^ 2) + (k / m) * x = 0

Weight units should be converted to mass units as follows:

m = W / g = 24 lb / (32 ft / s ^ 2) = 3/4 slug

We also need to convert inches to feet, to know the stretch, we know that 1 foot is twelve inches, therefore:

4 in * 1 ft / 12 in = 0.33 ft

With Hooke's law we proceed to calculate the spring constant k:

k = W / s = 24 lb / 0.33 ft = 72 lb / ft

Knowing then that m is equal to 3/4 and that k is equal to 72, we can replace in the initial equation:

(d ^ 2x) / (dt ^ 2) + (72 / (3/4)) * x = 0

(d ^ 2x) / (dt ^ 2) + 96x = 0

We know that the solution of a differential equation of the form a (d ^ 2x) / (dt ^ 2) + (w ^ 2) * x = 0 is equal to:

x (t) = C1 * cos (wt) + C2 * sin (wt)

Let w ^ 2 = 96, then w = 96 ^ (1/2) = 9.8

Replacing

x (t) = C1 * cos (9.8 * t) + C2 * sin (9.8 * t)

We have that the initial conditions are x (0) = - 8 in, which is equal to -8/12 ft = -0.66 ft

x (0) = -0.66 and x '(0) = 0 ft / s

Replacing we have:

x (t) = -0.66 * cos (9.8 * t) + 0 * sin (9.8 * t)

Then the equation would be

x (t) = -0.66 * cos (9.8 * t)

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