A weight stretches a spring 1.5 inches in equilibrium. The weight is initially displaced 8 inches above equilibrium and given a downward velocity of 4 ft/s. Find its displacement for t > 0

 x = 0.667 cos 10.69 t + 0.383)

Explanation:

This is a simple harmonic movement exercise, the general solution is

          x = A cos (wt + Ф)

          w² = k / m

In this exercise, the initial displacement is that it corresponds to the amplitude of the movement.

         A = 8 in. (1 ft. 12 in.) = ⅔ ft. = 0.6667 ft.

         x = 1.5 in = 0.125 ft

We look for the constant k with Hooke's law

           F = -kx

the force applied is the weight of the body F = W = mg

          -mg = - k x

           k = m g / x

we substitute

           k = m 32.16 / 0.125

           k = 257.28 m

we substitute

           w = √257.28

           w = 16.04 rad / s

we substitute in the expression of the position

           x = 0.667 cos (16.04 t + Ф)

to find the constant fi we look for the velocity

           v = dx / dt

           v = - 0.667 16.04 sin (16.04 t +Ф)

indicate the values ​​of v = 4 ft / s for t = 0

          - 4 = -10.6934 sin Ф

remember angles are in radians

         Ф = sin⁻¹ (4/106934)

         Ф = 0.383

we substitute

         x = 0.667 cos 10.69 t + 0.383)

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small = small current large mag large current