Answer:
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)
