Question

a steel ball weighing 128 lb is suspended from a spring, whereupon the spring is stretched 2 ft from its natural length. the ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position. assuming no damping force, find an expression for
a)the position of the ball at any time.
b)the position of the bal at t=pai/12 sec
c)the circular frequency, natural frequency and period for this system

Answer 1

The question is really a physics question but uses old English units of feet which means to get mass (in slugs) you must divide pounds by g which is about 32 ft/s^2.

Answer 2

k = 64 lb/ft for spring

position above equilirium = x
at t = 0 x = .5 ft
so
x = .5 cos 2 pi t/T
we need to find T, the period
f = -kx = m a
but a = .5 (2 pi/T)^2(-cos2 pi/T) =-(2pi/T)^2 x
so
-k x = -(2 pi/T)^2 x
2 pi/T = sqrt(k/m)
m = 128/32 = 4
so
2 pi/T = sqrt (64/4) = 4
therefore
x = .5 cos 4 t

at t = pi/12
x = .5 cos pi/3 = .5 cos 60deg = .25

w = 2 pi f = 2 pi/T = 4 radians/sec
f = w/2pi = 2/pi
T = pi/2