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

This is a "spring-mass" mass problem.


m=128 lb
L=2 ft
g=32.2 ft/sec² (accel. due to gravity)
k=mg/L=128*32.2/2=2060.8
Let y=displacement (downwards) in the same direction as g, then the differential equation that governs the motion is:
acceleration = -ky, or
y"+ky=0 where y"=d²y/dt²
The solution of the auxiliary equation is
(m^2+k)=0, or m=±i√k
and the solution to the equation is
y=C1 cos(√k t) + C2 sin(√k t)
The initial conditions are:
y(0)=-0.5, y'(0)=0.
Since
y(0)=-0.5=C1 cos(0) + C2 sin(0)
C2=0, and C1=-0.5

y'(t)=-C2√k sin(√k t) + C2 √k cos(√k t)
From
y'(0)=0=-C2√k sin(0) + C2 √k cos(0)
we conclude
C2=0 since sin(0)=0, √k≠0 and cos(0)≠0.

So the equation of motion is:
y=-(1/2)cos(√ t)

Answers to (b) and (c) can be deduced from the equation of motion.