Posted by **anisah** on Sunday, April 22, 2012 at 3:45am.

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

- math -
**Damon**, Sunday, April 22, 2012 at 5:30am
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

- actually ancient physics -
**Damon**, Sunday, April 22, 2012 at 5:32am
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.

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