Answer to Question #135838 in Differential Equations for Nikhil Singh

Question

A steel Ball weighing 39.2 kg is supposed from a spring due to which the spring is stretched 2m from its natural length. The ball is started in motion with no initial velocity by displacing it through 0.5m above the equilibrium position. Assuming no air resistance, find an expression for the position of the Ball at any time t.

Answer 1

Mass m=39.2kg

stretched length in spring x=2m

for ball to be in equlibrium

mg=kx (k=spring constant)

$39.2\times 9.8=k\times2$

k$=39.2\times 4.9$

=192.08 N/m

Force on the ball by spring for displacement of 0.5m

f=kx$=192.08\times0.5$

=96.04N

since f=mass$\times accelaration$

$96.04=39.2\times a$

a=2.45m/s$^2$

Initialy velocity u=0

distance covered by ball

$s=ut+\frac{1}{2}at^2$

$=\frac{1}{2}\times 2.45\times t^2$

s =$1.225t^2$ m