Answer to Question #249204 in Differential Equations for Micau

Question

A steel ball weighing 128 Ib 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 air resistance, find (a) an expression for the position of the ball at any time t, and (b) the position of the ball at t=pi/4 sec

Answer 1

Steel ball weight=128lb

∆x=2ft

At equilibrium x=0

Position above equilibrium, t=0,x=6in=0.5ft

x=ACos wt=$\frac{1}{2}Cos (\frac{2πt}{T})$

f=ma=-kx

$\frac{dx}{dt}=\frac{d}{dt}(\frac{1}{2}Cos(\frac{2πt}{T}))$

$=\frac{2π}{2T}(-Sin\frac{2πt}{T})$

a=$\frac{d^{2}x}{dt^{2}}=\frac{d}{dt}(-\frac{2π}{2T}Sin\frac{2πt}{T})$

$=\frac{1}{2}(\frac{2π}{T})^{2}Cos \frac{2πt}{T}$

$K=\frac{F}{∆x}$ =$\frac{128}{2}=64lb/ft$

m=$\frac{128}{32}=4$

a)

$x=\frac{1}{2}Cos (\frac{2πt}{\frac{π}{2}})$

$=0.5 Cos(4t)$

b)

t=$\frac{π}{4}$

$x=0.5Cos(4*\frac{π}{4})$

=0.5Cosπ

=-0.5