Answer to Question #249212 in Differential Equations for Micau

Question

A ball weighing 32 pounds is attached to a spring whose constant is 5 lb/ft. Initially, the mass is released 1 foot below the equilibrium position with a downward velocity of 5 ft/s, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 2 times the instantaneous velocity. (a) Find the equation of motion if the ball is driven by an external force equal to f(t) = 12 cos 2t + 3 sin 2t. (b) Determine the position and velocity of the ball at t = pi/4 sec.

Answer 1

a)

$mx''+kx+2v=f(t)$

$mx''+2x'+kx=12 cos 2t + 3 sin 2t$

$32x''+2x'+5x=12 cos 2t + 3 sin 2t$

b)

$32k^2+2k+5=0$

$k=\frac{-2\pm \sqrt{4-640}}{64}=-0.031\pm 0.394i$

$x_c=e^{-0.031t}(c_1cos(0.394t)+c_2sin(0.394t))$

$x_p=Acos2t+Bsin2t$

$32(-4Acos2t-4Bsin2t)+2(-2Asin2t+2Bcos2t)+5(Acos2t+Bsin2t)=$

$=12 cos 2t + 3 sin 2t$

$-128A+4B+5A=12\implies -123A+4B=12$

$-128B-4A+5B=3\implies -4A-123B=3$

$B=(12+123A)/4$

$-16A-123(12+123A)=12$

$-15145A=1488$

$A=-0.098$

$B=-0.014$

$x=x_c+x_p=e^{-0.031t}(c_1cos(0.394t)+c_2sin(0.394t))-0.098cos2t-0.014sin2t$

$x(0)=c_1-0.098=1\implies c_1=1.098$

$x'(0)=-0.031\cdot1.098+0.394c_2-0.028=5\implies c_2=12.848$

$x=e^{-0.031t}(1.098cos(0.394t)+12.848sin(0.394t))-0.098cos2t-0.014sin2t$

$x(\pi/4)=0.976(1.098+0.027)-0.014=1.084$ ft

$x'=-0.031e^{-0.031t}(1.098cos(0.394t)+12.848sin(0.394t))+$

$+e^{-0.031t}(-0.433sin(0.394t)+5.062cos(0.394t))+0.196sin2t-0.028cos2t$

$x'(\pi/4)=-0.031\cdot 0.976(1.098+0.069)+0.976(-0.003+5.062)+0.196=$

$=5.098$ ft/s