Answer to Question #125358 in Differential Equations for Jessica

Question

1. A mass weighing 8 lb.stretches a spring 3 inches. The mass is attached to a viscous damper with a damping constant of 2 lb - s/ ft. If the mass is set in motion from its equilibrium position with a downward velocity of 2 in/s , find its position u at any time t . Assume the acceleration of gravity g = 32 ft/s.

Answer 1

Given $w = 8 \ lb., L = 3 \ in. = \frac{3}{12} feet = \frac{1}{4} feet$ , $\gamma = 2 \ lbs/feet$ .

$u(0 ) = 0, u'(0) = 2 \ in/s = \frac{1}{6} feet/s$.

Now, differential equation is $mu'' + \gamma u' + ku = 0 : u(0)=0,u'(0) = \frac{1}{6}$

where $m = \frac{w}{g} = \frac{1}{4} \ lbs^2/feet, k = \frac{w}{L} = 32 \ lb/feet$ .

$\implies \frac{1}{4}u''+2u'+32u = 0 :u(0)=0,u'(0)=\frac{1}{6}$

$\implies u''+8u'+128u = 0$

$\implies D = \frac{-8\pm \sqrt{64-512}}{2} = -4\pm i \sqrt{112}$ where D $= \frac{d}{dt}$

Hence, $u = e^{-4t} (c_1 cos(\sqrt{112}t)+c_2 sin(\sqrt{112}t))$ .

Now using $u(0)=0 \implies c_1 = 0$ .

So, $u = c_2 e^{-4t} sin(\sqrt{112}t)$

Now, $u'(0)=\frac{1}{6} \implies \frac{1}{6} = c_2 (-4)(\sqrt{112}) \implies c_2 = -\frac{1}{24\sqrt{112}}$

Hence, $u(t) = -\frac{1}{24\sqrt{112}} e^{-4t} sin(\sqrt{112} \ t)$