Question

A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive direction and then released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.

Answer 1

Answer:

$\frac{1}{8} y'' + 2y' + 24y=0$

Explanation:

The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by

$my'' + \zeta y' + ky=0$

Where m is the mass, ζ is the damping constant, and k is the spring constant.

The spring constant k can be found by

$w - kL=0$

$mg - kL=0$

$4 - k\frac{1}{6}=0$

$k = 4\times 6 =24$

The damping constant can be found by

$F = -\zeta y'$

$6 = 3\zeta$

$\zeta = \frac{6}{3} = 2$

Finally, the mass m can be found by

$w = 4$

$mg=4$

$m = \frac{4}{g}$

Where g is approximately 32 ft/s²

$m = \frac{4}{32} = \frac{1}{8}$

Therefore, the required differential equation is

$my'' + \zeta y' + ky=0$

$\frac{1}{8} y'' + 2y' + 24y=0$

The initial position is

$y(0) = \frac{1}{2}$

The initial velocity is

$y'(0) = 0$