Suppose a 32-Pound Weight Is Suspended on a Spring $\frac { 1 } { 3 } \text { foot }$

Suppose a 32-pound weight is suspended on a spring. The weight is pulled $\frac { 1 } { 3 } \text { foot }$ below the equilibrium position and released. The motion takes place in a Med that furnishes a damping force of magnitude $\frac { 1 } { 8 }$ speed at all times. Assume that the weight stretches the spring $\frac { 2 } { 3 }$ foot from its natural position. Find a formula for the position of the weight as a function of time $t$ .

A) $y ( t ) = 3 \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$

B) $y ( t ) = \frac { e ^ { - t / 16 } } { 12 } \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$

C) $y ( t ) = \frac { e ^ { - t / 16 } } { 3 } \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$

D) $y ( t ) = 6 \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$

E) $y ( t ) = \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$