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A mass weighing 4 kg stretches a spring by 6 centimeters. The damping constant is c = 0.8.
External vibrations create a force of $F(t) = 42 \sin(5t)$ kg. Find the steady-state solution.
$\qquad u_p = \frac{10,752}{2,545} \cos(5t) + \frac{13,104}{2,545} \sin(5t)$
$\qquad u_p = -\frac{10,752}{2,545} \cos(5t) - \frac{13,104}{2,545} \sin(5t)$
$\qquad u_p = \frac{10,752}{2,545} \cos(5t) - \frac{13,104}{2,545} \sin(5t)$
$\qquad u_p = -\frac{10,752}{2,545} \cos(5t) + \frac{13,104}{2,545} \sin(5t)$
