If the mass in the previous problem is pulled down two centimeters and released, the solution for the position is
A) $x = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }$
B) $x = 2 e ^ { - 2 t } + 4 t e ^ { - 2 t }$
C) $x = 0.02 e ^ { 2 t } - 0.04 t e ^ { 2 t }$
D) $x = e ^ { - 2 t } \sin t$
E) $x = 0.02 e ^ { - 2 t } \cos t$

Question

Quizplus · Accepted Answer

The correct solution involves a decaying exponential function due to the damping effect (indicated by the negative exponent), and the initial displacement of 2 cm (0.02 m) from equilibrium. Option A correctly represents this scenario with both the initial displacement and the effect of damping over time.

If the Mass in the Previous Problem Is Pulled Down x=0.02e−2t+0.04te−2tx = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }x=0.02e−2t+0.04te−2t

If the Mass in the Previous Problem Is Pulled Down $x = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }$