If the Mass in the Previous Problem Is Pulled Down $x = 2 \cos ( 4 \sqrt { 2 } t ) + 2 \sin ( 4 \sqrt { 2 } t )$

If the mass in the previous problem is pulled down two feet and released, the solution for the position is

A) $x = 2 \cos ( 4 \sqrt { 2 } t ) + 2 \sin ( 4 \sqrt { 2 } t )$
B) $x = 2 \sin ( 4 \sqrt { 2 } t )$
C) $x = 2 \cos ( 4 \sqrt { 2 } t )$
D) $x = 2 \sin ( 4 t )$
E) $x = 2 \cos ( 4 t )$

Correct Answer:

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