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) $x = 2 \cos ( 4 t )$

Question

Quizplus · Accepted Answer

The mass is pulled down two feet, indicating an initial displacement, which corresponds to a cosine function with an amplitude of 2. The angular frequency is given by $4 \sqrt{2}$, matching the cosine term in option C.

If the Mass in the Previous Problem Is Pulled Down x=2cos⁡(42t)+2sin⁡(42t)x = 2 \cos ( 4 \sqrt { 2 } t ) + 2 \sin ( 4 \sqrt { 2 } t )x=2cos(42​t)+2sin(42​t)

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