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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 )

Question 3

Multiple Choice

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


A) x=2cos(42t) +2sin(42t) x = 2 \cos ( 4 \sqrt { 2 } t ) + 2 \sin ( 4 \sqrt { 2 } t )
B) x=2sin(42t) x = 2 \sin ( 4 \sqrt { 2 } t )
C) x=2cos(42t) x = 2 \cos ( 4 \sqrt { 2 } t )
D) x=2sin(4t) x = 2 \sin ( 4 t )
E) x=2cos(4t) x = 2 \cos ( 4 t )

Correct Answer:

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