Solve the problem.
-A spring with a spring constant of 6 and a 1-unit mass attached to it is stretched $2 \mathrm { ft }$ and released. What is the equation for the resulting oscillatory motion?
A) $s ( t ) = 2 \sin \sqrt { 6 } t$
B) $s ( t ) = \frac { 1 } { 2 } \sin \sqrt { \frac { 1 } { 12 } } t$
C) $s ( t ) = 2 \sin \sqrt { \frac { 1 } { 6 } } t$
D) $s ( t ) = 6 \sin \sqrt { 2 } t$

Question

Quizplus · Accepted Answer

The equation for the oscillatory motion of a spring-mass system is $s(t) = A \sin(\sqrt{\frac{k}{m}} t)$, where $A$ is the amplitude, $k$ is the spring constant, and $m$ is the mass. Here, $A = 2$, $k = 6$, and $m = 1$, so the equation becomes $s(t) = 2 \sin(\sqrt{\frac{6}{1}} t) = 2 \sin(\sqrt{6} t)$.

Solve the Problem 2ft2 \mathrm { ft }2ft And Released s(t)=2sin⁡6ts ( t ) = 2 \sin \sqrt { 6 } ts(t)=2sin6​t

Solve the Problem $2 \mathrm { ft }$ And Released $s ( t ) = 2 \sin \sqrt { 6 } t$