Solve the problem.
-The formula for the up-and-down motion of a weight on a spring is given by $S = a \sin \left( \frac { \sqrt { k } } { m } \right) t$
where $a$ is the radius of the circle, $\mathrm { k }$ is the spring constant, $\mathrm { m }$ is the mass, and $\mathrm { t }$ is the time. Solve the equation for $t$.
A) $t = \frac { \sqrt { \mathrm { k } } } { \mathrm { m } } \arcsin \left( \frac { \mathrm { s } } { \mathrm { a } } \right)$
B) $t = \frac { m } { \sqrt { k } } \arcsin \left( \frac { S } { a } \right)$
C) $t = \frac { \sqrt { k } } { m } \sin \left( \frac { a } { S } \right)$
D) $t = \frac { m } { \sqrt { k } } \sin \left( \frac { \mathrm { S } } { \mathrm { a } } \right)$

Question

Quizplus · Accepted Answer

To solve for $t$, you need to isolate $t$ on one side of the equation. Starting with $S = a \sin \left( \frac { \sqrt { k } } { m } \right) t$, you first divide both sides by $a$ to get $\frac{S}{a} = \sin \left( \frac { \sqrt { k } } { m } \right) t$. Then, to isolate $t$, you apply the inverse sine ($\arcsin$) to both sides, resulting in $\arcsin\left(\frac{S}{a}\right) = \left( \frac { \sqrt { k } } { m } \right) t$. Finally, to solve for $t$, you multiply both sides by the reciprocal of $\frac { \sqrt { k } } { m }$, which is $\frac { m } { \sqrt { k } }$, resulting in $t = \frac { m } { \sqrt { k } } \arcsin \left( \frac { S } { a } \right)$.

Solve the Problem S=asin⁡(km)tS = a \sin \left( \frac { \sqrt { k } } { m } \right) tS=asin(mk​​)t

Solve the Problem $S = a \sin \left( \frac { \sqrt { k } } { m } \right) t$