A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this Motion is modeled by where is the distance from equilibrium (in feet) and is the time (in seconds).
$y = \frac { 1 } { 8 } \sin 2 t + \frac { 1 } { 6 } \cos 2 t$
Use the identity
$a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$
where $C = \arctan ( b / a ) , a 0$, to write the model in the form $y = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B t + C )$.
A) $y = \frac { 5 } { 24 } \sin ( 2 t + 0.9273 )$
B) $y = \frac { 24 } { 5 } \sin ( 2 t - 0.9273 )$
C) $y = \frac { 24 } { 5 } \sin ( 2 t + 0.9273 )$
D) $y = \sin ( 2 t - 0.9273 )$
E) $y = \sin ( 2 t + 0.9273 )$

Question

A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this Motion is modeled by where is the distance from equilibrium (in feet) and is the time (in seconds). 
$y = \frac { 1 } { 8 } \sin 2 t + \frac { 1 } { 6 } \cos 2 t$ 
Use the identity
$a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$
where $C = \arctan ( b / a ) , a > 0$, to write the model in the form $y = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B t + C )$. 
A) $y = \frac { 5 } { 24 } \sin ( 2 t + 0.9273 )$
B) $y = \frac { 24 } { 5 } \sin ( 2 t - 0.9273 )$
C) $y = \frac { 24 } { 5 } \sin ( 2 t + 0.9273 )$
D) $y = \sin ( 2 t - 0.9273 )$
E) $y = \sin ( 2 t + 0.9273 )$

Quizplus · Accepted Answer

Given $a = \frac{1}{8}$ and $b = \frac{1}{6}$, using the identity, we find $\sqrt{a^2 + b^2} = \sqrt{\left(\frac{1}{8}\right)^2 + \left(\frac{1}{6}\right)^2} = \frac{5}{24}$ and $C = \arctan\left(\frac{b}{a}\right) = \arctan\left(\frac{1/6}{1/8}\right) = \arctan\left(\frac{4}{3}\right) \approx 0.9273$.

A Weight Is Attached to a Spring Suspended Vertically from a Ceiling