Question

The equation of a simple harmonic motion is given by $y=6\sin 10\pi t+8\cos 10\pi t$. The initial phase of this motion is _____

• A. ${\cot }^{-1}\left(1/3\right)$
• B. ${\tan }^{-1}\left(1.3\right)$
• C. ${\cot }^{-1}\left(2\right)$
• D. ${\tan }^{-}\left(\frac{4}{3}\right)$

Answer 1

The correct option is D ${\tan }^{-}\left(\frac{4}{3}\right)$

$\begin{array}{c}y=6\sin \left(10\pi t\right)+8\cos \left(10\pi t\right)\\ let6=R\cos \varphi \\ 8=R\sin \varphi \\ y=R\cos \varphi \sin \left(10\pi t\right)+R\sin \varphi \cos \left(10\pi t\right)\\ y=R\sin \left(10\pi t+\varphi \right)\\ WhereRistheamplitudeofthewave\varphi istheinitialphase\\ R=\sqrt{6^2+8^2}=\sqrt{36+64}=10\\ \frac{R\sin \varphi }{R\cos \varphi }=\frac{8}{6}\\ \tan \varphi =\frac{4}{3}\\ \varphi ={\tan }^{-}\left(\frac{4}{3}\right)\end{array}$