Question

The initial phase angle for $i=10\sin \omega t+8\cos \omega t$ is
(A) ${\tan }^{-1}\left({\displaystyle \frac{4}{5}}\right)$
(B) ${\tan }^{-1}\left({\displaystyle \frac{5}{4}}\right)$
(C) ${\sin }^{-1}\left({\displaystyle \frac{4}{5}}\right)$
(D) ${90}^0$

Answer 1

Hint: We are given with an equation and are asked to find the initial phase angle for the same. Thus, we will firstly evaluate the equation at time $t=0$ . Then, we will use some basic trigonometric ideas to manipulate the evaluated value and then come up with an answer.

Complete Step By Step Solution
Here, The given equation is,
 $i=10\sin \omega t+8\cos \omega t$
Now, For the initial value, we take time $t=0$
Taking here, we get
 $i=10\sin \left(0\right)+8\cos \left(0\right)$
We know,
 $\sin \left(0\right)=0$ And $\cos \left(0\right)=1$
Thus, we get
 $i=8\left(1\right)$
Further, we get
 $i=8$
Now,
 $i_o=\sqrt{{\left(10\right)}^2+{\left(8\right)}^2}$
Further, we get
 $i_o=\sqrt{164}$
Where, $i_o$ is the amplitude of the motion.
Now,
As per the generic equation of such motion,
 $i=i_o\sin \left(\omega t+\varphi \right)$
For time $t=0$ ,
 $i=i_0\sin \varphi$
Then, we get
 $\sin \varphi ={\displaystyle \frac{i}{i_o}}$
Thus, we get
 $\sin \varphi ={\displaystyle \frac{8}{\sqrt{164}}}$
Thus,
 $\tan \varphi ={\displaystyle \frac{8}{\sqrt{164-64}}}$
Thus,
 $\tan \varphi ={\displaystyle \frac{8}{10}}$
Thus,
 $\tan \varphi ={\displaystyle \frac{4}{5}}$
Hence, we get
 $\varphi ={\tan }^{-1}\left({\displaystyle \frac{4}{5}}\right)$
Hence, the correct option is (A).

Note
We have converted the sine function to a tangent one as all the given options are in the same format. We used basic trigonometry for conversion. One should not confuse it to be a given parameter.