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

seo-qna
Last updated date: 23rd Jun 2024
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Answer
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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,
$\Rightarrow$ $i = 10\sin \omega t + 8\cos \omega t$
Now,
For the initial value, we take time $t = 0$
Taking here, we get
$\Rightarrow$ $i = 10\sin \left( 0 \right) + 8\cos \left( 0 \right)$
We know,
$\sin \left( 0 \right) = 0$ And$\cos \left( 0 \right) = 1$
Thus, we get
$\Rightarrow$ $i = 8\left( 1 \right)$
Further, we get
$i = 8$
Now,
$\Rightarrow$ ${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,
$i = {i_o}\sin \left( {\omega t + \phi } \right)$
For time$t = 0$,
$i = {i_0}\sin \phi $
Then, we get
$\sin \phi = \dfrac{i}{{{i_o}}}$
Thus, we get
$\sin \phi = \dfrac{8}{{\sqrt {164} }}$
Thus,
$\Rightarrow$ $\tan \phi = \dfrac{8}{{\sqrt {164 - 64} }}$
Thus,
$\Rightarrow$ $\tan \phi = \dfrac{8}{{10}}$
Thus,
$\Rightarrow$ $\tan \phi = \dfrac{4}{5}$
Hence, we get
$\Rightarrow$ $\phi = {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right)$

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.