Question

Four simple harmonic vibrations, $ {x_1} = 8\sin wt $ , $ {x_2} = 6\sin \left( {wt + \dfrac{\pi }{2}} \right) $ , $ {x_3} = 4\sin \left( {wt + \pi } \right) $ , $ {x_4} = 2\sin \left( {wt + \dfrac{{3\pi }}{2}} \right) $ are superimposed on each other. The resulting amplitude and its phase difference with $ {x_1} $ are,
(A) $ 20,{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) $
(B) $ 4\sqrt 2 ,\dfrac{\pi }{2} $
(C) $ 20,{\tan ^{ - 1}}\left( 2 \right) $
(D) $ 4\sqrt 2 ,\dfrac{\pi }{4} $

Answer 1

Hint To solve this question we need to first calculate the resultant of the vibrations using the superposition principle. Then by using the trigonometric relations between sine and cosine, the relation between their angles and conversion of sine to cosine and vice-versa, we can calculate the resulting amplitude and the required phase difference.

Formula Used: The formula used here is given as,
 $\Rightarrow \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $

Complete step by step answer
From the principle of superposition, we get,
 $\Rightarrow x = {x_1} + {x_2} + {x_3} + {x_4} $
So, we get,
 $\Rightarrow x = (8\sin wt) + \left( {6\sin \left( {wt + \dfrac{\pi }{2}} \right)} \right) + \left( {4\sin \left( {wt + \pi } \right)} \right) + \left( {2\sin \left( {wt + \dfrac{{3\pi }}{2}} \right)} \right) $
From our knowledge of trigonometry and angles we know that,
 $\Rightarrow - \sin wt = \sin (wt + \pi ) $
And, $ - \cos wt = \sin \left( {\dfrac{\pi }{2} + wt} \right) $
Also we know that, $ \cos wt = \sin \left( {\dfrac{{3\pi }}{2} + wt} \right) $
So, we can write the superimposed wave as,
 $\Rightarrow x = 8\sin wt - 6\cos wt - 4\sin wt + 2\cos wt \\
   \Rightarrow x = 4\sin wt - 4\cos wt \\
 $
This can be simplified and written as,
 $\Rightarrow x = 4(\sin wt - \cos wt) \\
   \Rightarrow x = 4\left( {\sin wt + \sin \left( {wt + \dfrac{\pi }{2}} \right)} \right) \\
 $
Now using the trigonometric relation,
 $\Rightarrow \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $
We can write the equation of superimposed wave as,
 $ \Rightarrow x = 4 \times 2\left( {\sin \left( {\dfrac{{wt + \left( {wt + \dfrac{\pi }{2}} \right)}}{2}} \right)\cos \left( {\dfrac{{wt - \left( {wt + \dfrac{\pi }{2}} \right)}}{2}} \right)} \right) \\
   \Rightarrow x = 8\left( {\sin \left( {\dfrac{{2wt + \dfrac{\pi }{2}}}{2}} \right)\cos \left( {\dfrac{{\left( { - \dfrac{\pi }{2}} \right)}}{2}} \right)} \right) \\
 $
Now simplifying the above equation, we get,
 $\Rightarrow x = 8\left( {\sin \left( {wt + \dfrac{\pi }{4}} \right)\cos \left( { - \dfrac{\pi }{4}} \right)} \right) $
Now using the relation,
 $\Rightarrow \cos ( - x) = \cos (x) $
We get,
 $\Rightarrow x = 8\left( {\sin \left( {wt + \dfrac{\pi }{4}} \right)\cos \left( {\dfrac{\pi }{4}} \right)} \right) \\
   \Rightarrow x = 8\sin \left( {wt + \dfrac{\pi }{4}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
 $
Thus we get the superimposed wave as,
 $\Rightarrow x = 4\sqrt 2 \sin \left( {wt + \dfrac{\pi }{4}} \right) $
This means that the amplitude of the superimposed wave is $ 4\sqrt 2 $ and the phase difference with $ {x_1} $ is $ \dfrac{\pi }{4} $
$ \therefore $ Option (D) is the correct option.

Note
In order to remember in which quadrant which trigonometric function will be positive, we can use the ASTC mnemonic which can be expanded into ‘Add Sugar To Coffee’. According to this, A stands refers to the first quadrant telling you that ALL the trigonometric function will be positive in the first quadrant, S refers to the second quadrant telling you that only SINE will be positive in the second quadrant, T refers to the third quadrant telling you that only TAN will be constant in the third quadrant, and C refers to the fourth quadrant telling you that only COSINE will be positive in the fourth quadrant. Here we move anticlockwise through the quadrants.