Question

Four simple harmonic vibrations are:
$y_1=8\cos \omega t,y_2=4\cos (\omega t+\frac{\pi }{2}),y_3=2\cos (\omega t+\pi ),y_4=\cos (\omega t+\frac{3\pi }{2}),$ magnitude and direction of resultant is:

• A. $20$
• B. $8\sqrt{2}$
• C. $4\sqrt{2}$
• D. $4$

Answer 1

The correct option is A $20$

$y_1=8\cos \omega t{y}_2=4\cos (\omega t+\frac{\pi }{2})=-4\sin \omega t{y}_3=2\cos (\omega t+\pi )=-2\cos \omega t{y}_4=\cos (\omega t+\frac{3\pi }{2})=\sin \omega t$

lets say resultant of these four SHM's is $y$

$y=y_1+y_2+y_3+y_4=6\cos \omega t-3\sin \omega t$

let $x_0\cos \varphi =6,x_0\sin \varphi =3$

$\Rightarrow y=x_0(\cos \varphi \cos \omega t-\sin \varphi \sin \omega t)=x_0\cos (\omega t+\varphi )$

where $x_0=\sqrt{{\left(x_0\sin \varphi \right)}^2+{\left(x_0\cos \varphi \right)}^2}=\sqrt{{\left(3\right)}^2+{\left(6\right)}^2}=\sqrt{45}$

$\tan \varphi =\frac{1}{2}\Rightarrow \varphi ={\tan }^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow y=x_0\cos (\omega t+\varphi )=\sqrt{45}\cos (\omega t+{\tan }^{-1}\left(\frac{1}{2}\right))$

so the correct option is 'A'