Question

Find the amplitude and phase of the vibrations given by $x=A\cos \omega t+\frac{A}{2}\cos (\omega t+\frac{\pi }{2})+\frac{A}{4}\cos (\omega t+\pi )+\frac{A}{8}\cos (\omega t+\frac{3\pi }{2})$

• A. $\frac{3\sqrt{5}A}{8},{\tan }^{-1}\left(\frac{1}{2}\right)$
• B. $\frac{\sqrt{5}}{8}A,{\tan }^{-1}\left(\frac{1}{4}\right)$
• C. $\frac{\sqrt{5}}{8}A,{\tan }^{-1}\left(\frac{3}{4}\right)$
• D. $\frac{3\sqrt{5}A}{8},{\tan }^{-1}\left(\frac{1}{4}\right)$

Answer 1

The correct option is A $\frac{3\sqrt{5}A}{8},{\tan }^{-1}\left(\frac{1}{2}\right)$

$x=A\cos \omega t+\frac{A}{2}\cos (\omega t+\frac{\pi }{2})+\frac{A}{4}\cos (\omega t+\pi )+\frac{A}{8}\cos (\omega t+\frac{3\pi }{2})$
From the vector diagram, we can see


So, we can infer that
$A_R=\sqrt{{(A-\frac{A}{4})}^2+{(\frac{A}{2}-\frac{A}{8})}^2}$
$\Rightarrow A_R=A\sqrt{{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{8}\right)}^2}$
$\Rightarrow A_R=\frac{A\times 3\sqrt{5}}{8}$
Also, $\tan \varphi =\frac{(\frac{A}{2}-\frac{A}{8})}{\left(A\right)-\left(\frac{A}{4}\right)}=\frac{1}{2}$
or $\varphi ={\tan }^{-1}\frac{1}{2}$