Question

For simple harmonic vibrations $y_1=8cos\omega t$

$y_2=4cos(\omega t+\frac{\pi }{2})$

$y_3=2cos(\omega t+\pi )$

$y_4=cos\left(\omega t+\frac{3\pi }{2}\right)$ are superimposed on one another. The resulting amplitude and phase are respectively

• A. $\sqrt{45}$ and ${\tan }^{-1}\left(\frac{1}{2}\right)$
• B. $\sqrt{45}$ and ${\tan }^{-1}\left(\frac{1}{3}\right)$
• C. $\sqrt{75}$ and ${\tan }^{-1}(2)$
• D. $\sqrt{75}$ and ${\tan }^{-1}\left(\frac{1}{3}\right)$

Answer 1

Answer: (A)

Resultant displacement along $X$ - axis is $x=y_1-y_3=8-2=6$
Resultant displacement along $Y$ - axis is $y=y_2-y_4=4-1=3$
Net displacement,
$r=\sqrt{x^2+y^2}=\sqrt{6^2+3^2}=\sqrt{45}$
Also, $tan\theta =\frac{y}{x}=\frac{3}{6}=\frac{1}{2}$