Q. For simple harmonic vibrations $y_{1} = 8\ cos\omega t$ $y_{2} = 4cos(\omega t + \frac{\pi}{2})$ $y_{3} = 2cos(\omega t + \pi)$ $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 $\tan^{- 1}{(\frac{1}{2})}$
B) $\sqrt{45\ }$and $\tan^{- 1}\left( \frac{1}{3} \right)$
C) $\sqrt{75\ }$and $\tan^{- 1}(2)$
D) $\sqrt{75\ }$and $\tan^{- 1}{(\frac{1}{3})}$
Explanation
(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, $\tan{\theta = \frac{y}{x} = \frac{3}{6} = \frac{1}{2}}$
$\theta = \tan^{- 1}{(1/2)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.