Question

For simple harmonic vibrations $y_{1}=8 \, cos\omega t$

$y_{2}=4cos \left(\right.\omega t+\frac{\pi }{2}\left.\right)$

$y_{3}=2cos \left(\right.\omega t+\pi \left.\right)$

$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, $tan \theta = \frac{y}{x} = \frac{3}{6} = \frac{1}{2}$