Question

The minimum phase difference between two SHMs, $y_{1}=\frac{1}{2}\sin\omega t+\frac{\sqrt{3}}{2}\cos\omega t$ and $y_{2}=\sin\omega t+\cos\omega t$ is,

• A. $\frac{\pi }{6}$
• B. $\frac{- \pi }{6}$
• C. $\frac{\pi }{12}$
• D. $\frac{7 \pi }{12}$

Answer 1

Answer: (C)

$y_1=A_{\text {net }} \sin \omega t+\phi_1$
$A_{\text {net }}=\sqrt{\frac{1}{2}^2+{\frac{\sqrt{3}^2}{2}}^2}=1$
$\Rightarrow \tan \phi_1=\frac{b}{a}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} $
$\Rightarrow \tan \phi_1=\sqrt{3} $
$\Rightarrow \phi_1=\frac{\pi}{3}$
$y_2=A_{\text {net }} \sin \omega t+\phi_2$
$\Rightarrow \tan \phi_2=\frac{1}{1}=45^{\circ}$
$\Rightarrow \phi_2=\frac{\pi}{4}$
$\Delta \phi=\phi_1-\phi_2$
$\Rightarrow \Delta \phi=\frac{\pi}{3}-\frac{\pi}{4}$
$\Delta \phi=\frac{\pi}{12}$