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)

$y_1=A_{\text{net }}\sin \omega t+{\varphi }_1$
$A_{\text{net }}=\sqrt{{\frac{1}{2}}^2+{\frac{{\sqrt{3}}^2}{2}}^2}=1$
$\Rightarrow \tan {\varphi }_1=\frac{b}{a}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$
$\Rightarrow \tan {\varphi }_1=\sqrt{3}$
$\Rightarrow {\varphi }_1=\frac{\pi }{3}$
$y_2=A_{\text{net }}\sin \omega t+{\varphi }_2$
$\Rightarrow \tan {\varphi }_2=\frac{1}{1}=45^{\circ }$
$\Rightarrow {\varphi }_2=\frac{\pi }{4}$
$\Delta \varphi ={\varphi }_1-{\varphi }_2$
$\Rightarrow \Delta \varphi =\frac{\pi }{3}-\frac{\pi }{4}$
$\Delta \varphi =\frac{\pi }{12}$