• # question_answer The minimum phase difference between two simple harmonic oscillations. ${{y}_{1}}=\frac{1}{2}\sin \omega \,t+\frac{\sqrt{3}}{2}\cos \omega \,t$ ${{y}_{2}}=\sin \omega \,t+\cos \omega \,t$, is A)  $\frac{7\pi }{12}$                                B)  $\frac{\pi }{12}$ C)  $-\frac{\pi }{6}$                                   D)  $\frac{\pi }{6}$

${{y}_{1}}=\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t$ $=\cos \,\frac{\pi }{3}\,\sin \omega t\,+\frac{\sqrt{3}}{2}\,\cos \omega t$ Similarly, ${{y}_{2}}=\sqrt{2}\,\sin \,\left( \omega t\,+\frac{\pi }{4} \right)$ $\therefore$ Phase difference $\Delta f=\frac{\pi }{3}\,-\frac{\pi }{4}\,=\frac{\pi }{12}$