• question_answer Two waves are represented as ${{y}_{1}}=2a\,\sin \,\left( \omega t+\frac{\pi }{6} \right)$and${{y}_{2}}=-2a\,\cos \,\left( \omega t-\frac{\pi }{6} \right)$.The phase difference between the two waves is A)  $\frac{\pi }{3}$                                 B)  $\frac{4\pi }{3}$ C)  $\frac{3\pi }{3}$                               D)  $\frac{5\pi }{6}$

${{y}_{1}}=2a\,\sin \,\left( \omega t+\frac{\pi }{6} \right)=-2a\,\cos \,\left( \frac{\pi }{2}+\omega t+\frac{\pi }{6} \right)$ $\Rightarrow$            ${{\phi }_{1}}=\frac{\pi }{2}+\omega t+\frac{\pi }{6}$ and      ${{y}_{2}}=-2a\,\cos \,\left( \omega t-\frac{\pi }{6} \right)$ $\Rightarrow$            ${{\phi }_{2}}=\omega t-\frac{\pi }{6}$ $\therefore$ Phase difference $={{\phi }_{1}}-{{\phi }_{2}}$ $=\frac{\pi }{2}+\omega t+\frac{\pi }{6}-\left( \omega t-\frac{\pi }{6} \right)$ $=\frac{\pi }{2}+\frac{\pi }{3}=\frac{5\pi }{6}$