• 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)  $\frac{5\pi }{6}$

Idea we can compare the phase of two waves only when they both are represented by 'sin' or 'cosine' functions. ${{y}_{1}}=2a\sin \left( \omega t+\frac{\pi }{6} \right)=-2a\cos \left( \frac{\pi }{2}+\omega t+\frac{\pi }{6} \right)$ $\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 $={{\phi }_{1}}-{{\phi }_{2}}=\frac{\pi }{2}+\omega t+\frac{\pi }{6}-\left( \omega t-\frac{\pi }{6} \right)$ $=\frac{\pi }{2}+\frac{\pi }{3}=\frac{5\pi }{6}$ TEST Edge For example the phase of a wave can be represented by$\left( \omega t+\frac{\pi }{3} \right)$. Here$\frac{\pi }{3}$is 'phase constant. The value of phase at t = 0 is known as phase constant.
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