JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • 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)  \[\frac{5\pi }{6}\]

    Correct Answer: D

    Solution :

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

You need to login to perform this action.
You will be redirected in 3 sec spinner