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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    Two waves are given by \[{{y}_{1}}=a\,\sin \,\]\[(\omega t-kx)\]and \[{{y}_{2}}=a\,\cos \]\[(\omega t-kx).\]The phase difference between the two waves is

    A)  \[\text{ }\!\!\pi\!\!\text{ /4}\]                                 

    B) \[\text{ }\!\!\pi\!\!\text{ }\]

    C)  \[\text{ }\!\!\pi\!\!\text{ /8}\]                                 

    D)  \[\text{ }\!\!\pi\!\!\text{ /2}\]

    Correct Answer: D

    Solution :

    Waves equations are \[{{y}_{1}}=a\sin (\omega t-kx)\]                 and        \[{{y}_{2}}=a\cos \,(\omega t-kx)\]                                 \[{{y}_{2}}=a\cos \,\left( \omega t-kx+\frac{\pi }{2} \right)\] \[\therefore \] Phase difference between the waves \[=\frac{\pi }{2}\]


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