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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 36

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

    A) \[\pi /4\]                 

    B)        \[\pi \]

    C) \[\pi /8\]                 

    D)        \[\pi /2\]

    Correct Answer: D

    Solution :

    [d] \[{{y}_{1}}=a\,sin\left( \omega t-kx \right)\] and \[{{y}_{2}}=a\,cos\left( \omega t-kx \right)=a\,sin\left( \omega t-kx+\frac{\pi }{2} \right)\] Hence phase difference between these two is \[\frac{\pi }{2}\]


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