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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    If x = \[\alpha \]sin \[\left( \omega t+\frac{\pi }{6} \right)\]and\[x'=\alpha \,cos\omega t,\]then the resultant phase difference between both will be

    A)  \[\pi /3\]               

    B)  \[\pi /6\]

    C)  \[\pi /2\]           

    D)  \[\pi \]

    Correct Answer: A

    Solution :

     \[x=a\sin \left( \omega t+\frac{\pi }{6} \right)\] \[x'=a\sin \left( \omega t+\frac{\pi }{2} \right)\] Phase difference\[={{\phi }_{2}}-{{\phi }_{1}}\] \[=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}\]


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