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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    The displacement of interference waves respectively are\[{{y}_{1}}=4sin\]\[\omega t\]and \[{{y}_{2}}=3\,\sin \left( \omega t+\frac{\pi }{2} \right).\] The amplitude of resultant wave will be

    A)  5                   

    B)  7

    C)  1                  

    D)  zero

    Correct Answer: A

    Solution :

     \[y=\sqrt{a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos ({{\phi }_{2}}-{{\phi }_{1}})}\] \[=\sqrt{16+9+2\times 4\times 3\cos \frac{\pi }{2}}=5\]


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