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JEE Main & Advanced Physics Wave Mechanics Question Bank Progressive Waves

  • question_answer
    The  phase  difference  between  two  waves  represented  by \[{{y}_{1}}={{10}^{-6}}\sin [100\,t+(x/50)+0.5]m\]            \[{{y}_{2}}={{10}^{-6}}\cos \,[100\,t+(x/50)]m\] where x is expressed in metres and t is expressed in seconds, is approximately                 [CBSE PMT 2004]

    A)            1.5 rad                                    

    B)            1.07 rad

    C)            2.07 rad                                  

    D)            0.5 rad

    Correct Answer: B

    Solution :

                       \[{{y}_{1}}={{10}^{-6}}\sin \,[100\,t+(x/50)+0.5]\]                    \[{{y}_{2}}={{10}^{-6}}\sin \,\left[ 100\,t+\left( \frac{x}{50} \right)+\left( \frac{\pi }{2} \right) \right]\]                    Phase difference f                      \[=[100t+(x/50)+1.57]-[100t+(x/50)+0.5]\]               \[=1.07\]radians.


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