JEE Main & Advanced Physics Wave Mechanics Question Bank Stationary Waves

  • question_answer
    In stationary waves, distance between a node and its nearest antinode is 20 cm. The phase difference between two particles having a separation of 60 cm will be [CMEET Bihar 1995]

    A)            Zero                                         

    B)            p/2

    C)            p    

    D)            3p/2

    Correct Answer: D

    Solution :

                         \[\frac{\lambda }{4}=20\Rightarrow \lambda =\,80\,cm\], also \[\Delta \varphi =\frac{\lambda }{2\pi }.\,\Delta x\]            Þ \[\Delta \varphi =\] \[\frac{60}{80}\times 2\pi =\frac{3\pi }{2}\]


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