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JEE Main & Advanced Physics Wave Mechanics Question Bank Progressive Waves

  • question_answer
    Equation of a progressive wave is given by \[y=0.2\cos \pi \left( 0.04t+.02x-\frac{\pi }{6} \right)\]            The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of \[\pi /2\]

    A)            4 cm                                         

    B)            8 cm

    C)            25 cm                                      

    D)            12.5 cm

    Correct Answer: C

    Solution :

               Comparing with \[y=a\cos (\omega t+kx-\varphi )\],                     We get \[k=\frac{2\pi }{\lambda }=0.02\,\Rightarrow \,\lambda =100\,cm\] Also, it is given that phase difference between  particles \[\Delta \varphi =\frac{\pi }{2}.\]\[\]Hence path difference between them \[\Delta =\frac{\lambda }{2\pi }\times \Delta \varphi =\frac{\lambda }{2\pi }\times \frac{\pi }{2}=\frac{\lambda }{4}=\frac{100}{4}=25\,cm\]


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