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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    A progressive wave is represented as \[y=0.2\cos \pi \left( 0.04t+0.2x-\frac{\pi }{6} \right)\] where distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of\[\frac{\pi }{2}\]?

    A)  4 cm                                     

    B)  8 cm

    C)  25 cm                  

    D)         12.5 cm

    Correct Answer: C

    Solution :

    On comparing the given equation with \[y=a\cos (\omega t+kx-o|)\] we get,\[k=\frac{2\pi }{\lambda }=0.02\] and  \[\lambda =100\,cm\] It is given that phase difference between particles,\[\Delta \phi \text{=}\frac{\pi }{2}\] So, the path difference between them, \[\Delta p=\frac{\lambda }{2\pi }\times \Delta \phi \] \[=\frac{\lambda }{2\pi }\times \frac{\pi }{2}=\frac{\lambda }{4}=\frac{100}{4}=25\,cm\]


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