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BHU PMT BHU PMT Solved Paper-2003

  • question_answer
    When a wave travels in a medium the particles displacement is given by the equation \[y=0.03\,\,\sin \pi \left( 2t-0.01\,x \right)\]where x and y are in seconds. The wavelength of the wave is: [BHU M-2003]

    A)  \[200\,\,m\]                                     

    B)  \[100\,\,m\]

    C)  \[20\,\,m\]                                       

    D)  \[10\,\,m\]

    Correct Answer: A

    Solution :

                     Key Idea: Compare the given wave equation with its similar standard wave equation. The standard wave equation is                                 \[y=a\,\sin \,2\pi \left( \frac{t}{T}-\frac{x}{\lambda } \right)\]                     ?(1) Where \[a\] is amplitude, \[T\] is periodic time and \[\lambda \]  is displacement. Given, equation is                                 \[y=0.03\,\,\sin \,\,\pi \left( 2t-0.01x \right)\]     ?(2) Comparing Eqs. (1) and (2), we get                                                 \[\frac{2}{\lambda }=0.01\]                 \[\Rightarrow \]                               \[\lambda =\frac{2}{0.01}=200\,m\]


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