Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    A wave equation is \[y=0.1\sin [100\pi t-kx]\]and wave velocity is 100 m/s, its wave number is equal to

    A)  \[1{{m}^{-1}}\]                                

    B) \[2{{m}^{-1}}\]

    C) \[\pi {{m}^{-1}}\]            

    D)        \[2\pi {{m}^{-1}}\]

    Correct Answer: C

    Solution :

    The standard equation of wave is \[y=a\sin (\omega t-kx)\]? (i) where \[a\]is amplitude,\[\omega \]the angular velocity and\[x\]the displacement at instant\[t\]. Given equation is \[y=0.1\,\,\sin (100\pi t-kx)\]                                     ? (ii) Comparing Eq. (i) with Eq. (ii), we get \[\omega =100\pi \] \[\therefore \]  Wave number\[=\frac{\omega }{v}=\frac{100\pi }{100}=\pi \,\,{{m}^{-1}}\]


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