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BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    The maximum velocity of a simple harmonic motion represented by \[y=3\sin \left( 100\,t+\frac{\pi }{6} \right)m\]  is given by:

    A)  300 m/s                              

    B)         \[\frac{\text{3}}{\text{6}}\text{m/s}\]

    C)         100 m/s                              

    D)        \[\frac{}{\text{6}}\text{m/s}\]

    Correct Answer: A

    Solution :

    Key Idea: Equating the given equation with general equation of SHM. The given equation is written as \[y=3\sin \left( 100t+\frac{\pi }{6} \right)\]                                          ?(i) The general equation of simple harmonic motion is written as \[y=a\sin (\omega t+o|)\]                           ?(ii) Equating Eqs. (i) and (ii), we get \[a=3,\,\,\omega =100\] Maximum velocity, \[v=a\omega \]                                 \[=3\times 100\]                                 \[=300\,m/s\,\]


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