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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    The velocity of a particle in S.H.M. at displacement y from mean position is (\[a=\]amplitude\[\omega =\] angular frequency)

    A)  \[\omega \sqrt{{{a}^{2}}+{{y}^{2}}}\]                                   

    B)  \[\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\]

    C)  \[\omega y\]                   

    D)         \[{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\]

    Correct Answer: B

    Solution :

    In \[S.H.M.,\] velocity of particle at displacement v from mean position is\[\upsilon =\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\]


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