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CMC Medical CMC-Medical VELLORE Solved Paper-2013

  • question_answer
    The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin\[(0.2\,t)\]in SI units. The kinetic energy of the particle at a points which is at a distance equal to \[\frac{1}{3}\] of its amplitude from its mean position is

    A)  \[12\times {{10}^{3}}J\]                              

    B)  \[25\times {{10}^{-3}}J\]

    C)  \[0.48\times {{10}^{-3}}J\]                         

    D)  \[0.24\times {{10}^{-3}}J\]

    Correct Answer: C

    Solution :

                    Equation of SHM \[Y=3\,\sin \,(0.2t)\] Comparing with \[Y=a\,\,\sin \,\omega t,\]we have \[a=3\,m,\,\]\[\omega =0.2\,{{s}^{-1}}\] Mass of the particle \[=3g=3\times {{10}^{-3}}kg\] Therefore, kinetic energy of the particle is \[K=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})\] \[=\frac{1}{2}\times 3\times {{10}^{-3}}\times {{(0.2)}^{2}}({{3}^{2}}-{{1}^{2}})\] \[\left( \because \,x=\frac{a}{3} \right)\] \[=0.48\times {{10}^{-3}}J\]


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