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Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer
    The displacement of a particle of mass\[3g\] executing simple harmonic motion is given by \[Y=3\,\,\sin (0.2t)\] in SI units. The kinetic energy of the particle at a point 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\[\text{SHM}\,\,Y=3\sin (0.2t)\] Comparing with\[Y=a\sin \omega t\], we have                 \[a=3m,\]         \[\omega =0.2{{s}^{-1}}\]               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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