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BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    The displacement of a particle of mass 3 g 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)  \[4.8\times {{10}^{-3}}J\]                           

    D)  \[0.24\times {{10}^{-3}}J\]

    Correct Answer: C

    Solution :

                     Equation of SHMY =3 sin (0.20 t) Comparing with\[Y=a\text{ }sin\,\omega t,\] of, 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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