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J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    The displacement of a particle of mass \[3\text{ }g\] executing simple harmonic motion is given by \[Y=3\text{ }sin\text{ (}0.2\text{ t)}\]in SI units. The KE of the particle at a point which is at a distance equal to \[1/3\] 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 :

    Displacement of particle in the case of SHM \[y=A\,\sin \,(\omega t+\phi )\] ?.(i) \[y=3\,\,\,\sin \,(0.2t)\] ...(ii) (given) Comparing Eqs. (i) and (ii), we get \[A=3,\] \[\omega =0.2\] Now, particle distance \[x=\frac{A}{3}=1\] Kinetic energy in SHM \[=\frac{1}{2}m\,\,\,{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] \[=\frac{1}{2}\times 3\times {{10}^{-3}}{{(0.2)}^{2}}[{{3}^{2}}-{{1}^{2}}]\] \[=0.48\times {{10}^{-3}}J\]


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