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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer
    A particle executes simple harmonic scillation with an amplitude \[a\]. The period of oscillation is T. The minimum time taken by the particle  to travel half of the amplitude from the equilibrium position is                   

    A)  T/4                                       

    B)  T/8            

    C)  T/12                                     

    D)  T/2            

    Correct Answer: C

    Solution :

    Let displacement equation of particle executing SHM is \[y=a\,\sin \,\omega t\] As particle travels half of the amplitude from the equilibrium position, so                 \[y=\frac{a}{2}\] Therefore,     \[\frac{a}{2}=a\,\sin \,\omega t\] or            \[\,\sin \,\omega t=\frac{1}{2}=\sin \frac{\pi }{6}\] or            \[\,\omega t=\frac{\pi }{6}\] or            \[t=\frac{\pi }{6\omega }\] or            \[t=\frac{\pi }{6\left( \frac{2\pi }{T} \right)}\]      \[\left( as\,\,\omega =\frac{2\pi }{T} \right)\] or            \[t=\frac{T}{12}\] Hence, the particle travels half of the amplitude from the equilibrium in  \[\frac{T}{12}s\].


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