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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    A particle executes simple harmonic oscillation 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) \[\frac{T}{4}\]                                   

    B) \[\frac{T}{8}\]

    C) \[\frac{T}{12}\]                                

    D)  \[\frac{T}{2}\]

    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                            \[t=\frac{\pi }{6\left( \frac{2\pi }{T} \right)}\]      \[\left( \text{as}\,\,\omega =\frac{2\pi }{T} \right)\] or                            \[t=\frac{T}{12}\] Hence, the particle travels half of the amplitude from the equilibrium in\[\frac{T}{12}\sec \].


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