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JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer
    The amplitude and the periodic time of \[\text{a}\] \[SHM\] are \[5\,\,cm\] and \[6\,\,s\] respectively. At a distance of \[2.5\,\,cm\] away from the mean position, the phase will be

    A) \[\frac{\pi }{3}\]                                              

    B) \[\frac{\pi }{4}\]

    C)  \[\frac{\pi }{6}\]                                             

    D)  \[\frac{5\pi }{12}\]

    Correct Answer: C

    Solution :

    The equation of motion is given as                 \[y=5\sin \frac{2\pi t}{6}\] Here,     \[y=2.5\,\,cm\] \[\therefore \]  \[2.5=5\sin \frac{2\pi t}{6}\] \[\Rightarrow \]               \[\frac{\pi }{6}=\frac{2\pi t}{6}\Rightarrow t=\frac{1}{2}s\] \[\therefore \] The phase\[=\frac{2\pi t}{6}=\frac{2\pi }{6}\times \frac{1}{2}=\frac{\pi }{6}\]


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