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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    The motion of a particle executing SHM in one dimension is described by\[x=-0.23\sin \left( t+\frac{\pi }{4} \right)\]where, \[x\] is in metre and t in second. The frequency of oscillation in \[Hz\] is

    A) \[3\]                                     

    B) \[\frac{1}{2\pi }\]

    C) \[\frac{\pi }{2}\]                                              

    D)  \[\frac{1}{\pi }\]

    Correct Answer: B

    Solution :

    \[x=-0.3\sin \left( t+\frac{\pi }{4} \right)\]                 \[x={{x}_{0}}\sin (\omega t+\phi )\]                 \[{{x}_{0}}=0.3,\,\,\omega =1\]and\[\phi =\frac{\pi }{4}\]                 \[2\pi f=1\]                 \[f=\frac{1}{2\pi }\]


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