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JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    A point particle of mass 0.1 kg is executing SHM of amplitude of 0.1m. When the particle passes through the mean position, its kinetic energy is \[18\times {{10}^{-3}}\text{J}\text{.}\]The equation of motion of this particle when the initial phase of oscillation is \[45{}^\circ \] can be given by

    A)  \[y=0.1\cos \left( 6t+\frac{\pi }{4} \right)\]

    B)  \[y=0.1\sin \left( 6t+\frac{\pi }{4} \right)\]

    C)  \[y=0.4\sin \left( t+\frac{\pi }{4} \right)\]

    D)  \[y=0.2\sin \left( \frac{\pi }{2}+2t \right)\]

    Correct Answer: B

    Solution :

     \[A=0,1m,m=0.1kg,K{{E}_{\max }}=18\times {{10}^{-3}}J,\] \[\phi =\frac{\pi }{4}\] \[k=\frac{36\times {{10}^{-3}}}{{{(0.1)}^{2}}}=3.6;\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{3.6}{0.1}}=6\,\text{red/s}\] \[\therefore \]Eqn. \[y=0.1\sin \left( 6t+\frac{\pi }{4} \right)\]


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