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JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

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

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

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

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

    D)  \[y=0.2sin\left( \frac{\pi }{4}+2t \right)\]

    Correct Answer: B

    Solution :

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


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