JEE Main & Advanced Sample Paper JEE Main Sample Paper-32

  • question_answer
    Two simple harmonic motions are represented by the equations \[{{y}_{1}}=0.1\,\sin \,(100\pi t+\pi /3)\] and \[\,(100\pi t+\pi /3)\]. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2is:

    A) \[-\pi /3\]     

    B)                    \[\pi /6\]

    C) \[-\pi /6\]

    D)    \[\pi /3\]

    Correct Answer: C

    Solution :

    \[{{V}_{1}}\,=\frac{d{{y}_{1}}}{dt}=0.1\times 100\pi \cos \,(100\pi t+\pi /3)\]                 \[=10\pi \sin \,\left( 100\pi t+\frac{\pi }{2}\,+\frac{\pi }{3} \right)\]                 \[{{V}_{2}}=\frac{d{{y}_{2}}}{dt}\]                 \[=-0.1\times 100\pi \sin (100\pi t)\,=10\pi \sin (100\pi t+\pi )\] Argument of sign function is called phase. Required phase difference = \[\left( 100\pi t+\frac{\pi }{2}\,+\frac{\pi }{3} \right)\,-(100\pi t+\pi )\,=-\frac{\pi }{6}\]


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