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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    Two particles \[P\] and \[Q\] describe \[SHM\] of same amplitude\[a\], frequency \[v\] along the same Straight line. The maximum distance between the two particles is\[a\sqrt{2}\]. The initial phase difference between the particles is

    A) \[zero\]                                               

    B) \[\pi /2\]

    C) \[\pi /6\]                                             

    D)  \[\pi /3\]

    Correct Answer: B

    Solution :

    \[{{y}_{1}}=a\sin 2\pi \,\,vt\] and        \[{{y}_{2}}=a\sin (2\pi vt+\phi )\] \[y={{y}_{2}}-{{y}_{1}}=a[\sin (2\pi \,\,vt+\phi )-\sin 2\pi \,\,vt]\]                 \[=2a\sin \frac{\phi }{2}\cos \left[ 2\pi \,\,vt+\frac{\phi }{2} \right]\] \[\therefore \]Maximum value of\[y=2a\sin \frac{\phi }{2}\] Now,\[2a\sin \frac{\phi }{2}=a\sqrt{2}\]


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