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JEE Main & Advanced Sample Paper JEE Main Sample Paper-2

  • question_answer
    Two particles move parallel to x-axis about the origin with the same amplitude and frequency. At a certain instant they are found at distance A/3 from the origin on opposite sides but their velocities are found to be in the same direction. The phase difference between the two particles is

    A)  \[\phi =\frac{\pi }{2}\]                 

    B)  \[\phi ={{\cos }^{-1}}\left( \frac{7}{9} \right)\]

    C)  \[\phi ={{\cos }^{-1}}\left( \frac{3}{7} \right)\]  

    D)  \[\phi =\pi \]

    Correct Answer: C

    Solution :

    Let the equation of SHM be \[{{x}_{1}}=A\,\sin (\omega t)\] \[{{x}_{2}}=A\sin (\omega t+\phi )\] According to contrition, \[\frac{A}{3}=A\sin (\omega t)-\frac{A}{3}=A\sin (\omega t+\phi )\] By solving above equation, \[\phi =\pi \] or \[\cos \phi =\frac{7}{9}\] Also, \[{{v}_{1}}=A\omega \cos (\omega t),\] \[{{v}_{2}}=A\omega \cos (\omega t+\phi )\] For velocities in same direction, \[\phi ={{\cos }^{-1}}\left( \frac{7}{9} \right)\]


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