JEE Main & Advanced AIEEE Solved Paper-2011

  • question_answer
    Two particles are executing simple harmonic motion of the same amplitude A and frequency \[\omega \] along the x-axis. Their mean position is separated by distance \[{{X}_{0}}({{X}_{0}}>A)\]. If the maximum separation between them is \[({{X}_{0}}+A)\], the phase difference between their motion is   AIEEE  Solved  Paper-2011

    A) \[\frac{\pi }{6}\]                                  

    B) \[\frac{\pi }{2}\]

    C)              \[\frac{\pi }{3}\]                                  

    D)              \[\frac{\pi }{4}\]

    Correct Answer: C

    Solution :

                 \[{{x}_{1}}=A\sin \omega \,t\] \[{{x}_{2}}={{x}_{0}}+A\sin (\omega t+\phi )-A\sin \omega t\] Separation \[{{x}_{0}}+A\sin (\omega t+\phi )-A\sin \omega t\]                    \[={{x}_{0}}+2A\sin i\left( \frac{\phi }{2} \right)\cos \omega t\] Maximum separation \[={{x}_{0}}+2A\sin \frac{\phi }{2}={{x}_{0}}+A\] \[\Rightarrow \,\,\phi =\frac{\pi }{3}\]


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