JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    The minimum phase difference between two simple harmonic oscillations.
    \[{{y}_{1}}=\frac{1}{2}\sin \omega \,t+\frac{\sqrt{3}}{2}\cos \omega \,t\]
    \[{{y}_{2}}=\sin \omega \,t+\cos \omega \,t\], is

    A)  \[\frac{7\pi }{12}\]                               

    B)  \[\frac{\pi }{12}\]

    C)  \[-\frac{\pi }{6}\]                                  

    D)  \[\frac{\pi }{6}\]

    Correct Answer: B

    Solution :

    \[{{y}_{1}}=\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t\] \[=\cos \,\frac{\pi }{3}\,\sin \omega t\,+\frac{\sqrt{3}}{2}\,\cos \omega t\] Similarly, \[{{y}_{2}}=\sqrt{2}\,\sin \,\left( \omega t\,+\frac{\pi }{4} \right)\] \[\therefore \] Phase difference \[\Delta f=\frac{\pi }{3}\,-\frac{\pi }{4}\,=\frac{\pi }{12}\]

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