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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Oscillations

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

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

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

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

    D) \[\frac{-\pi }{3}\]

    Correct Answer: B

    Solution :

    [b] \[{{v}_{1}}=\frac{d{{y}_{1}}}{dt}=0.1\times 100\pi \cos \left( 100\pi t+\frac{\pi }{3} \right)\] \[{{v}_{2}}=\frac{d{{y}_{2}}}{dt}=-0.1\pi \sin \pi t\] \[=0.1\pi \cos \left( \pi t+\frac{\pi }{2} \right)\] \[\therefore \,\,Phase\,diff.\,={{\phi }_{1}}-{{\phi }_{2}}=\frac{\pi }{3}-\frac{\pi }{2}=\frac{2\pi -3\pi }{6}=\frac{\pi }{6}\]


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