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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    The displacement of a particle varies according to the relation \[x=4\,(\cos \pi \,t+\sin \pi t)\]. The amplitude of the particle is     AIEEE  Solved  Paper-2003

    A) \[-4\]    

    B)                       4                             

    C) \[4\sqrt{2}\]

    D) 8

    Correct Answer: C

    Solution :

            Displacement of the particle is given as                     \[x=4\,(\cos \pi t+\sin \pi t)\]                     \[=\frac{4}{\sqrt{2}}\times \sqrt{2}[\cos \pi t+\sin \pi t]\]                     \[=\left[ \frac{1}{\sqrt{2}}\cos \pi +\frac{1}{\sqrt{2}}\sin \pi t \right]4\sqrt{2}\] \[=\left[ \sin \frac{\pi }{4}\cos \pi t+\cos \frac{\pi }{4}\sin \pi t \right]4\sqrt{2}\] \[x=4\sqrt{2}\sin \left[ \pi t+\frac{\pi }{4} \right]\]     \[[\because \sin A\cos B+\cos A\sin B=\sin (A+B)]\] So, amplitude \[=4\sqrt{2}\]                


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