2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • 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

    A)  -4                                          

    B)  4

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

    D)         8

    Correct Answer: C

    Solution :

    \[x=4(\cos \pi t+\sin \pi t)\]                 \[=\frac{4}{\sqrt{2}}\times \sqrt{2}[\cos \pi t+\sin \pi t]\]                 \[x=4\sqrt{2}\left[ \pi t+\frac{\pi }{4} \right]\] So, amplitude\[=4\sqrt{2}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner