2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-1998

  • question_answer
    The displacement equation of a particle is \[x=3\,\sin \,2t\,+4\,\cos \,2t.\] The amplitude and maximum velocity will be respectively:

    A)  5, 10     

    B)                         3, 2                       

    C)  4, 2                       

    D)         3, 4

    Correct Answer: A

    Solution :

    Displacement equation \[x=3\,\sin \,2t+4\,\cos \,2t\] Comparing this equation with                 \[x={{a}_{1}}\,\sin \omega t+{{a}_{2}}\,\cos \,\omega t\]                 \[{{a}_{1}}=3,\,{{a}_{2}}=4,\,\omega =2\]                 \[a=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}}\]                                 \[=\sqrt{{{3}^{2}}+{{4}^{2}}}=5\]                                 \[{{\upsilon }_{\max }}=a\omega =5\times 2=10\]


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