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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
     he displacement of a particle executing simple harmonic motion, from its mean position is given by \[x=0.5\,\sin \,(10\pi l+1.5x)\cos (10\pi l+1.5\pi ).\] The ratio of the maximum velocity to the maximum acceleration of the body is given by:

    A)  \[20\pi \]                                           

    B)  \[\frac{1}{20\,\pi }\]

    C)  \[\frac{1}{10\,\pi }\]                                     

    D)  \[10\,\pi \]

    Correct Answer: C

    Solution :

                     From the relation, \[{{v}_{\max }}=A\omega \]`                                      ?(i)                 and        \[{{a}_{\max }}=A{{\omega }^{2}}\]                                        ?(ii) The ratio is obtained or dividing Eqs. (i) and (ii) \[\frac{{{v}_{\max }}}{{{a}_{\max }}}=\frac{A\omega }{A{{\omega }^{2}}}=\frac{1}{\omega }\]                                   ?(iii) Equation of SHM is given \[x=0.5\sin (10\pi t+1.5x).\cos (10\pi t+1.5x)\] So, from above equation \[\omega =10\pi \] Now, putting the value of \[\omega =10\pi \]in Eq.                                                                          (iii) \[\frac{{{v}_{\max }}}{{{a}_{\max }}}=\frac{1}{10\pi }\]


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