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J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    A particle moves so that its acceleration a is given by a = - bx, where x is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is :

    A) \[2\pi \sqrt{b}\]                                               

    B) \[\frac{2\pi }{b}\]

    C) \[\frac{2\pi }{\sqrt{b}}\]                                              

    D) \[2\sqrt{\frac{\pi }{b}}\]

    Correct Answer: C

    Solution :

                    The relation between acceleration  and displacement\[(x)\]for a body in SHM is \[a=-{{\omega }^{2}}x\] Given,                   \[a=-bx\] On comparing the two equations, we get                 \[{{\omega }^{2}}=b\] \[\therefore \]  \[\omega =\sqrt{b}\] Since,    \[\omega =\frac{2\pi }{T}\] \[\therefore \]  \[\frac{2\pi }{T}=\sqrt{b}\Rightarrow T=\frac{2\pi }{\sqrt{b}}\]


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