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NEET Sample Paper NEET Sample Test Paper-23

  • question_answer
    If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are related through the expression \[4{{v}^{2}}=25-{{x}^{2}},\] then its time period is

    A) \[\pi \]                           

    B) \[2\pi \]

    C) \[4\pi \]                         

    D) \[6\pi \]  

    Correct Answer: C

    Solution :

    Given, \[4{{v}^{2}}=25-{{x}^{2}}\] Differentiating with respect to time t \[8v\frac{dv}{dt}=-2x\frac{dx}{dt}\] But \[\frac{dv}{dt}=a\](acceleration) and \[\frac{dx}{dt}=v\] \[\therefore \]\[8va=-2xv\]or \[a=-\frac{1}{4}x\] For SHM \[a=-{{\omega }^{2}}x\] \[\therefore \]\[{{\omega }^{2}}=\frac{1}{4}\]or \[\omega =\frac{1}{2}\] \[\therefore \]Period of Oscillation \[T=\frac{2\pi }{\omega }=4\pi \] Hence, the correction option is [c].


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