VMMC VMMC Medical Solved Paper-2012

  • question_answer
    The displacement of a particle along the \[x-\]axis is given by \[x=a{{\sin }^{2}}\omega t.\]The motion of the particle corresponds to

    A)  simple harmonic motion of frequency\[\omega /\pi \]

    B)  simple harmonic motion of frequency\[3\omega /2\pi \]

    C)  non simple harmonic motion

    D)  simple harmonic motion of frequency\[\omega /2\pi \]

    Correct Answer: C

    Solution :

     For a particle executing SHM acceleration \[(a)\propto -{{\omega }^{2}}\]displacement \[(x)\]? (i) Given     \[x=a{{\sin }^{2}}\omega t\]               ...(ii) Differentiating the above equation w.r.t., we get \[\frac{dx}{dt}=2a\omega (sin\omega t)(cos\omega t)\] Again differentiating, we get \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=a=2a{{\omega }^{2}}[co{{s}^{2}}\omega t-si{{n}^{2}}\omega t]\] \[=2a{{\omega }^{2}}\cos 2\,\omega t\] The given equation does not satisfy the condition for SHM [Eq. (i)]. Therefore, motion is not simple harmonic.


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