NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • question_answer
    A particle is executing a simple harmonic motion. Its maximum acceleration is \[\alpha \] and maximum velocity is \[\beta \]. Then, its time period of vibration will be                                                                             [NEET (Re) 2015]

    A)  \[\frac{{{\beta }^{2}}}{{{\alpha }^{2}}}\]                

    B)      \[\frac{\alpha }{\beta }\]

    C)  \[\frac{{{\beta }^{2}}}{\alpha }\]                    

    D)       \[\frac{2\pi \beta }{\alpha }\]

    Correct Answer: D

    Solution :

    For a particle executing SHM, we have maximum acceleration,
    \[\alpha =A{{\omega }^{2}}\]               …(i)
    where, A is maximum amplitude and co is angular velocity of a particle.
    Maximum velocity,
                \[\beta =A\omega \,\]                  …(ii)
    Comparing Eq. (i) and Eq. (ii), we get
    \[\frac{\alpha }{\beta }=\frac{A{{\omega }^{2}}}{A\omega }\Rightarrow \frac{\alpha }{\beta }=\omega =\frac{2\pi }{T}\]
    i.e.,       \[T=\frac{2\pi \beta }{\alpha }\]
    Thus, its time period of vibration, \[T=\frac{2\pi \beta }{\alpha }\].


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