JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Time Period and Frequency

  • question_answer
    If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to \[2.0m{{s}^{-2}}\] at any time, the angular frequency of the oscillator is equal to [CBSE PMT 1992; RPMT 1996]

    A)            \[10\,rad\,{{s}^{-1}}\]      

    B)            \[0.1\,rad\,{{s}^{-1}}\]

    C)            \[100\,rad\,{{s}^{-1}}\]    

    D)            \[1\,rad\,{{s}^{-1}}\]

    Correct Answer: A

    Solution :

               \[\omega =\sqrt{\frac{Acceleration}{Displacement}}\]\[=\sqrt{\frac{2.0}{0.02}}\]\[=10\,rad\ {{s}^{-1}}\]


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