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

  • question_answer
    The acceleration of a particle performing S.H.M. is \[12cm/se{{c}^{2}}\] at a distance of 3 cm from the mean position. Its time period is                                 [MP PET 1996; MP PMT 1997]

    A)            0.5 sec                                     

    B)            1.0 sec

    C)            2.0 sec                                     

    D)            3.14 sec

    Correct Answer: D

    Solution :

               \[T=2\pi \sqrt{\frac{\text{Displacement}}{\text{Acceleration}}}\]\[=2\pi \sqrt{\frac{3}{12}}=\pi =3.14\ \sec \]


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