JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Spring Pendulum

  • question_answer
    A particle of mass 200 gm executes S.H.M. The restoring force is provided by a spring of force constant 80 N / m. The time period of oscillations is                                   [MP PET 1994]

    A)            0.31 sec                                  

    B)            0.15 sec

    C)            0.05 sec                                  

    D)            0.02 sec

    Correct Answer: A

    Solution :

               \[T=2\pi \sqrt{\frac{m}{K}}=2\pi \sqrt{\frac{0.2}{80}}=0.31\]sec


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