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

  • question_answer
    A spring is stretched by 0.20 m, when a mass of 0.50 kg is suspended. When a mass of 0.25 kg is suspended, then its period of oscillation will be \[(g=10m/{{s}^{2}})\]

    A)            0.328 sec                                

    B)            0.628 sec

    C)            0.137 sec                                

    D)            1.00 sec

    Correct Answer: B

    Solution :

                       Force constant \[k=\frac{F}{x}=\frac{0.5\times 10}{0.2}=25\ N/m\]            Now \[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.25}{25}}=0.628\]sec


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