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

  • question_answer
     A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are k1 and k2 respectively. When set into vertical vibrations, the period will be [MP PMT 2001]

    A)            \[2\pi \sqrt{\left( \frac{m}{{{k}_{1}}{{k}_{2}}} \right)}\]     

    B)            \[2\pi \sqrt{m\left( \frac{{{k}_{1}}}{{{k}_{2}}} \right)}\]

    C)            \[2\pi \sqrt{\left( \frac{m}{{{k}_{1}}-{{k}_{2}}} \right)}\]    

    D)            \[2\pi \sqrt{\left( \frac{m}{{{k}_{1}}+{{k}_{2}}} \right)}\]

    Correct Answer: D

    Solution :

                       Given spring system has parallel combination, so            \[{{k}_{eq}}={{k}_{1}}+{{k}_{2}}\] and time period \[T=2\pi \sqrt{\frac{m}{({{k}_{1}}+{{k}_{2}})}}\]


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