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

  • question_answer
    Two masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are suspended together by a massless spring of constant k. When the masses are in equilibrium, \[{{m}_{1}}\] is removed without disturbing the system. Then the angular frequency of oscillation of \[{{m}_{2}}\] is

    A)            \[\sqrt{\frac{k}{{{m}_{1}}}}\]

    B)            \[\sqrt{\frac{k}{{{m}_{2}}}}\]

    C)            \[\sqrt{\frac{k}{{{m}_{1}}+{{m}_{2}}}}\]                      

    D)            \[\sqrt{\frac{k}{{{m}_{1}}{{m}_{2}}}}\]

    Correct Answer: B

    Solution :

               \[\omega =\sqrt{\frac{k}{m}}\]


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