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VMMC VMMC Medical Solved Paper-2014

  • question_answer
    Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants \[{{k}_{1}}\]and \[{{k}_{2}}\] respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of vibrations of M to that of N is

    A) \[\frac{{{k}_{1}}}{{{k}_{2}}}\]                                     

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

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

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

    Correct Answer: D

    Solution :

    Maximum velocities are equal. Hence,\[{{a}_{1}}{{\omega }_{1}}={{a}_{2}}{{\omega }_{2}}\] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}}\] \[=\frac{2\pi /{{n}_{2}}}{2\pi /{{n}_{1}}}\] \[=\frac{{{n}_{1}}}{{{n}_{2}}}\] \[=\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{2\pi \sqrt{\frac{m}{{{k}_{1}}}}}{2\pi \sqrt{\frac{m}{{{k}_{2}}}}}=\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}}\]


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